## A Brief Introduction to Categories, Part 2: Functors

This post is a continuation of a previous post and part of a mini-series on category theory. As before, knowing the background of every single example is not necessary (nor sufficient) to grasp the general concept.

## Introduction

Previously, we have defined and discussed categories. However, studying categories as isolated objects with no relation to other categories would be a fruitless toil. It would also not be in the spirit of category theory, which studies everything through its relationship to other things. Yet the necessity to bring together distinct instances of the objects one is studying is not specific to category theory: For example, it’s hard to imagine studying group theory without studying group homomorphisms.

It should by now be clear that we are in dire need of a notion allowing us to connect different categories and to express relationships between them. It seems natural to let the definition of a category guide us to the definition of things that relate them. This plan comes to fruition in the conception of functors. These should be thought of as (homo)morphisms of categories, mappings preserving the structure categories have, i.e. identities and composition, just like a monoid homomorphism preserves the neutral element and the monoid operation.

The ubiquity with which functors arise is remarkable, and after some experience in thinking categorically it will become second nature to ask whether a construction behaves “functorially”.

## Functors

Definition C2.1 Let $\mathcal C$ and $\mathcal D$ be categories, then a (covariant) functor $F$ consists of the following data:

• A map $\mathrm{Obj}(\mathcal C) \to \mathrm{Obj}(\mathcal D), X \mapsto F(X)$
• For any pair of objects $X, Y \in \mathrm{Obj}(\mathcal C)$ a map $\mathrm{Hom}_{\mathcal C}(X,Y) \to \mathrm{Hom}_{\mathcal D}(F(X),F(Y)), f \mapsto F(f)$

such that the following conditions hold:

• For every object $X \in \mathrm{Obj}(\mathcal C)$ we have $F(\mathrm{id}_X)=\mathrm{id}_{F(X)}$
• For any three objects $X,Y,Z \in \mathrm{Obj}(\mathcal C)$ and morphisms $f:X \to Y, g:Y \to Z$ we have $F(g \circ f)= F(g) \circ F(f)$.

Let us ponder for a moment how this definition generalizes familiar concepts.

Example C2.2 Let $X$ and $Y$ be sets, then we can consider them as discrete categories (cf. C1.4), so that the elements of each set are the objects of the associated discrete category and the only morphisms are the identity morphisms. Then any map $f:X \to Y$ uniquely determines a functor $F$ between the associated discrete categories restricting to $f$ on the objects, as the only morphisms are identities and the definition of a functor requires sending identity morphisms to the corresponding identity morphisms in the target category. To summarize, functors between discrete categories correspond to maps between the underlying sets.

Example C2.3 Let $M$ and $N$ be monoids, considered as one-object categories. Then a functor $F:M \to N$ has to send the unique object of $M$ to the unique object of $N$. The action on morphisms is more interesting: from the definition of a functor, we see that a map of the underlying sets $f:M \to N$ determines a functor if and only if it preserves the monoid operation and the identity, i.e. if and only if it is a monoid homomorphisms. Thus for monoids (and hence for groups), we can regard homomorphisms as functors.

Example C2.4 Let $G$ be a group, considered as a one-object category and let us consider functors to the category of sets (cf. C1.2), $F:G \to \mathbf{Set}$. Since $G$ has one proxy object $x$, the value of $F(x)$ corresponds to a choice of a particular set $X$. As for the morphisms, we have for each $g \in G$ a map $F(g):X \to X$ such that $F(e)=\mathrm{id}_X$ and $F(gh)=F(g) \circ F(h)$. These are precisely the axioms for a group action on $X$, such that functors $G \to \mathbf{Set}$ correspond to sets with a (left) $G$-action, i.e. $G$-Sets.

Example C2.5 By reasoning as in the last example, one can show that if $K$ is a field and we consider the category $K\mathbf{-Vect}$ of vector spaces over $K$ with linear maps as morphisms, then functors $G \to K\mathbf{-Vect}$ correspond to representations of $G$ on vector spaces over $K$.

Now we shall look into examples of particular functors.

Example C2.6 There are a plethora of so-called “forgetful” functors. These arise when you “forget” part of the structure of an object. In this case, the target category consists of objects with less structure than the source category.
For example, we can forget about the multiplication in a ring and are left with an abelian group, this gives rise to a forgetful functor from the category of rings to the category of abelian groups. We can then proceed on our path to oblivion and forget that an abelian group is abelian, yielding a forgetful functor from abelian groups to groups. Finally, every group is a set, giving rise to yet another forgetful functor. It should be noted that it is important to consider also the morphisms for these forgetful functors to work: every ring homomorphism is a homomorphism of abelian groups on the additive groups, every homomorphism of abelian groups is a group homomorphism and every group homomorphism is a map of sets. Another chain of forgetful functors goes from smooth complex varieties to complex manifolds, from complex manifolds to smooth manifolds, from smooth manifolds to topological manifolds, from topological manifolds to topological spaces and from topological spaces to sets, ending another journey of obliviousness in the same destination.
Forgetful functors are often implicitly applied, especially if one is not doing category theory. And indeed, it should deserve no particular mention that for instance, any group is a set. It can be quite helpful, however, to explicitly state when you apply them, for example to clarify in which category two objects are isomorphic or if you want to apply concepts or theorems on functors to a forgetful functor.

Example C2.7 One could say that the first example of functoriality is encountered in an introductory calculus class, although that would be a bit of a stretch.
Let $\mathbf{Diff}_{*}$ be the category of pointed smooth manifolds, i.e. objects are pairs $(M,x)$ where $M$ is a smooth manifold and $x \in M$ and morphisms $(M,x) \to (N,y)$ are smooth maps $f:M \to N$ such that $f(x)=y$. Then there’s the tangent space functor $T_{-}: \mathbf{Diff}_* \to \mathbb{R}\mathbf{-Vect}$ to the category of $\mathbb{R}$-vector spaces with linear maps that sends a pair $(M,x)$ to the tanget space $T_x M$ of $M$ at $x$ and a morphism $f:(M,x) \to (N,y)$ to the differential $D_x(f):T_xM \to T_y N$. The fact that this behaves functorially is an abstract version of the chain rule: $D_x(g \circ f)=D_{f(x)}(g) \cdot D_x(f)$, justifying the assertion on functoriality in calculus above.

Example C2.8 Let $\mathbf{Ring}$ be the category of rings with ring homomorphisms as morphisms, then for any ring $A$, the units $A^\times$ form a group and ring homomorphisms preserve units so that a ring homomorphism $f: A \to B$ restricts to a group homomorphism $A^\times \to B^\times$. We obtain a functor from $\mathbf{Ring}$ to $\mathbf{Grp}$, the category of groups with group homomorphisms as morphisms.

Example C2.9 Let $A$ be an abelian group and $n \in \mathbb{N}$, then there’s a functor from the category $\mathbf{Top}$ of topological spaces to the category of abelian groups that sends a space $X$ to the singular homology group $H_n(X,A)$. The fact that this is a functor is even part of the Eilenberg-Steenrod axioms for homology theories.

Example C2.10 Let $\mathbf{Top}$ be the category of topological spaces with continuous maps and let $\mathcal C$ be the category of measurable spaces with measurable maps as morphisms. There’s a functor $\mathbf{Top} \to \mathcal C$ that equips a topological space with its Borel $\sigma$-algebra, turning it into a measurable space. Every continuous map is measurable with respect to the corresponding Borel $\sigma$-algebras, ensuring that this construction yields a functor.

Example C2.11 Not every constructions that seems in some sense “natural” can be turned into a functor. For example, we can associate to each group $G$ its center $Z(G)$ which is an abelian group. One could hope that this can be made into a functor. The problem is that central elements don’t necessarily get sent to central elements under a group homomorphism. Indeed, it is impossible to define any action on morphisms such that the assignment on objects $G \mapsto Z(G)$ becomes a functor, for the identity on the cyclic group on two elements $C_2$ factors through the inclusion $\iota$ of $C_2$ to $S_3$ as a transposition and the sign homomorphism $\sigma:S_3 \to C_2$. We have $\mathrm{id}_{C_2}=\sigma \circ \iota$. However, the center of $C_2$ is itself, whereas the center of $S_3$ is trivial. If there was any functor $F$ that restricts to $G \mapsto Z(G)$ on objects, then we would get $\mathrm{id}_{C_2}=F(\mathrm{id}_{C_2})=F(\sigma \circ \iota)=F(\sigma) \circ F(\iota)$ implying that the identity on $C_2$ factors over the trivial group, which is impossible.

