Skeletons of small categories and the axiom of choice. 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!
 A: 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 $R$.
