Proving there exists a category corresponding to a group (Clara Loeh pg-14)

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.

Answer 2

I'm not sure what exactly your confusion is, and to be honest, the "More details" section confused me...

Let's talk groups and vector spaces as a quick table setting.

What is a group? There are a couple of ways of defining a group, but the most common is to say something like:

A group is an ordered pair $(G,\cdot)$, where $G$ is a set, and $\cdot\colon G\times G\to G$ is a binary operation on the set $G$, such that the operation is associative, has an identity element, and every element has an inverse.

So, how do we describe a group, or how do we specify or "construct" a group? We specify a set, and we specify a binary operation on the set; if necessary, we verify that the operation has the desired properties. Once I've told you the set and the operation, I have "constructed" the group (provided it satisfies the properties it needs to satisfy).

If $K$ is a field, what is a "vector space over $K$"? Again, the precise wording may vary but you usually have something like

A vector space over $K$ is an ordered triple, $(V,+,\cdot)$, where $V$ is a set, $+\colon V\times V\to V$ is a binary operation on $V$, and $\cdot\colon K\times V\to V$ is a way associate to each element $k\in K$ and $v\in V$ an element $k\cdot v\in V$, such that list of 10 axioms.

How do I construct a vector space over $K$? I specify a set $V$ of "vectors", I tell you how to add vectors (elements of $V$), I tell you how to multiply scalars (elements of $K$) by vectors. Once I've specified those three things, if they satisfy the 10 axioms, then I have "constructed a vector space".

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What is a category? Again, the precise definition varies (though they all end up defining equivalent structures). A very common way to define a category $\mathscr{C}$ is to say that it consists of:

1. A collection $\mathrm{Ob}(\mathscr{C})$ of "objects". (I use "collection" because sometimes in categories we work with proper classes rather than sets).

2. For each ordered pair $(X,Y)$ with $X,Y\in\mathrm{Ob}(\mathscr{C})$, a collection $\mathscr{C}(X,Y)$ of "morphisms from $X$ to $Y$".

3. For each ordered triple $(X,Y,Z)$ with $X,Y,Z\in\mathrm{Ob}(\mathscr{C})$, a function $\circ\colon \mathscr{C}(Y,Z)\times\mathscr{C}(X,Y)\to\mathscr{C}(X,Z)$, called "composition".

And then we require that it satisfy certain properties/axioms: for each object $X$ there must exist an element $\mathrm{id}_X\in\mathscr{C}(X,X)$ with certain properties relative to composition, $\circ$ is associative, etc.

So, how do we "construct" or specify a category? I need to tell you what the objects are, I need to tell you what the morphisms are, and I need to tell you how to compose morphisms. Once I specify those things, I have "constructed" a category, provided it satisfies the necessary properties/axioms. Note that the "objects" don't need to be sets themselves; not that the morphisms don't need to be "functions" in the usual sense. They are just "things" we label as 'objects' and 'morphisms' (just like a "vector" in a vector space doesn't actually have to be a tuple, and "vector addition" doesn't have to have anything to do with addition in the usual sense).

Just as we can talk about vector spaces in general ("Let $V$ be a vector space; then any spanning set of $V$ contains a basis for $V$", for example), or in particular ("The row space of the matrix $A$ has dimension $3$"), and similarly for groups, we can talk about categories in general (without constructing a specific one; just talking about any category, or any category satisfying certain properties), or in particular (by constructing/specifying a category and saying things about it).

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The statement at hand here is: for every group $G$, there is a category $\mathscr{C}$ and an object $X\in\mathrm{Ob}(\mathscr{C})$ such that $\mathrm{Aut}_{\mathscr{C}}(X)\cong G$.

So, let $G$ be a group. We will construct a category and specify an object that satisfy the conclusion.

To specify the category $\mathscr{C}$, I need to tell you: the objects of $\mathscr{C}$; the morphisms between any two objects of $\mathscr{C}$; and how to compose morphisms that can be composed. Let us do so:

1. Pick a one element set, $\{\star\}$. This will be my set of objects of $\mathscr{C}$. That is, the only object in this category $\mathscr{C}$ I am constructing is $\star$.

2. Since there is only one object, I only need to tell you what $\mathscr{C}(\star,\star)$ is. Well, it will be the underlying set of the group $G$.

3. Since there is only one object, I just need to tell you how to compose elements of $\mathscr{C}(\star,\star) = G$ in the category. If $g,h\in\mathscr{C}(\star,\star)$, then they are "really" elements of $G$. So I will define $g\circ h$ to be the element $gh\in G$.

Is this a category? Yes! Composition is associative (because $G$ is a group). The identity morphism on $\star$ is $e_G$, which satisfies the needed condition: for every $g\in\mathscr{C}(\star,\star)$, $e\circ g = eg = g$, $g\circ e = ge = g$.

What object is $\star$? It's just the name of the one element in the one element set you chose. If you aren't sure there is such an object, remember that we know the empty set exists, so we can just take the set $\{\varnothing\}$ (which exists by the Axiom of the Power set). It doesn't matter, because this object is just a placeholder, a scaffolding on which we are constructing our collection of "morphisms".

Now, in this category, I claim that $\mathrm{Aut}_{\mathscr{C}}(\star)$, which is a group under $\circ$, is isomorphic to $G$. Indeed, in fact $\mathrm{Aut}_{\mathscr{C}}(\star)=\mathscr{C}(\star,\star) = G$, and the operation $\circ$ in $\mathrm{Aut}_{\mathscr{C}}(\star)$ is the same as the operation in $G$. So it is not just "isomorphic to $G$", it is in a sense identical to $G$.

This shows the claimed result is true.