Example C2.12 Let $C$ be a category and let $X \in \mathrm{Obj}(\mathcal C)$, then for any object $Y \in \mathrm{Obj}(\mathcal C)$, we have a set $\mathrm{Hom}_{\mathcal C}(X,Y)$ and for a morphism $f:Y \to Y'$, we can postcompose with $f$, yielding a map $f \circ -: \mathrm{Hom}_{\mathcal C}(X,Y) \to \mathrm{Hom}_{\mathcal C}(X,Y')$. By associativity of composition and the definition of the identity elements, this construction yields a functor $\mathcal C \to \mathbf{Set}$. These functors – affectionally called Hom-functors – are of paramount importance, as we shall see in future posts.

Example C2.13 We have said that functors can be thought of as morphisms between categories. This can be taken quite literally. To avoid Russel’s paradox and to ensure that our sets of morphisms will actually be sets, let’s restrict our attention to categories whose class of objects is actually a set, so-called “small” categories. There’s a (non-small!) category $\mathbf{Cat}$, in which the objects are small categories and morphisms are functors between small categories. Note that we have implicitly said that one can compose functors and that there’s an identity functor for each category.

Example C2.14 Let $\mathbf{Cat}$ be the category from the previous example. Fix a commutative ring $R$. For any small category $\mathcal C$, let $R[\mathcal C]$ be the free $R$-module with a basis given by the set of all morphisms in $\mathcal C$. (Note for the set-theoretically careful: since the objects form a set and the morphisms between two fixed objects form a set, the set of all morphisms is a union of sets indexed by a set (twice), which is thus again a set)
This means that elements of $R[\mathcal C]$ are formal linear combinations of morphisms in $\mathcal C$. For two morphisms $f,g$ in $\mathcal C$, considered as basis elements of $R[\mathcal C]$, define their product to be $f \circ g$, if $f$ and $g$ can be composed or $0$ otherwise. This extends bilinearly to a mapping $R[\mathcal C] \times R[\mathcal C] \to R[\mathcal C]$ that makes $R[\mathcal C]$ into an associative, generally non-commutative and non-unital $R$-algebra. If $F: \mathcal C \to \mathcal D$ is a functor of small categories, then $F$ extends linearly to a homomorphism of $R$-algebras $R[\mathcal C] \to R[\mathcal D]$.
Thus we obtain a functor from the category of small categories to the category of associative $R$-algebras. If $G$ is a group, considered as a one-object category (cf. example C1.7), then $R[G]$ coincides with the group algebra associated to $R$ and $G$ (for more details on how this is useful for representation theory, see this post).
The same construction also works if $R$ is not commutative, yielding a non-unital ring, but then it wouldn’t make sense to call the result an $R$-algebra.

## Contravariance

All the previous examples of functors had one thing in common: they didn’t reverse the orientation of the arrows. Even so, there are other instances of functoriality. The following construction allows us to subsume those instances under the previous definition:

Definition C2.15 Let $\mathcal C$ be a category, then the opposite category $\mathcal C^{op}$ has the same objects. For $X,Y \in \mathrm{Obj}(\mathcal C^{op})$, we set $\mathrm{Hom}_{\mathcal C^{op}}(X,Y)=\mathrm{Hom}_{\mathcal C}(Y,X)$ composition is defined like the composition in $\mathcal C$ except in the opposite order.

Intuitively, we just reverse all the arrows.

Definition C2.16 Let $\mathcal C$ and $\mathcal D$ be categories, then a contravariant functor $F: \mathcal C \to \mathcal D$ is a covariant functor $\mathcal C^{op} \to \mathcal D$.

For clarification, regular functors (as in definition C2.1) are called covariant, but if “functor” is used without specification, they are always covariant by default.

From this definition, it follows that a contravariant functor reverses the direction of morphisms when it is applied. (Hence the picture in the beginning of this section)

Example C2.16 Consider the category of sets $\mathbf{Set}$. The assignment on objects $X \mapsto \mathcal{P}(X)$ can be made into a functor in two different ways, a covariant one and a contravariant one. For the covariant one, let $f:X \to Y$ be any map. Then we can define a map $\mathcal{P}(X) \to \mathcal{P}(Y)$ via $X \supset U \mapsto f(U)$, giving rise to the covariant powerset functor. For the contravariant one, we can define for a map $f:X \to Y$ a map $\mathcal{P}(Y) \to \mathcal{P}(X)$ via $Y \supset V \mapsto f^{-1}(V)$. In both cases the verification of the functorial properties is easy.

Example C2.17 Let $k$ be a field. Consider the category $\mathcal C$ of finite separable extensions of $k$ with $k$-linear field homomorphisms as morphisms. We can define a contravariant functor $F$ from $\mathcal C$ to $\mathbf{Grp}$, the category of groups with group homomorphisms as morphisms, in the following manner: For an object $L \in \mathrm{Obj}(\mathcal C)$, set $F(L)=L^\times$ and for a (necessarily injective) morphism $f:L \to M$, define $F(f)$ to be the field norm $N_{M/L}:M^\times \to L^\times$. The transitivity formula for norms in towers translates into functoriality of this construction: For a tower of extensions $L \subset M \subset N$ we have $N_{N/M} \circ N_{M/L}=N_{N/L}$.
A similar construction works for the field trace.

Example C2.18 Consider the category of compact Hausdorff spaces with continuous maps as morphisms. For any compact Hausdorff space $X$, the continuous functions $X \to \mathbb{C}$ form a C*-algebra with pointwise operations (i.e. addition, multiplication, scalar multiplication and conjugation), denoted by $C(X)$. If $X$ and $Y$ are compact Hausdorff spaces and $f:X \to Y$ is a continuous map, then we can define a homomorphism of C*-algebras by pulling back continuous functions along $f$: for a continuous function $g: Y \to \mathbb C$, $g \circ f$ is a continuous function $X \to \mathbb C$. We obtain a contravariant functor from the category of compact Hausdorff spaces to the category of commutative unital C*-algebras.

Example C2.19 As in the examples C2.4 and C2.5, if $G$ is a group, considered as a one-object category and we let $\mathcal C$ be the category of sets or the category of $k$-vector spaces for a field $k$, then contravariant functors correspond to right $G$-sets or right representations on vector spaces over $k$, respectively.

Example C2.20 The construction from C2.12 has a contravariant sibling. Let $\mathcal C$ be any category and $x \in \mathrm{Obj}(\mathcal C)$. Then we have for any object $y$ in $\mathcal C$, a set $\mathrm{Hom}_{\mathcal C}(y,x)$ and for a morphism $f:y \to y'$, we get a map $\mathrm{Hom}_{\mathcal C}(y',x) \to \mathrm{Hom}_{\mathcal C}(y,x), g \mapsto g \circ f$. This construction is a functor, called the contravariant Hom functor.

## A Brief Introduction to Categories, Part 1: Categories

The purpose of this post is to establish some basic language of categories. This can be read independently of the ongoing series on representation theory, although concepts and results from this post will be used in the representation theory series. So with respect to the representation theory series, one can think of this post (and the follow-up posts) as the analog of an appendix.

The notions of category theory can be encountered in many different mathematical contexts. To illustrate the concepts, a range of examples will be given, but I will go quickly over standard examples – For common examples in greater detail and multitude, see the references below. It is not necessary to know the background of every single example to grasp the concepts.

To readers who wish to read a more detailed introduction, I recommend Tom Leinster’s “Basic Category Theory”  (available as a free pdf, but also published in book form) or in case you know German Martin Brandenburg’s “Einführung in die Kategorientheorie”. If you simply cannot get enough, Francis Bourceux’ three-volume Handbook of Categorical Algebra is the right place to get lost in the realm of categories and functors.

The views expressed on category theory are naturally (no pun intended) my own highly subjective ones and others may differ, especially those with a categorical disapproval of categorical things.

## Introduction

The meta-mathematical notions of category theory provide a unifying framework for a lot of different mathematical theories. (Merely having a framework, however, can’t always replace delving into the specifics of the particular objects that one is interested in.) We can not only translate statements into categorical language, but we can also use categorical language to make things precise that might otherwise be vague, e.g. what it means for two different kinds of mathematical structures to be “equivalent” or what it means for a map to be “natural”.

Thinking categorically, we can economize on routine verifications by doing them in the utmost generality. More importantly, category theory can expose formal similarities between seemingly disconnected phenomena. These analogies might only become apparent through the high level of abstraction category-theoretic concepts provide, allowing one to transfer intuition, techniques and concepts in ways not available otherwise, at least not as easily or as precisely-defined.

