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I'm doing a problem to show that any functor $F: \mathcal{A} \to \mathcal{C}$ between categories can be factorized as

$F_L: \mathcal{A} \to \mathcal{B},\,F_R: \mathcal{B} \to \mathcal{C}$, where $F_L$ is a bijection on the objects of $\mathcal{A}$, and $F_R$ is full and faithful.

To some extent, I can sort of see what I'm meant to do, and am trying to use each of $F_L$ and $F_R$ in order to ensure the other has the required properties: for example, if $F$ is not faithful, say $F(f)=F(g)$, then define $F_L(f)=F_L(g)$ so that $F_R F_L(f) = F(f) = F(g) = F_R F_L(g)$ does not contradict faithfulness of $F_R$. However, the only way I could see to make $F_R$ full would be to 'insert' morphisms in $\mathcal{B}$ wherever we have a morphism $c$ in $\mathcal{C}$ which is not mapped onto by $F$, and then map this new morphism under $F_R$ to $c$.

However, doing this raises a number of new problems: first of all, we don't yet know what the objects are in $\mathcal{B}$, though given $F_L$ is bijective we may presumably treat them as the objects of $\mathcal{A}$. Also, we can't just say 'let's add a morphism between any 2 objects in $\mathcal{B}$ wherever $F_R$ needs to map something to another morphism which $F$ is not surjective on', because for one thing we don't know where $F_R$ maps objects to yet, and for another I'm not convinced we can just freely insert morphisms wherever we fancy without problems arising (though I'm new to category theory, so maybe I'm wrong!). Could anyone help? Many thanks - J

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Just a bit of a side comment here... isn't this related to the fact the category of small categories has a model structure on it or something? Also, @JSp, this is going to be pretty unnatural, in the sense that you just construct B to be what you need, you know. – Jon Beardsley Oct 22 '11 at 0:28
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Construct $B$ with objects those of $A$ and for $a$, $a'\in A$, define $\hom_B(a,a')=\hom_C(Fa,Fa')$, with composition that of $C$.

Then you can consider the functor $F_L$ which is the identity on the objects and $F$ on the morphisms, and $F_R$ which is $F$ on the objects and the identity on the morphisms.

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I'm just going to delete all of the above comments, I'm not making any sense. – Jon Beardsley Oct 22 '11 at 0:43
Ok, I understand all that except for how we can "define $hom_B(a,a') = hom_C(Fa,Fa')$": surely we can't just 'equate' the morphisms in the former to those in the latter? I mean, if we were in the category of sets for example, then our morphisms would be functions mapping elements of a to a' and elements of Fa to Fa': it isn't clear to me in what sense you can just equate the 2 groups of morphisms - sorry I'm new to all this! – JSp Oct 22 '11 at 1:28
@JSp: I am defining a category $B$. In the process of doing so, I have to decide, given two objects of $B$, what the morphisms in $B$ from one of them to the other are going to be. And I decided that the set of morphisms from an object $a$ to an object $a'$ is going to be the set of morphisms from $Fa$ to $Fa'$ in the category $C$. – Mariano Suárez-Alvarez Oct 22 '11 at 12:58
@Mariano: so if I'm understanding correctly, even if your objects were (e.g.) sets, your morphisms in B wouldn't necessarily have to be functions, say: you can just define a morphism from a to a' to "exist if it is a morphism from Fa to Fa'". I guess my confusion was that I felt a morphism had to have more structure to it than just us saying "there is a morphism between 2 objects" (e.g. functions for sets have to act on individual elements of an object (set)), but given we define B we can make a morphism between objects to be no more than that, as long as it obeys the usual axioms. Right? – JSp Oct 22 '11 at 15:40
The morphisms between two objects in a category can be the chairs in my house, if you want. There is no reason for them to be functions, or anything. – Mariano Suárez-Alvarez Oct 22 '11 at 20:29

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