# $\operatorname{Hom}$ functor is projective?

Consider $A$ a small category and consider $\operatorname{Hom}(-,X) \in \operatorname{HOM}(A^\text{op},Ab)$ the category of contravariant functors from $A$ to $Ab$. It is true that the $\operatorname{Hom}$ functor is a projective object and a faithful functor?

I can't solve this problem my idea was to use Yoneda's lemma, some suggestions? Thank you!

• Are we assuming $A$ is enriched over $\mathbf{Ab}$? Because generally $\mathrm{Hom}(-,X)$ is read as a $\mathbf{Set}$-valued functor. – Malice Vidrine Sep 9 '18 at 21:20

Yoneda’s lemma says $X\to Hom(-,X)$ is faithful. There is no reason that for fixed $X$, the functor $Y\to Hom(Y,X)$ should be faithful. For instance, $X$ might be a terminal object. Sometimes it is, such as when $A$ is a connected groupoid.
Edit: The rest is wrong, the Yoneda embedding doesn’t preserve epimorphisms, as pointed out in the comment. The reason $Hom(-,X)$ is projective is explained the comment too.
[Wrong: Yoneda’s lemma suggests $Hom(-,X)$ should only be projective if $X$ is projective in $A^{op}$ (ie injective). This is just because it lives in a strictly bigger category, so there are at least as many maps to lift. $X$ being injective is presumably not necessarily sufficient, but I don’t know.]