Peter Freyd wrote once that “Perhaps the purpose of categorical algebra is to show that which is trivial is trivially trivial.”
Peter May commented: “I prefer an update of that quote: ‘ Perhaps the purpose of categorical algebra is to show that which is formal is formally formal’. It is by now abundantly clear that mathematics can be formal without being trivial.”

Now then! After this informal introduction, let us delve into the formal details of why that which is formal is formally formal.

## Categories

Definition C1.1 A category $\mathcal{C}$ consists of the following data:

• A class of objects $\mathrm{Obj}(\mathcal{C})$
• For every pair of objects $X,Y \in \mathrm{Obj}(\mathcal{C})$, a set of morphisms $\mathrm{Hom}_{\mathcal{C}}(X,Y)$
• For every triple of objects $X,Y,Z \in \mathrm{Obj}(\mathcal{C})$, a map $\mathrm{Hom}_{\mathcal{C}}(Y,Z) \times \mathrm{Hom}_{\mathcal{C}}(X,Y) \to \mathrm{Hom}_{\mathcal{C}}(X,Z)$ called composition and denoted by $(f,g) \mapsto f \circ g$

Such that the following conditions hold:

• For any object $X \in \mathrm{Obj}(\mathcal{C})$, there exists an identity $\mathrm{id}_X \in \mathrm{Hom}_{\mathcal C}(X,X)$ such that for all objects $Y,Z \in \mathrm{Obj}(\mathcal{C})$ and all morphisms $f \in \mathrm{Hom}_{\mathcal{C}}(Y,X), g \in \mathrm{Hom}_{\mathcal{C}}(X,Z)$ we have $\mathrm{id}_X \circ f= f$ and $g \circ \mathrm{id}_X = g$
• Composition is associative, that means for all objects $X,Y,Z,W \in \mathrm{Obj}(\mathcal{C})$ and $f \in \mathrm{Hom}_{\mathcal{C}}(X,Y), g \in \mathrm{Hom}_{\mathcal{C}}(Y,Z), h \in \mathrm{Hom}_{\mathcal{C}}(Z,W)$, we have $h \circ (g \circ f) = (h \circ g) \circ f$
• $\mathrm{Hom}_{\mathcal{C}}(X,Y) \cap \mathrm{Hom}_{\mathcal{C}}(X',Y') = \varnothing$ unless $X=X'$ and $Y=Y'$

Remark The last condition is just a technical one that is not very interesting and is mostly omitted from verifications that things are categories. You can always replace the Hom-sets by disjoint copies by set-theoretic juggling, so it’s not much of an issue. The advantage of this convention is that source and target of a morphism are uniquely determined.

One should think of a category as a big directed graph in which objects are represented by nodes and morphisms are represented by arrows and where we have a special operation called composition that takes two consecutive arrows and produces a third arrow representing going along one arrow first and then along the other one.
Compared with set theory, which can be called “materialistic”, in the sense that it is about what objects actually are made out of, category theory can be considered “behavioristic”, in the sense that it’s more about how objects behave, i.e. relate to each other. Relations between objects are conceptualized by the arrows also known as morphisms. But for morphisms, just like for the objects themselves, it’s more about how they relate each other and less about what they actually represent. The relations between morphisms are given by composition.
From a linguistic point of view, category theory is purely syntax, whereas semantics come into play only in the interpretation of particular examples.

Example C1.2 We can form the category of sets $\mathbf{Set}$, where $\mathrm{Obj}(\mathbf{Set})$ is the class of all sets and for $X,Y \in \mathrm{Obj}(\mathbf{Set})$, $\mathrm{Hom}_{\mathbf{Set}}(X,Y)$ is the set of all maps from $X$ to $Y$ and the composition of morphisms is just the usual composition of maps. Many other examples (but not all of them) can be formed as “subcategories” of this example in the sense that the objects are sets with additional structure and morphisms are mappings which preserve that structure in some way, such as topological spaces and continuous maps or algebraic structures of a specific type and homomorphisms or smooth manifolds and smooth maps or complex manifolds and holomorphic maps or varieties over an algebraically closed fields and regular maps etc.

Example C1.3 There is another category with the class of all sets as objects, but with more morphisms. Recall that a mapping is a particular kind of relation. $\mathbf{Rel}$ is the category consisting of all sets as objects, and for two objects $X,Y \in \mathrm{Obj}(\mathbf{Rel})$, $\mathrm{Hom}_{\mathbf{Rel}}(X,Y)$ is the set of all relations from $X$ to $Y$, i.e. subsets $R \subset X \times Y$. Given three sets $X,Y,Z$ and relations $R \subset X \times Y, S \subset Y \times Z$, we can form the composite relation $S \circ R := \{(x,z) \in X \times Z \mid \exists y \in Y: ((x,y) \in R \land (y,z) \in S) \}$. (If $R$ and $S$ are functions, this reduces to regular function composition.) With this composition, $\mathbf{Rel}$ satisfies the axioms for a category.

Example C1.4 In the previous examples for categories, the objects were given by the class of all sets. But there’s also the possibility to fix a particular set $X$ and consider $X$ as a category $\mathcal{C}$ in the following way: $\mathrm{Obj}(\mathcal{C})=X$ and the only morphisms are the identity morphisms required by the definition of categories. Due to the striking similarity with discrete topological spaces (compare for instance the set of morphisms $\mathrm{Hom}_{\mathcal{C}}(x,y)=\varnothing$ if $x \neq y$ and $\mathrm{Hom}_{\mathcal{C}}(x,y)=\mathrm{id}_x$ if $x=y$ with the discrete metric), this construction is called the discrete category on the set $X$.

Example C1.5 Instead of only allowing for the minimum amount of morphisms, we can take a set $X$ and consider such categories with $X$ as their class of objects that have at most one morphism between two objects. In such a case, the only interesting information is whether there is a morphism between two objects or not. Consequently, such a category corresponds to a relation $R \subset X \times X$. But the requirements for compositions and identities force a relation to have some properties if it is supposed to define a category in this manner: the existence of identity morphisms in the supposed category is equivalent to the relation being reflexive and the existence of compositions translates to transivity. A reflexive and transitive relation is called “preorder”. For instance, all equivalence relations and all partial orderings are preorders. An example of a preorder that is neither can be formed by taking a ring, say, an integral domain $A$ such as $\mathbb{Z}$ and considering the divisibility relation on elements of $A$. Here the failure to be a partial order is due to the existence of non-trivial units and the failure to be an equivalence relation is to due the existence of non-invertible elements.

Definition C1.6 Speaking of invertible things, just like for an element in a ring, we can define when a morphism in a category is invertible. Let $f:X \to Y$ a morphism in a category. Then $f$ is called invertible, or an isomorphism, if there is a morphism $g:Y \to X$ such that $g \circ f = \mathrm{id}_X$ and $f \circ g = \mathrm{id}_Y$. If there is such a morphism, then $X$ and $Y$ are called isomorphic. Note that the notations $f: X \to Y$ and $g \circ f$, albeit reminiscent of the situation in the category $\mathbf{Set}$, don’t necessarily refer to mappings or function composition, but to the particular morphisms and composition in an arbitrary category.

Example C1.7 Other than restricting the size of morphisms as in example C1.5, one can also restrict the size of the class of objects. For this, let’s consider categories with only one object. In general, if we consider composition as a function on all morphisms of a category, it is only partially defined, since source and target need to match up correctly. But this doesn’t arise if we have only one object. Let $\mathcal C$ be a category with one object $x$ and let $\mathrm{Hom}_{\mathcal C}(x,x)=M$. Then composition is a (totally defined) mapping $M \times M \to M$, i.e. a binary operation.
The existence of the identity $\mathrm{id}_x:=e$ implies that $e$ is a unit with respect to this binary operation and the associativity of the composition implies that the binary operation is, well, associative. Thus composition $\circ:M \times M \to M$ makes $M$ into a monoid.
Conversely, given a monoid $M$, we can turn $M$ into a one-object category by taking a proxy object $x$ with no particular meaning, setting $\mathrm{Hom}(x,x)=M$ and defining the composition via the monoid operation. These constructions are inverse to each other, such that monoids correspond to one-object categories.
A particular frequently occuring and much beloved class of monoids consists of groups, i.e. monoids in which every element is invertible. By a pleasant convergence of terminology, an element in a monoid being invertible (in the classical sense) is the same as being an invertible morphism if we consider the monoid as a one-object category. Thus we can say that a monoid is a group if and only if every morphism is an isomorphism. This leads to the following notion:

## Groupoids

Definition C1.8 A groupoid is a category such that every morphism is an isomorphism.

If we think of groups as encoding the symmetries of one object, then groupoids can capture the symmetries of a configuration with many objects.

