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 $\mathcal{D}$)

  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 $$\hom_{\mathcal E}(Fa,Fb)\ \cong\ \hom_{\mathcal D}(a,b)\,.$$

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  • $\begingroup$ Ok, thanks! Could you please clarify a bit what you mean when you say "[...] and the current definition of category you are using assumes that $\hom(A,B)$ is disjoint from all other homsets [...]"? How is it possible that a morphism belongs to two different homsets? Does it not depend on the objects $A$ and $B$ being its source and target (and hence lies necessarily in $\hom(A,B)$)? $\endgroup$ – Sun_seeker May 11 '14 at 8:17
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    $\begingroup$ Well, mostly categories are defined so that every morphism $f$ indeed has a unique source and target, i.e. we have global maps ${\rm Mor}\mathcal A \to {\rm Ob}\mathcal A$. However, in some constructions -like this one- one could apply a weaker definition where we have objects and for every pair of objects $A,B$ there is the homset $\hom(A,B)$ assigned, with further structures and properties, but it is not stated that these homsets should be disjoint. $\endgroup$ – Berci May 12 '14 at 16:19
  • $\begingroup$ In the weaker setting, it is legal to define a category with two objects, $A,\ B$ and with $\hom(A,A):=\hom(A,B):=\hom(B,B):=\hom(B,A):=\{*\}\ $ and with the unique possible composition. This has exactly $4$ different morphisms, $2$ of them are identities and the others go $A\to B$ and $B\to A$, however, seeing these as elements in the underlying homsets, all $4$ morphism is the same element (namely $*$). $\endgroup$ – Berci May 12 '14 at 16:29

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