Factorization of functors

Question

I would like to prove the following claim :

Any functor $F : C \rightarrow D$ can be factorized as $C \overset{L} {\longrightarrow} E \overset{R}{\longrightarrow} D$, where $L$ is bijective on objects and full and $R$ is faithful.

$L$ is bijective on objects if the map $A \mapsto F(A)$ is bijective for all $A \in Ob(C)$. $L$ is full if $Hom_C(A, B) \ni f \mapsto L(f) \in Hom_E(L(A), L(B))$ is a surjection, where $A, B \in Ob(C)$ and $R$ is faithful if $Hom_E(A, B) \ni f \mapsto R(f) \in Hom_D(R(A), R(B))$ is an injection, where $A, B \in Ob(E)$.

Unfortunately, I do not understand how to show the above claim. Can someone explain me how to proceed ?

Thanks for your help.

Answer 1

The given conditions are practically an explicit recipe for constructing the factorization.

Since $L$ is bijective on objects, then (up to isomorphism) the objects of $E$ are the objects of $C$. It's obvious what $L$ and $R$ must do on objects.

For every pair of objects $x,y$ of $C$, $F$ gives a function $\hom_C(x, y) \to \hom_D(F(x), F(y))$. The condition that $L$ is full and $R$ is faithful means $\hom_C(x, y) \to \hom_E(x, y) \to \hom_D(F(x), F(y))$ is an epi-mono factorization of that function. (Up to isomorphism) that too is unique.

Since $R$ is faithful, that tells us what composition needs to be in $E$.

So that tells us exactly what everything is:

• The objects of $E$ are the objects of $C$.
  • $L(x) = x$
  • $R(x) = F(x)$
• $\hom_E(x,y)$ is the image of $\hom_C(x,y) \to \hom_D(F(x), F(y))$.
  • $L(f) = F(f)$
  • $R(f) = f$
• $f \circ_E g = f \circ_D g$

All that's left is to check that $E$ is actually a category and that $L$ and $R$ are actually functors.