I am trying to learn Category Theory and using some lecture notes I found online. They mentioned in passing:
The statement that "Every small category is equivalent to a skeletal category" is equivalent to the axiom of choice.
I am trying to see why this is true and how I can prove it, any help would be much appreciated!
Thinking about my comment, I'm fairly confident it can be regarded as an answer, so here goes.
If the axiom of choice holds, then any small category $\mathcal{C}$ has a skeleton. To see this, we simply note that every isomorphism class in $\mathcal{C}$ is a set. Hence we may use $AC$ to choose a single object from each isomorphism class. The resulting full subcategory $\mathcal{D}$ of $\mathcal{C}$ is skeletal. Since we are assuming $AC$ and every object in $\mathcal{C}$ is isomorphic to one in $\mathcal{D}$, the inclusion $\iota: \mathcal{D} \to \mathcal{C}$ is an equivalence.
For the other direction, let $A$ and $B$ be sets, and let $R \subseteq A \times B$. We need to show that $R$ contains a function with the same domain as $R$. Define an equivalence relation $\sim$ on $R$ by declaring $(a,b) \sim (c,d)$ if and only if $a=c$. This is an equivalence relation since it's clearly symmetric, transitive, and reflexive. As such, it defines a groupoid $R$. Choose a skeletal category $\mathcal{C}$ such that there is an equivalence $F: \mathcal{C} \to R$. I claim that the image $F(\mathcal{C}) \subseteq R$ is a function. Let $A,B \in \mathcal{C}$ and suppose $FA=(a,b)$ and $FB=(c,d)$. If $FA \cong FB$, then $a=c$. Since $F$ is an equivalence of categories, there exists a functor $G: R \to \mathcal{C}$ such that $$A \cong GF(A) \cong GF(B) \cong B.$$ But $\mathcal{C}$ is skeletal, so if $A \cong B$, then $A=B$. Hence $F(id): FA \to FB$ is the identity, and thus the image of $\mathcal{C}$ is a function. Since $FG(a,b) \cong (a,b)$, that means the first component of $FG(a,b)$ is $a$, so the image of $F$ has the same domain as $R$.