Example C1.9 Let $R$ be a preorder on a set $X$, then we can form a category from these data as described in example C1.8. One can ask when one obtains groupoid in this way. It turns out that the corresponding category is a groupoid if and only if $R$ is symmetric, i.e. an equivalence relation.

Remark From this observation we can extract an intuition for general groupoids: a groupoid is like a set with an equivalence relation, except that two objects can be equivalent in more than one way and we’re keeping track of different ways of being equivalent. From this viewpoint a group, being a one-object groupoid, is like the various ways a particular object is equivalent to itself, following the narrative that groups encode the symmetries of an object.

Example C1.10 Let $G$ be a group acting on a set $X$, then one can form a groupoid encoding the group action as follows: Let $X // G$ be the category having elements from $X$ as its objects and for every $x \in X$ and $g \in G$, there’s a morphism $x \to gx$, such that $\mathrm{Hom}_{X // G}(x,y)$ is in bijection with $\{g \in G:gx=y\}$. This groupoid played an important role in a previous post.

Example C1.11 Let $\mathcal C$ be any category. Then there is a maximal subcategory that is a groupoid: simply take the subcategory of $\mathcal C$ with the same objects as $\mathcal C$ and all the isomorphisms of $\mathcal C$ as morphisms. This is called the core of the category $\mathcal C$ and it is a groupoid. If $\mathcal C$ has only one object, i.e. is a monoid, then the core of $\mathcal C$ is just the group of invertible elements, also known as units.
As an exercise, one can check that the categories $\mathbf{Set}$ and $\mathbf{Rel}$ (cf. C1.2 and C1.3) have the same core, i.e. any invertible relation is already a function and in fact a bijection.

Example C1.12 Let $X$ be a topological space. Then the fundamental groupoid $\Pi_1(X)$ has as its objects all elements of $X$. For $x,y \in X$, $\mathrm{Hom}_{\Pi_{\leq 1}(X)}(x,y)$ consists of all homotopy classes of paths from $x$ to $y$. Composition is given by concatenation of paths. $\Pi_{\leq 1}(X)$ contains important homotopy-theoretic information about $X$, such as the path-components and the fundamental group of each path-component. For more information on this very geometric construction and groupoids in algebraic topology, I heartily recommend Ronald Brown’s “Topology and Groupoids”.

This concludes the first part of the mini-series on introductory categories. Next time we’ll look into functors.

## Tensor Products for Group Actions, Part 2

In this previous post, tensor products of $G$-sets were introduced and some basic properties were proved, this post is a continuation, so I’ll asume that you’re familiar with the contents of that post.

In this post, unless specified otherwise, $G,H,K,I$ will denote groups. Groups will sometimes be freely identified with the corresponding one-object category. Left $G$-sets will sometimes just be referred to as $G$-sets.

After some lemmas, we will begin this post by some easy consequences of the Hom-Tensor adjunction, which is the main result of the previous post.

Lemma The category $G\textrm{-}\mathbf{Set}$ is complete and cocomplete and the limits and colimits look like in the category of sets (i.e. the forgetful functor $G\textrm{-}\mathbf{Set} \to \mathbf{Set}$ is continuous and cocontinuous) with the “obvious” actions from $G$. Similarly for $\mathbf{Set}\textrm{-}G$.

Proof This is not too difficult to prove directly (you can reduce the existence of (co)limits to (co)products and (co)equalizers by general nonsense), but it also follows directly from the fact that $G\textrm{-}\mathbf{Set}$ is the functor category $[G, \mathbf{Set}]$. The reason is that if $\mathcal{C}$ and $\mathcal{D}$ are categories and $\mathcal{D}$ is (co)complete (and $\mathcal{C}$ is small to avoid any set-theoretic trouble), then the functor category $[\mathcal{C},\mathcal{D}]$ is also (co)complete and the (co)limits may be computed “pointwise”. In the case of $[G, \mathbf{Set}]$, $G$ has only one object, so the (co)limits look like they do in $\mathbf{Set}$.

Lemma If $X$ is a $(G,H)$-set, $Y$ is a $(H,K)$-set and $Z$ is a $(K,I)$-set, then we have a natural isomorphism of $(G,I)$-sets
$(X \otimes_H Y) \otimes_K Z \cong X \otimes_H (Y \otimes_K Z)$

Proof The proof is the same as the proof for modules, mutatis mutandis. Use the universal property of tensor products a lot to get well-defined maps $(X \otimes_H Y) \otimes_K Z \to X \otimes_H (Y \otimes_K Z) , (x \otimes y) \otimes z \mapsto x \otimes (y \otimes z)$ and  $X \otimes_H (Y \otimes_K Z) \to (X \otimes_H Y) \otimes_K Z, x\otimes (y\otimes z) \mapsto (x \otimes y) \otimes z$.

Lemma If $H \leq G$ is a subgroup, and we regard $G$ as a $(G,H)$-set via left and right multiplication, then $\mathrm{Hom}_{G\textrm{-}\mathbf{Set}}(G,-): G\textrm{-}\mathbf{Set} \to H\textrm{-}\mathbf{Set}$ is naturally isomorphic to the restriction functor $\mathrm{res}_H^G: G\textrm{-}\mathbf{Set} \to H\textrm{-}\mathbf{Set}$ (this functor takes any $G$-set, which we may think of as a group homomorphism or a functor and restricts it to the subgroup/subcategory given by $H$.)

Proof Define a (natural) map $\varphi: \mathrm{Hom}_{G\textrm{-}\mathbf{Set}}(G,X) \to \mathrm{Res}_H^G(X)$ via $\varphi(f)=f(1)$. This is $H$-equivariant, because $\varphi(hf)(1)=f(1h)=f(h)=hf(1)=h\varphi(f)$. On the other hand, given $x \in \mathrm{Res}_H^G(X)$ (which is just $X$ as a set), we can define $f \in \mathrm{Hom}_{G\textrm{-}\mathbf{Set}}(G,X)$ via $f(g)=gx$. This defines an inverse for $\varphi$.

Corollary The restriction functor $\mathrm{Res}_H^G$ has a left adjoint $G \otimes_H - := \mathrm{Ind}_H^G$.

The notation $\mathrm{Ind}$ is chosen because we can think of this functor as an analog to the induced representation from linear representation theory, where we think of group actions as non-linear represenations. (Similar to the induced representation, one can give an explicit description of $\mathrm{Ind}_H^G$ after choosing coset representatives for $G/H$ etc.)
In linear represenation theory, the adjunction between restriction and induction is called Frobenius reciprocity, so if we wish to give our results fancy names (as mathematicians like to do) we can call this corollary “non-linear Frobenius reciprocity”.

If we take $H$ to be the trivial subgroup, we obtain a corollary of the corollary:

Corollary The forgetful functor $G\textrm{-}\mathbf{Set}\to \mathbf{Set}$ has a left adjoint, the “free $G$-set functor”.

Proof If $H$ is the trivial group, then $H$-sets are the same as sets and the restriction functor $G\textrm{-}\mathbf{Set}\to H\textrm{-}\mathbf{Set}$ is the same as the forgetful functor. Since $G \otimes_H$ commutes with coproducts and $H$ is a one-point set, we can also describe this more explicitly: for a set $X$, we have $G \otimes_H X \cong G \otimes_H \coprod_{x \in X} H \cong \coprod_{x \in X} G \otimes_H H \cong \coprod_{x \in X} G := G^{(X)}$

We can also use the Hom-Tensor adjunction to get a description of some tensor products. Let $1$ denote a one-point set (simultanously the trivial group), considered as a $(1,G)$-set with (necessarily) trivial actions.

Lemma For a $G$-set $X$, $1 \otimes_G X$ is naturally isomorphic to the set of orbits $X/G$ and both are left adjoint to the functor $\mathbf{Set} \to G\textrm{-}\mathbf{Set}$ which endows every set with a trivial $G$-action.

Proof Let $Y$ be a set and $X$ be a $G$-set. Denote $Y^{triv}$ the $G$-set with $Y$ as its set and a trivial action. If we have any $f \in \mathrm{Hom}_{G\textrm{-}\mathbf{Set}}(X,Y^{triv})$, then $f$ must be constant on the orbits, since $f(gx)=gf(x)=f(x)$, so $f$ descends to a map of sets $X/G \to Y$. Conversely, if we have any map $h: X/G \to Y$, then we can define a $G$-equivariant map $f:X \to Y^{triv}$ by setting $f(x)=h([x])$, where $[x]$ denotes the orbit of $x$. These maps are mutually inverse natural bijection which shows that “set of orbits”-functor is left adjoint to $Y \mapsto Y^{triv}$. On the other hand, we can identify $Y^{triv}$ with $\mathrm{Hom}_{\mathbf{Set}}(1,Y)$ (where the $G$ action is induced from the trivial right $G$-action on $1$), so the left adjoint must be given by $X \mapsto 1 \otimes_G X$. Since adjoints are unique (by a Yoneda argument), we have a natural bijection $1 \otimes_G X \cong X/G$

The set of orbits $X/G$ carries some information about the $G$-set, but we can do a more careful construction which also includes $X/G$ in a natural way as part of the information.

