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I'm working on a problem in which a have two categories $\mathcal{C}$ and $\mathcal{D}$ and I must show they are equivalent.

I make a functor $T:\mathcal{C}\longrightarrow \mathcal{D}$ using the axiom of choice as follows: For each object $C$ of $\mathcal{C}$ I choose, by the axiom of choice, a $\sigma$ (it does not matter what it is to understand the problem) and this allows me to associate an object $C^{\sigma}$ of $\mathcal{D}$. On the the other hand I can define can easily a functor $S:\mathcal{D}\longrightarrow \mathcal{C}$.

Question. When I take an object $D$ of $\mathcal{D}$ then $S(D)$ has a prefered $\sigma$ associated to it. Can I use this $\sigma$ when I apply $T$ on $S(D)$?

Thanks.

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    Don't you need that $S$ is essentially surjective to get the prefered $\sigma$ for every $C\in \mathcal{C}$? Other than that I see no problem with the reasoning. – 2017-01-14
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    You mean, to get the prefered $\sigma$ for every $S(D)$? I hadn't thought of it, but I seems the construction works fine as it is done. – 2017-01-14
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    Only if $S$ in injective on objects, certainly! – 2017-01-15

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