On recognizing equality against bijection of hom-sets (in a locally small category).

Question

I am a bit confused about the use of the bijection symbol against the equality symbol when dealing with hom-sets. I will give an example. Suppose you want to prove that every functor $F:\mathcal{C}\to\mathcal{E}$ factors as $F^*\circ F_*$, where the functors $F_*:\mathcal{C}\to\mathcal{D}$ and $F^*:\mathcal{D}\to\mathcal{E}$ satisfy $F_*$ bijective on objects and $F^*$ full and faithful (for some category $\mathcal{D}$). What you do is, define the objects of $\mathcal{D}$ to be the objects of $\mathcal{C}$ and $\text{Hom}_{\mathcal{D}}(a,b)=\text{Hom}_\mathcal{E}(Fa,Fb)$. You then define $F_*:\mathcal{C}\to\mathcal{D}$ by $F_*a=a$, $F_*f=Ff$ and $F^*:\mathcal{D}\to \mathcal{E}$ by $F^*a=Fa$, $F^*f=f$. To prove that $F^*$ is full and faithful, it is enough to prove, by the definition of "full and faithful", that $$\text{Hom}_{\mathcal{E}}(F^*a,F^*b)\cong \text{Hom}_{\mathcal{D}}(a,b)$$ My question is, in the following sequence of "relations" (word used here in the non-mathematical context) between the hom-sets ("relations" symbolized by a square), which squares are equalities and which are bijections? $$\text{Hom}_{\mathcal{E}}(F^*a,F^*b)\; \square \;\text{Hom}_{\mathcal{E}}(Fa,Fb)\;\square\;\text{Hom}_{\mathcal{D}}(a,b)$$ (of course the first "relation" comes from the definition of $F^*$ and the second by the definition of the category $\mathcal{D}$)

Answer 1

1. By the construction written above, there is equality in both places, exactly because of the definitions of $F^*$ and $\mathcal D$, as you say.

2. However, if you want to be extra precise, and the current definition of category you are using assumes that $\hom(A,B)$ is disjoint from all other homsets, then you have to pay more attention at the definition of $\mathcal D$, and define e.g. $$\hom_{\mathcal D}(a,b):=\left\{\,\langle a,\varphi,b\rangle\ \mid\ \varphi\in\hom_{\mathcal E}(Fa,Fb)\,\right\}\,.$$ All the rest is OK, so we will have $F^*a=Fa\ $ (and $F^*(\langle a,f,b\rangle ):=Ff$), $\,$hence $\hom_{\mathcal E}(F^*a,F^*b)\ =\ \hom_{\mathcal E}(Fa,Fb)$, but the other one is only $$\hom_{\mathcal E}(Fa,Fb)\ \cong\ \hom_{\mathcal D}(a,b)\,.$$