Definition If $X$ is a $G$-set, then the action groupoid $X//G$ is the category with $\mathrm{Obj}(X//G) := X$ and $\mathrm{Hom}_{X//G}(x,y):= \{g \in G \mid gx=y\}$. Composition is given by $\mathrm{Hom}_{X//G}(y,z) \times\mathrm{Hom}_{X//G}(x,y) \to \mathrm{Hom}_{X//G}(x,z), (h,g) \mapsto hg$.

The fact that this is called a groupoid is not important here, one can think of that as just a name (it means that every morphism in $X//G$ is an isomorphism).
The set of isomorphism classes of $X//G$ correspond to the orbits $X/G$. For $x \in X//G$, the endomorphisms $\mathrm{End}_{X//G}$ is the stabilizer group $G_x$. The following lemma shows how to reconstruct a $G$-set $X$ from $X//G$, assuming that we know how all the Hom-sets lie inside $G$.

Lemma (“reconstruction lemma”) If $X$ is a $G$-set, then we define the functor $F: (X//G)^{op} \to G\textrm{-}\mathbf{Set}$ with $F(x)=G$ for all $x \in X//G$ and for $g \in \mathrm{Hom}_{(X//G)}(x,y)$, we define the map $F(g): F(y) \to F(x)$ via $a \mapsto ag$. Then we have $\varinjlim\limits_{x \in (X//G)^{op}}F(x) \cong X$

Proof For $x \in X//G$, define a map $F(x)=G \to X$ via $g \mapsto gx$. This defines a cocone over $G(.)$, so we get an induced map $\varphi: \varinjlim\limits_{x \in (X//G)^{op}}F(x) \to X$.  $\varinjlim\limits_{x \in (X//G)^{op}}F(x)$ can be described explicitly as $\coprod_{x \in (X//G)^{op}} F(x)/\sim$, where the equivalence relation $\sim$ is generated by $ga \in F(x) \sim a \in G(gx)$. To see that $\varphi$ is surjective, note that $x \in X$ is the image of $1 \in F(x)$. To see that $\varphi$ is injective, suppose $g \in F(x)$ and $h \in G(y)$ are sent to the same element, i.e. $gx=hy$, then we have $(h^{-1}g)x=y$, so that we may assume $h=1$. Then $gx=y$ implies that $1 \in G(y)=G(gx) \sim g \in F(x)$, so the two elements which map to the same element are already equal in $\varinjlim\limits_{x \in (X//G)^{op}}F(x)$.

The previous lemma can be thought of as a generalization of the orbit-stabilizer theorem.  (The proof has strong similarities as well.) For illustration, let us derive the usual orbit-stabilizer theorem from it.

Lemma Let $X$ be a $G$-set, then we have an isomorphism of $G$-sets $G/G_x \cong Gx$, where $Gx$ is the orbit of $x$ (with the restricted action) and $G/G_x$ is the coset space of the stabilizer subgroup with left multiplication as the action.

Proof We may replace $X$ with $Gx$ so that we have a transitive action. Then the previous lemma gives us an isomorphism $X \cong \varinjlim\limits_{x \in (X//G)^{op}}F(x)$.
Consider the one-object category $(G_x)$. This can be identified with a full subcategory of $X//G$ corresponding to the object $x$. Because we have a transitive action, all objects in $X//G$ are isomorphic (isomorphism classes correspond to orbits), so that the inclusion functor $G_x \to X//G$ is also essentially surjective, so it is a category equivalence.
We may thus replace the colimit by the colimit $\varinjlim\limits_{x \in (G_x)^{op}}F(x)$. As $(G_x)^{op}$ has just one object, this colimit is a colimit over a bunch of parallel morphism $F(x) \to F(x)$, so it is the simultanous coequalizer of these morphisms. We know how to compute coequalizers in $G\textrm{-}\mathbf{Set}$: the same way that we compute coequalizers in $\mathbf{Set}$. So we have the families of maps $\cdot g: F(x)=G \to G, a \mapsto ag$, where $g$ varies over $G_x$. The coequalizer is the quotient $G/\sim$, where $\sim$ is generated by $a \sim ag$ for each $a \in G$ and $g \in G_x$. But this is exactly the equivalence relation that defines $G/G_x$.

There is another case where the colimit takes a simple form after replacing $X//G$ with an equivalent category.

Lemma A $G$-set $X$ is free in the sense that it is in the essential image of the “free $G$-set functor” $Y \mapsto G \otimes_{1} Y$ or equivalently it is a coproduct of copies of $G$ with the standard action iff the action of $G$ on $X$ is free in the sense that $\forall x \in X \forall g \in G: (gx=x \Rightarrow g=1)$.

Proof It’s clear that if we have a disjoint union $X= \coprod_{i \in I} G$, then no element of $G$ other than $1$ can fix an element in $X$. For the other direction, suppose that we have the condition $\forall x \in X \forall g \in G: (gx=x \Rightarrow g=1)$. This implies that the morphism sets in the action groupoid are really small: Suppose $g,h \in \mathrm{Hom}(x,y)$ such that $gx=y=hx$, which implies that $h^{-1}gx=x$, so $h^{-1}g=1$ by assumption, thus $h=g$. This means that for any pair of objects in $X//G$, there is at most one morphism between them. So if we consider the set $X/G$ as a discrete category (i.e. the only morphisms are the identities), then if we take a representative for each orbit $X/G$, this defines an inclusion of categories $X/G \to X//G$. As elements in $X/G$ represent isomorphism classes in $X//G$, this inclusion is always essentially surjective. By our computations of the Hom-sets, it is also fully faithful if the action of $G$ on $X$ is free. So if we apply the “reconstruction lemma” we get $X \cong \varinjlim\limits_{x \in (X//G)^{op}}G(x) \cong \varinjlim\limits_{x \in X/G} G$. But a colimit over a discrete category is just a coproduct, so this is isomorphic to $\coprod_{x \in X/G} G$ which shows that $X$ is free.

After some further lemmas, we will come to the main result of this post, which is also an application of the reconstruction lemma.

In the previous post, I described $G$-sets in different ways, among them as functors $G \to \mathbf{Set}$, but I didn’t do the same for $(G,H)$-sets. The following lemma remedies this deficiency.

Lemma $(G,H)$-sets may be identified with left $G \times H^{op}$-sets or with functors $G \to \mathbf{Set}\textrm{-}H$ or with functors $H^{op} \to G\textrm{-}\mathbf{Set}$. In other words, we have equivalences of categories $G\textrm{-}\mathbf{Set}\textrm{-}H \cong G\times H^{op}\textrm{-}\mathbf{Set} \cong [G,\mathbf{Set}\textrm{-}H] \cong [H^{op},G\textrm{-}\mathbf{Set}]$.

The proof of this lemma is a lot of rewriting of definitions, not more difficult than proving the corresponding statements for one-sided $G$-sets.

This lemma has a useful consequence, which one could also verify by hand:

Observation If $F: G\textrm{-}\mathbf{Set} \to H\textrm{-}\mathbf{Set}$ is a functor and $X$ is a $(G,K)$-set, then $F(X)$ is a $(H,K)$-set in a “natural” way.

Proof Think of $X$ as functor $X:K^{op} \to G\textrm{-}\mathbf{Set}$, composing with $F$, gives us a functor $F(X): K^{op} \to H\textrm{-}\mathbf{Set}$, which we may also think of as a $(H,K)$-set.
More explicitly, the action of $K$ on $F(X)$ can be described as follows: for $k \in K$, the right-multiplication-map $X \to X, x \mapsto xk$ is left $G$-equivariant, so it induces a left $H$-equivariant map $F(X) \to F(X)$, we can define the action of $k$ on $F(X)$ via this map.

The following lemma is an analog of the classical Eilenberg-Watts theorem from homological algebra which describes colimit-preserving functors $R\textrm{-}\mathbf{Set} \to S\textrm{-}\mathbf{Set}$ as tensor products with a $(S,R)$-bimodule.

