I'm reading through the article Full and faithful functors on Wikipedia and am struggling with this point:
A full and faithful functor is necessarily injective on objects up to isomorphism. That is, if F : C → D is a full and faithful functor and $F(X) \cong F(Y)$ then $X \cong Y$.
Consider the category with two objects X and Y, and one morphism f from X to Y. If we map f to the identity on X, we almost have a functor that is full and faithful but not injective on objects up to isomophism. The visual version:
The map $F_{X,Y}$ sends $f$ to $id_X$ (both hom-sets have one element). The map $F_{X,X}$ sends an identity to an identity, and similarly for $F_{Y,Y}$. Notice the map $F_{Y,X}$ is not surjective, however, because it sends zero objects to one (the empty function). Technically this last map is injective though (see When is the empty function injective? surjective? bijective? for details).
So I've not found a counterexample, but is there a simple proof of the original statement from Wikipedia?
This is pretty much a "face the right way and follow your nose" argument. It's mainly about applying the definitions, which is probably why Wikipedia doesn't go into the details.
Suppose $F$ is full and faithful. Let $F(X)\cong F(Y)$ hold. Then there are morphisms $f\colon F(X)\to F(Y)$ and $g\colon F(Y)\to F(X)$ such that $gf=i_{F(X)}$ and $fg=i_{F(Y)}$.
Since $F$ is full, there are morphisms $f'\colon X\to Y$ and $g'\colon Y\to X$ such that $F(f')=f$ and $F(g')=g$. Then $F(f'g')=F(f')F(g')=fg=i_{F(Y)}=F(i_Y)$. Since $F$ is faithful, this means $f'g'=i_Y$.
Symmetrically, $F(g'f')=F(g')F(f') = gf = i_{F(X)}= F(i_X)$, so $g'f'=i_X$.
Thus, $f'$ and $g'$ witness that $X\cong Y$, which is what we wanted to show.