I need to show that the assertion 'If $\mathcal{C}_{0}$ is a skeleton of a small category $\mathcal{C}$, then $\mathcal{C}\simeq \mathcal{C}_{0}$' implies that, given a family $(A_{i}\mid i\in I)$ of nonempty sets, we can find a family $(A_{i}^\prime \mid i\in I)$ where each $A_{i}^\prime$ is a nonempty finite subset of $A_{i}$.
For that I have used the category whose objects are the members of $I\times \{0,1\}$ and the morphisms $(i,j)\longrightarrow (i,k)$ are finite sums $\sum_{s=1}^{t}n_{s}a_{s}$ where $a_{s}\in A_{i}$ and $n_{s}$ integers with $\sum n_{s}=k-j$.
The objects of $\mathcal{C}_{0}$ have to be $(i,j)$ for each $i$, and $j=0$ or $j=1$ ($(i,0)$ and $(i,1)$ are in the same isomorphism class).
Since $\mathcal{C}_{0}\simeq \mathcal{C}$, we have functors $F\colon \mathcal{C}_{0} \longrightarrow \mathcal{C}$ and $G \colon \mathcal{C} \longrightarrow C_{0}$ such that there is a natural isomorphism from $1_{\mathcal{C}}$ to $FG$ and another one from $GF$ to $1_{\mathcal{C}_{0}}$.
Then, by taking any morphism $f\colon (i,j) \longrightarrow (i,k)$ in $\mathcal{C}$, $G(f)\colon (i,j) \longrightarrow (i,j)$ is a morphism in $\mathcal{C}_{0}$, and I can take the set $A_{i}^\prime$ to have the elements of the summand.
However, I am not using any of my hypothesis (the functor $F$ and the two natural isomorphisms), so something has to be wrong. Can anyone help me, please? Thanks.
We may assume that the objects of $\mathcal{C}_{0}$ are $(i,0)$ for $i\in I$.
If $F\colon \mathcal{C}_{0} \longrightarrow \mathcal{C}$ and $G\colon \mathcal{C}\longrightarrow \mathcal{C}_{0}$ are functors such that there is a natural isomorphism $\beta \colon FG\longrightarrow 1_{\mathcal{C}}$, one of $\beta_{(i,0)}$ or $\beta_{(i,1)}$ has a formal addition distinct to $0$, so we may take the elements of that finite sum.