Thereom (Eilenberg-Watts theorem for group actions) Every colimit-preserving functor $W: G\textrm{-}\mathbf{Set} \to H\textrm{-}\mathbf{Set}$ is naturally equivalent to $X \otimes_G$ for a $(H,G)$-set $X$. One can explicitly choose $X = W(G)$ (with the $(H,G)$-set structure from the previous observation, as $G$ is a $(G,G)$-set.)

Proof Let $X$ be a $G$-set and $Y$ be a $H$-set, then we have a natural bijection $\mathrm{Hom}_{H\textrm{-}\mathbf{Set}}(W(G) \otimes_G X, Y) \cong \mathrm{Hom}_{G\textrm{-}\mathbf{Set}}(X,\mathrm{Hom}_{H\textrm{-}\mathbf{Set}}(W(G),Y))$
Using the reconstruction lemma, we get $\mathrm{Hom}_{G\textrm{-}\mathbf{Set}}(X,\mathrm{Hom}_{H\textrm{-}\mathbf{Set}}(W(G),Y)) \cong \mathrm{Hom}_{G\textrm{-}\mathbf{Set}}(\varinjlim\limits_{x \in (X//G)^{op}}F(x),\mathrm{Hom}_{H\textrm{-}\mathbf{Set}}(W(G),Y)) \cong \varprojlim\limits_{x \in (X//G)^{op}}\mathrm{Hom}_{G\textrm{-}\mathbf{Set}}(F(x),\mathrm{Hom}_{H\textrm{-}\mathbf{Set}}(W(G),Y))$
For every $x \in (X//G)^{op}$, $F(x)=G$, so $\mathrm{Hom}_{G\textrm{-}\mathbf{Set}}(F(x),\mathrm{Hom}_{H\textrm{-}\mathbf{Set}}(W(G),Y)) \cong \mathrm{Hom}_{H\textrm{-}\mathbf{Set}}(W(G),Y))$ via the map $f \mapsto f(1)$. We need to consider how this identification behaves under the morphisms involved in the colimit. For $g \in G$, we have the map $F(gx) \to F(x), a \mapsto ag$, this induces a map $\varphi_g: \mathrm{Hom}_{G\textrm{-}\mathbf{Set}}(F(x),\mathrm{Hom}_{H\textrm{-}\mathbf{Set}}(W(G),Y)) \to \mathrm{Hom}_{G\textrm{-}\mathbf{Set}}(F(gx),\mathrm{Hom}_{H\textrm{-}\mathbf{Set}}(W(G),Y))$ given by $\varphi_g(f)(h)=f(hg)$. If we make the indentification described above by evaluating both sides at $1$, we get $\varphi_g(f)(1)=f(1g)=gf(1)$. Using the definition of the $G$-action on the Hom-set $\mathrm{Hom}_{H\textrm{-}\mathbf{Set}}(W(G),Y)$, this left multiplication translates to right multiplication on $W(G)$. Because of the construction of the right $G$-action on $W(G)$, this right multiplication is the map that is induced from right multiplication $G \to G$. We may summarize this computation by stating that $\varprojlim\limits_{x \in (X//G)^{op}}\mathrm{Hom}_{G\textrm{-}\mathbf{Set}}(F(x),\mathrm{Hom}_{H\textrm{-}\mathbf{Set}}(W(G),Y)) \cong \varprojlim\limits_{x \in (X//G)^{op}}\mathrm{Hom}_{H\textrm{-}\mathbf{Set}}(W(F(x)),Y)$
Using the assumption that $W$ preserves colimits, we get $\varprojlim\limits_{x \in (X//G)^{op}}\mathrm{Hom}_{H\textrm{-}\mathbf{Set}}(W(F(x)),Y) \cong \mathrm{Hom}_{H\textrm{-}\mathbf{Set}}(\varinjlim\limits_{x \in (X//G)^{op}}W(F(x)),Y) \cong \mathrm{Hom}_{H\textrm{-}\mathbf{Set}}(W(\varinjlim\limits_{x \in (X//G)^{op}} F(x)),Y) \cong \mathrm{Hom}_{H\textrm{-}\mathbf{Set}}(W(X),Y)$ where we used the reconstruction lemma again in the last step.
We conclude $W(X) \cong W(G) \otimes_G X$ by the Yoneda lemma.

This theorem (like the classical Eilenberg-Watts-theorem) is remarkable not only because it gives a concrete description of every colimit-preserving functor between certain categories, but also because it shows that such functors are completely determined by the image of one object $G$ and how it acts on the endomorphisms of that object (which are precisely the right-multiplications.)

It’s natural to ask at this point when two functors of the form $X \otimes_G$ and $Y \otimes_G$ for $(H,G)$-sets are naturally isomorphic. It’s not difficult to see that it is sufficient that $X$ and $Y$ are isomorphic as $(H,G)$-sets. The following lemma shows that this is also necessary, among other things.

Lemma For $(H,G)$-sets $X$ and $Y$, every natural transformation $\eta:X \otimes_G \to Y \otimes_G$ is induced by a unique $(H,G)$-equivariant map $f: X \to Y$

Proof Assume we have a natural transformation $\eta_A: X \otimes_G A \to Y \otimes_G A$, then we have in particular a left $H$-equivariant map $\eta_G: X \otimes_G G \to Y \otimes_G G$. We have $X \otimes_G G \cong X$ and $Y \otimes_G G \cong Y$, so this gives us a $H$-equivariant map $X \to Y$ which I call $f$. Clearly $f$ is uniquely determined by this construction. For a fixed $g \in G$, right multiplication by $g$ defines a left $G$-equivariant map $G \to G$. Under the isomorphism $X \otimes_G G \cong X$ these maps describe the right $G$ action on $X$. Naturality with respect to these maps implies that $f$ is right $G$-equivariant.

This lemma allows a reformulation of the previous theorem.

Theorem (Eilenberg-Watts theorem for group actions, alternative version)
The following bicategories are equivalent:
– The bicategory where the objects are groups, 1-morphisms between two groups $G, H$ are $(G,H)$-sets $X$, where the composition of 1-morphisms is given by taking tensor products and 2-morphisms between two $(G,H)$-sets are given by $(G,H)$-equivariant maps.
– The 2-subcategory of the 2-category of categories $\mathbf{Cat}$ where the objects are all the categories $G\textrm{-}\mathbf{Set}$ for groups $G$, 1-morphisms are colimit-preserving functors $G\textrm{-}\mathbf{Set} \to H\textrm{-}\mathbf{Set}$ and 2-morphisms are natural transformations between such functors.

This concludes my second blog post. If you want, please share or leave comments below.

## Tensor Products for Group Actions, Part 1

Suppose $R$ is a ring, $M$ is a right $R$-module and $N$ is a left $R$-module, then we can form the tensor product $M \otimes_R N$ that we all know and love. If we consider that a module is just an abelian group together with an action from a ring, one might ask the question: Can we imitate this construction for group actions?

It turns out that we can; this is what I’ll investigate in this post.

Recall that for a fixed group $G$ a left $G$-set is a set $X$ together with a map $G \times X \to X, (g,x) \mapsto gx$ that satisfies $\forall g,h \in G, x \in X: 1x=x, g(hx)=(gh)x$. An equivalent formulation is that a left $G$-set is a set $X$ together with a group homomorphism $G \to S_X$, where $S_X$ denotes the group of bijections $X \to X$. Yet another way to phrase this definition is that a left $G$-set is a functor from $G$ (regarded as a one-object category) to the category of sets.

There’s an obvious choice of morphisms for left $G$-sets $X$ and $Y$,  namely maps $f: X \to Y$ that satisfy $\forall x \in X, g \in G: f(gx)=gf(x)$. Traditionally called $G$-equivariant maps. If you take the definiton via functors, then these are exactly natural transformations. It follows that the left $G$-sets form a category, which we will denote by $G\textrm{-}\mathbf{Set}$.

In a similar way, one defines right $G$-sets as sets $X$ together with a map $X \times G \to X, (x,g) \mapsto xg$ that satisfies $\forall g,h \in G, x \in X: x1=x, (xg)h=x(gh)$, these can also be thought of as contravariant functors from $G$ to $\mathbf{Set}$ or equivalently as left $G^{op}$-sets, where $G^{op}$ denotes the opposite group of $G$. The category of right $G$-sets will be denoted by $\mathbf{Set}\textrm{-}G$.

