Category Theory – Proving Existence of Category Corresponding to a Group

Question

Context:

Definition 2.1.18 (Automorphism group). Let $C$ be a category and let $X$ be an object of $C$. Then the set ${\rm Aut}_C(X) $of all isomorphisms $ X \to X$ in $C$ is a group with respect to composition in $C$ (Proposition 2.1.19), the
automorphism group of $X$ in $C$.

Doubt:

The following statement is proved in the book:

2.1.19.2. Let $G$ be a group. Then there exists a category $C$ and an object $X$ in $C$ such that $G \cong {\rm Aut}_C(X)$

The following proof is given:

We consider the category $C$ that contains only a single object $X$. We set $\text{Mor}_C(X,X) := G$ and we define the composition in $C$ via the composition
in $G$ by

$$ \circ : \text{Mor}_C(X,X) \times \text{Mor}_C(X,X) \to \text{Mor}_C(X,X)$$

$$(g,h) \to g \cdot h$$

A straight forward computation proves the proposition.

In the above proof, how do we know there exists such an object $X$ such that we are allowed to set $\text{Mor}_C(X,X) :=G$?

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More details:

I can't really explain my doubt in this context itself because my knowlege in CT is not much but I will try using another issue as analogy.

So suppose we are talking about Groups. To define a group, we says it's a set with a binary operation. Now, but to say this, shouldn't we have defined what the set looks like (out of ZFC) before we talk about the binary operation?

Answer 1

Whether and how an object $X$ exists depends on your formalization (though formalizatoins in which it doesn't I suspect are not common if they're to be used for formalizing mathematics).

Let me use the standard first-order logic formalization, in which a category has the following structure:

1. a collection $O$ whose elements are called objects
2. a collection $M$ whose elements are called morphisms
3. functions $d,c\colon M\to O$ sending each morphism $f\in M$ to its domain object $d(f)\in O$ and to its codomain object $c(f)\in O$, respectively codomain
4. a function $e\colon O\to M$ sending an object to each identity morphism.
5. a composition morphism $m\colon M\times_O M\to M$ where $M\times_O M\subseteq M$ is the subcollection of compostable pairs of morphisms given by $M\times_O M=\{(f,g)\in M\times M: d(f)=c(g)\}$.

The axioms that make this structure a category are:

1. That the identity morphism of an object has domain and codomain that same object: it is expressed by the equation of function composites $d\circ e=c\circ e=\mathrm{id}_O\colon O\to M\rightrightarrows O$, where $\mathrm{id}_O\colon O\to O$ is the identity function on the collection $O$.
2. The composing a morphism with the identity morphism results in the same morphism. This is expressed by the equation of elements $m\circ(f,e(o))=m\circ(e(o),g)=f$ for each $f,g\in M$ with $d(f)=o=c(g)$. It can also be expressed as equations of functions: $m\circ(\mathrm{id}_M,e\circ d)=m\circ(e\circ c,\mathrm{id}_M)=\mathrm{id}_M$ where $(e\circ c,\mathrm{id}_M),(\mathrm{id}_M,e\circ d)\colon M\to M\times_OM$. Note that the functions have image in the correct subcollection of $M\times M$ due to 1.
3. The domain and codomain of a composite of a composable pair are the domain and codomain that were not required to match. Explicitly, $d(m(f,g))=d(g)$ and $c(m(f,g))=c(g)$. As equations of functions, this is $d\circ m=d\circ\pi_2$ and $c\circ m=c\circ\pi_1$ where $\pi_1,\pi_2\colon M\times_OM\to M$ are the functions given by $\pi_1(f,g)=f$ and $\pi_2(f,g)=g$.
4. That composition of morphisms is associative. This is expressed by the equation $m\circ(m\times\mathrm{id}_M)=m\circ(\mathrm{id}_M\times m)\colon M\times_OM\times_OM\to M$ where $M\times_OM\times_OM\subseteq M\times M\times M$ is the subcollection of composable triples of morphisms explicitly given by $M\times_OM\times_OM=\{(f,g,h)\in M\times M\times M: d(f)=c(g),d(g)=c(h)\}$, and $\mathrm{id}_M\times m,m\times\mathrm{id}_M\colon M\times_OM\times_OM\to M\times_OM$ are given by $\mathrm{id}_M\times m(f,g,h)=(f,m(g,h))$ and $m\times\mathrm{id}_M(f,g,h)=(m(f,g),h)$. Note that the functions have image in the correct subcollection of $M\times M\times M$ due to 3.

A more conceptual view is that the pair of functions $d,c\colon M\to O$ realize a family of sets of morphisms $\mathrm{Hom}(X,Y)=d^{-1}(X)\cap d^{-1}(Y)$ indexed by pairs of objects. Then the above definition amounts to a first-order encoding of the notion of a category as a collection of objects, equipped with a family of morphisms indexed by pairs of objects, and composition function defined on the family of compostable pairs, etc. However, the formal language of families isn't widely-used because one can get away with formalizing in first-order logic as above.

In any case, recall now that the structure of a monoid is a collection $M$ equipped with a binary function $m\colon M\times M\to M$ and an element $e\in M$. I claim that any choice of a singleton collection (collection with exactly one element) allows the construction of a category with one object (the element of the singleton) our of the monoid, and conversely that category with one object yields a monoid by forgetting its (singleton) collection of objects.

Let the singleton be $\{X\}$ and set $O=\{X\}$ (so that the category we are constructing has a single object the element $X$ of the singleton). Since functions $\{X\}\to M$ are in bijection with elements of $M$, the above element $e\in M$ corresponds to a function $e\colon\{X\}\to M$, i.e. a function $e\colon O\to M$ (I abuse notation and label both the same).

Moreover, there is a unique function $M\to\{X\}$ sending every element of $M$ to the element $X$. Let's use it twice with two distinct labels: $d,c\colon M\to O$. In that case $M\times_OM=\{(f,g)\in M\times M:d(f)=c(g)\}=M\times M$ since $d(f)=X=c(g)$ for all $(f,g)\in M\times M$. Note that we are interpreting the elements of the monoid as morphisms with domain and codomain $X$.

Thus our choice of a singleton for $O$ and identification of the unit of the monoid with a function out of $O$ and of the domain and codomain functions with the unique function from $M$ to $O$ augments the structure of a monoid into the structure of a category: $d,c:M\to O=\{X\}$, $e\colon O=\{X\}\to M$, and $m\colon M\times_O M=M\times M\to M$.

It is now straightforward to verify the category axioms hold for this structure if and only if the monoid axioms hold for the original structure, which are the associativity: $m(m(x,y),z)=m(x,m(y,z))$ for all $x,y,z$ and the unit axiom $x=m(e,x)=m(x,e)$ for every elements $x\in M$.

1. $d\circ e=\circ e=\{X\}$ since both have to be the unique function $O=\{X\}\to\{X\}=O$.
2. is literally the unit axiom of a monoid
3. $d\circ m=d\circ\pi_2$ since both have to be the unique function $M\times M=M\times_O M\to\{X\}$ and similarly $\circ m=c\circ\pi_1$.
4. $M\times_O\times M\times_O M=\{(f,g,h)\in M\times M\times M:d(f)=c(g),d(g)=c(h)\}=M\times M\times M$. Then this axiom is literally the associativty axiom for the monoid $M$.

Now as to the existence of a singleton $\{X\}$, in usual set theory any set is contained in a singleton. In more general first-order logic, any element of a collection determines a subcollection consisting of exactly that object. If there are no sets or elements of collections, then there are simply neither monoids nor categires with one object, so the result still works.