If we have two groups $G$ and $H$, then a $(G,H)$-set is a set $X$ that is simultaneously a left $G$-set and a right $H$-set, such that the actions are compatible in the sense that $\forall x \in X, g \in G, h \in H: (gx)h=g(xh)$. We get the category of $(G,H)$-sets $G\textrm{-}\mathbf{Set}\textrm{-}H$ where the morphisms are maps that are both $G$ and $H$-equivariant. Note that if we take $H$ or $G$ to be the trivial group, then $(G,H)$-sets are equivalent to left $G$-sets or right $H$-sets respectively, thus we may always assume to deal with $(G,H)$-sets and will cover all three cases. Also note that if we take both $G$ and $H$ to be the trivial group, then $(G,H)$-sets are just sets.

We are now ready to begin our renarration of the story of tensor products, where groups play the role of rings, sets play the role of abelian groups and left and right $G$-sets play the role of modules.

Since $\textrm{Hom}$-Sets are important in the theory of tensor products for modules, we will study these first.

In the following, $G, H$ and $K$ will denote groups.

Lemma If $X$ is a $(G,H)$-set and $Y$ is a $(G,K)$-set, then $\mathrm{Hom}_{G\textrm{-}\mathbf{Set}}(X,Y)$ is a $(H,K)$-set with the actions defined by $\forall h \in H \forall x \in X: (hf)(x) := f(xh)$ and $\forall k \in K, \forall x \in X (fk)(x):=f(x)k$. If $X$ is a $(H,G)$-set and $Y$ is a $(K,G)$-set, then $\textrm{Hom}_{\mathbf{Set}\textrm{-}G}(X,Y)$ is a $(K,H)$-set in the analogous way.

The proof is a routine verification.

As a special case, when $Y$ is just a set and $X$ is a $G$-set, then $\textrm{Hom}_{\mathbf{Set}}(X,Y)$ is a $G$-set, with the action on the opposite side. If $X$ is a $(G,H)$-set, then the two actions we get this way are compatible, so that $\textrm{Hom}_{\mathbf{Set}}(X,Y)$ is a $(H,G)$-set.

We can copy the following definition almost verbatim from the case for modules.

Definition If $X$ is a $(H,G)$-set, $Y$ is a $(G,K)$-set and $Z$ is a $(H,K)$-set, then a map $f: X \times Y \to Z$ is called $G$-balanced and $(H,K)$-equivariant if $\forall x \in X, y \in Y, g \in G: f(xg,y)=f(x,gy)$ and $\forall h \in H, k \in K: f(hx,yk)=hf(x,y)k$. We denote the set of all such maps by $\textrm{Bal}_G^{(H,K)}(X,Y;Z)$. If $H$ and $K$ are the trivial group, we drop them from the notation and just write $\textrm{Bal}_G(X,Y;Z)$.

Lemma If $X$ is a $(H,G)$-set, $Y$ is a $(G,K)$-set and $Z$ is any set, then $\textrm{Bal}_G(X,Y;Z)$ is a $(K,H)$-set with the actions defined by $\forall k \in K, x \in X, y \in Y: (kf) (x,y) := f(x,yk)$ and $\forall h \in H, x \in X, y \in Y: (fh) (x,y) :=f(hx,y)$.

This is again a routine verification, similar to the previous lemma.

We now come to the usual “Currying” argument.

Lemma If $X$ is a $(H,G)$-set and $Y$ is a $(G,K)$-set and $Z$ is any set, then there is a natural isomorphism of $(K,H)$-sets $\textrm{Bal}_G(X,Y;Z) \cong \textrm{Hom}_{\mathbf{Set}\textrm{-}G}(X, \textrm{Hom}_{\mathbf{Set}}(Y,Z))$

Proof We define the “Currying map” $\varphi: \textrm{Bal}_G(X,Y;Z) \to \textrm{Hom}_{\mathbf{Set}\textrm{-}G}(X, \textrm{Hom}_{\mathbf{Set}}(Y,Z))$ via $f \mapsto \varphi(f)$, where $\varphi(f)(x) = (y \mapsto f(x,y))$. Or more compactly $f \mapsto (x \mapsto (y \mapsto f(x,y)))$
Let us check that for $f \in \textrm{Bal}_G(X,Y;Z)$ the map $x \mapsto \varphi(f)(x)$ is $G$-equivariant, for $g \in G, y \in Y$ $\varphi(f)(x)g$ is defined as $(\varphi(f)(x)g)(y) = \varphi(f)(x)(gy)=f(x,gy)$. Because $f$ is $G$-balanced, this equals $f(xg,y)= \varphi(f)(xg)(y)$,  so $\varphi(f)(xg)=\varphi(f)(x)g$.
Now we check that $\varphi$ is $K$– and $H$-equivariant. Let $k \in K$, then for all $f \in \textrm{Bal}_G(X,Y;Z), x \in X, y \in Y$, we have $\varphi(kf)(x)(y)=kf(x,y)=f(x,yk)$, on the other hand we have $(k \varphi(f)(x))(y)=\varphi(f)(x)(yk)=f(x,yk)$, this shows that $\varphi(kf)=k\varphi(f)$.
Let $h \in H$, then for all $f \in \textrm{Bal}_G(X,Y;Z), x \in X, y \in Y$, we have $\varphi(fh)(x)(y)=f(hx,y)$, on the other hand $(\varphi(f)(x)h)(y)=f(hx,y)$, so $\varphi(fh)=\varphi(f)h$.
So we have shown that $\varphi$ is indeed a well-defined $(K,H)$-equivariant map $\textrm{Bal}_G(X,Y;Z) \to \textrm{Hom}_{\mathbf{Set}\textrm{-}G}(X, \textrm{Hom}_{\mathbf{Set}}(Y,Z))$.
To see that $\varphi$ is bijective, note that an inverse is given by the “Uncurry”-map $\psi: \textrm{Hom}_{\mathbf{Set}\textrm{-}G}(X, \textrm{Hom}_{\mathbf{Set}}(Y,Z)) \to \textrm{Bal}_G(X,Y;Z), \xi \mapsto \psi(\xi)$, where $\psi(\xi)(x,y)=\xi(x)(y)$.
We omit the verification that $\varphi$ is natural in $X$, $Y$ and $Z$.

We also give the following variants of the Currying isomorphism:

Lemma If $X$ is a $(H,G)$-set, $Y$ is a $(G,K)$-set and $Z$ is a $(H,K)$-set, then we have a natural bijections
$\mathrm{Bal}_G^{(H,K)}(X,Y;Z) \cong \mathrm{Hom}_{H\textrm{-}\mathbf{Set}\textrm{-}G}(X,(\mathrm{Hom}_{\mathbf{Set}\textrm{-}K}(Y,Z))$
and $\mathrm{Bal}_G^{(H,K)}(X,Y;Z) \cong \mathrm{Hom}_{G\textrm{-}\mathbf{Set}\textrm{-}K}(Y,\mathrm{Hom}_{H\textrm{-}\mathbf{Set}}(X,Z))$

Proof The proof is very similar to the last one, so we omit some steps.
For the first bijection, let us just show that the map $\varphi: \mathrm{Bal}_G^{(H,K)}(X,Y;Z) \to \mathrm{Hom}_{H\textrm{-}\mathbf{Set}\textrm{-}G}(X,(\mathrm{Hom}_{\mathbf{Set}\textrm{-}K}(Y,Z)), f \mapsto \varphi(f)$, where $\varphi(f)(x)(y)=f(x,y)$ is well-defined.
For a fixed $x \in X$, we have $\varphi(f)(x)(yk)=f(x,yk)=f(x,y)k=\varphi(f)(x)(y)k$, so $\varphi(f)(x)$ is $K$-equivariant.
Let $h \in H, g \in G$, then $\varphi(f)(hxg)(y)=f(hxg,y)=hf(xg,y)=hf(x,gy)=h\varphi(f)(x)(gy)= (h\varphi(f)g)(y)$.
So we get $\varphi(f)(hxg)=h\varphi(f)(x)g$, thus $\varphi(f)$ is $(H,G)$-equivariant.
For the second bijection, we use the map $\psi: \mathrm{Bal}_G^{(H,K)}(X,Y;Z) \cong \mathrm{Hom}_{G\textrm{-}\mathbf{Set}\textrm{-}K}(Y,\mathrm{Hom}_{H\textrm{-}\mathbf{Set}}(X,Z)), f \mapsto \psi(f)$, where $\psi(f)(y)(x) = f(x,y)$. The verification that this is a well-defined bijection is very similar to the previous computations.

We now come to the main definition of this post.

Definition Let $X$ be a right $G$-set, $Y$ be a left $G$-set, then a tensor product $X \otimes_G Y$ is a set together with a $G$-balanced map $\psi_{\otimes}: X \times Y \to X \otimes_G Y$ that satisfies the following universal property: for any set $Z$ and $G$-balanced map $f: X \times Y \to Z$, there is a unique map $\overline{f}: X \otimes_G Y \to Z$ such that $\overline{f} \circ \psi_{\otimes} = f$.

Lemma In the situation of the previous definition, the tensor product $(\psi_{\otimes}, X \otimes_G Y)$ exists and if $(\psi'_{\otimes}, (X \otimes_G Y)')$ is a second tensor product, then there exists a unique bijection $g: X \otimes_G Y \to (X \otimes_G Y)'$ such that $\psi'_{\otimes}=\psi_{\otimes} \circ f$.

Proof We first prove uniqueness. Suppose $(\psi_{\otimes}, X \otimes_G Y)$ and $(\psi'_{\otimes}, (X \otimes_G Y)')$, then because of the universal property of $X \otimes_G Y$, we get a unique map $f: X \otimes_G Y \to (X \otimes_G Y)'$ such that $\psi'_{\otimes}=f \circ \psi_{\otimes}$ and due to universal property of $(X \otimes_G Y)'$ we get a unique map $g:(X \otimes_G Y)' \to X \otimes_G Y$ such that $\psi_{\otimes}=g \circ \psi'_{\otimes}$, then $g \circ f$ satisfies $g \circ f \circ \psi_{\otimes} = g \circ \psi'_{\otimes} = \psi_{\otimes}$, but by the universal property of $X \otimes_G Y$, there is a unique map $h: X \otimes_G Y \to X \otimes_G Y$ that satisfies $h \circ \psi_{\otimes} = \psi_{\otimes}$. We have just computed that $g$ satisfies this, but $\textrm{id}_{X \otimes_G Y}$ does, too. Thus $g \circ f = \textrm{id}_{X \otimes_G Y}$. The proof that $f \circ g = \textrm{id}_{(X \otimes_G Y)'}$ follows analogously.

Now we show existence. Consider the equivalence relation $\sim$ on $X \times Y$ generated by $\forall g \in G, x \in X, y \in Y: (xg,y) \sim (x,gy)$. We set $X \otimes_G Y := (X \times Y)/\sim$ and let $\psi_{\otimes}: X \times Y \to X \otimes_G Y$ be the quotient map. By construction, $\psi_{\otimes}$ is $G$-balanced.
We denote the image $\psi_{\otimes}(x,y)$ by $x \otimes y$. We have the relation $gx \otimes y = x \otimes gy$. (Note that unlike in the case of modules, $\psi_{\otimes}$ is surjective. So every element is an “elementary tensor”.)
Suppose $Z$ is a set and $f:X \times Y \to Z$ is $G$-balanced, then the map $\overline{f}: X \otimes_G Y \to Z, x \otimes y \mapsto f(x,y)$ is well-defined and it is the unique map $X \otimes_G Y \to Z$ that satisfies $f = \overline{f} \circ \psi_{\otimes}$. (This follows just from the universal property of a quotient set.)
In the case of modules, when we have a $(R,S)$-bimodule $M$ and a $(S,T)$-bimodule $N$, then $M \otimes_S N$ is a $(R,T)$-bimodule. We know show the analogous result for $G$-sets.

Lemma If $X$ is a $(H,G)$-set and $Y$ is a  $(G,K)$-set, then $X \otimes_G Y$ is a $(H,K)$-set.

Proof Let $h \in H$, then the map $X \times Y \to X \otimes_G Y, (x,y) \mapsto hx \otimes y$ is $G$-balanced, because $X$ is a $(H,G)$-set, so this map descends to a well-defined map $X \otimes_G Y \to X \otimes_G Y, (x,y) \mapsto (hx,y)$, which shows that the map $H \times (X \otimes_G Y) \to X \otimes_G Y, (h, x \otimes y) \mapsto hx \otimes y$ is well-defined. It’s clear that this makes $X \otimes_G Y$ into a left $H$-set. In the same manner, we get a right action of $K$ on $X \otimes_G Y$ given by $(x \otimes y)k =x \otimes yk$. These actions are compatible, since $(h(x \otimes y))k = hx \otimes yk = h((x \otimes y)k)$.

One of the basic properties of the tensor product of modules is that it defines a functors between module categories, this works also for $G$-sets.

Lemma/Definition The tensor product $\otimes_G$ can be made into a bifunctor $H\textrm{-}\mathbf{Set}\textrm{-}G \times G\textrm{-}\mathbf{Set}\textrm{-}K \to H\textrm{-}\mathbf{Set}\textrm{-}K$

Proof Suppose $X$ and $X'$ are $(H,G)$-sets, and $\alpha: X \to X'$ is a $(H,G)$-equivariant map, $Y$ and $Y'$ are $(G,K)$-sets and $\beta:Y \to Y'$ is a $(G,K)$-equivariant map. Consider the map $f: X \times Y \to X' \otimes_G Y', (x,y) \mapsto \alpha(x) \otimes \beta(y)$. This is $G$-balanced, because $f(xg,y)=\alpha(xg) \otimes \beta(y) = \alpha(x)g \otimes \beta(y) = \alpha(x) \otimes g \beta(y) = \alpha(x) \otimes \beta(gy) = f(x,gy)$.
So we get a well-defined map $X \otimes_G Y \to X' \otimes_G Y', x \otimes y \mapsto \alpha(x) \otimes \beta(y)$, which we will denote by $\alpha \otimes \beta$. $\alpha \otimes \beta$ is $(H,K)$-equivariant because $\alpha$ ist left $H$-equivariant and $\beta$ is right $K$-equivariant and the $(H,K)$-set structure on a tensor product is defined by acting with $H$ on the left factor and with $K$ on the right factor.
If $X''$ is another $(H,G)$-set and $Y''$ is another $(G,K)$-set and $\alpha': X' \to X'', \beta': Y' \to Y''$ are morphisms of the respective type, then if we consider the map $f: X \times Y \to X'' \otimes Y'', (x,y) \mapsto \alpha'(\alpha(x)) \otimes \beta'(\beta(y))$ both $(\alpha' \otimes \beta') \circ (\alpha \otimes \beta)$ and $(\alpha' \circ \alpha) \otimes (\beta' \circ \beta)$ are maps $h:X \otimes_G Y \to X'' \otimes_G Y''$ which satisfy $h \circ \psi_{\otimes} = f$, thus by the uniqueness part of the universal property, they must be equal. That $\mathrm{id} \otimes \mathrm{id} = \mathrm{id}$ is clear from the definition.

Much of the utility of tensor products of modules lies in the Hom-Tensor-adjunction, which also has an analog for group actions. There’s not much left to do to prove it.

Theorem For a $(H,G)$-set $X$, the functor $G\textrm{-}\mathbf{Set}\textrm{-}K \to H\textrm{-}\mathbf{Set}\textrm{-}K$, $Y \mapsto X \otimes_G Y, (f: Y \to Y') \mapsto (\mathrm{id}_X \otimes f: (X \otimes_G Y \to X \otimes_G Y'))$ is left-adjoint to the $\mathrm{Hom}$-functor $\mathrm{Hom}_{H\textrm{-}\mathbf{Set}}(X,-)$.

Proof The universal property of the tensor product $(\psi_\otimes, X \otimes_G Y)$ can be reformulated in the form that the map $\mathrm{Hom}_{\mathbf{Set}}(X \otimes_G Y,Z) \to \mathrm{Bal}_G(X,Y;Z), f \mapsto f \circ \psi_\otimes$ is a bijection. It’s not difficult to check that this map is natural in $Y$ and $Z$ and that if $Z$ is a $(H,K)$-set, then it restricts to a bijection  $\mathrm{Hom}_{H\textrm{-}\mathbf{Set}\textrm{-}K}(X \otimes_G Y,Z) \cong \mathrm{Bal}_G^{(H,K)}(X,Y;Z)$. If we compose this bijection with the Currying bijection $\mathrm{Bal}_G^{(H,K)}(X,Y;Z) \cong \mathrm{Hom}_{G\textrm{-}\mathbf{Set}\textrm{-}K}(Y,(\mathrm{Hom}_{H\textrm{-}\mathbf{Set}}(X,Z))$, we get a natural bijection
$\mathrm{Hom}_{H\textrm{-}\mathbf{Set}\textrm{-}K}(X \otimes_G Y,Z) \cong \mathrm{Hom}_{G\textrm{-}\mathbf{Set}\textrm{-}K}(Y,(\mathrm{Hom}_{H\textrm{-}\mathbf{Set}}(X,Z))$

This concludes this first post on tensor products of $G$-sets, I will investigate more properties of this construction in future posts. Feel free to leave comments below.