# $\operatorname{Hom}$ 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

## 1 Answer

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.]

• This is incorrect. Representables are always projective. In fact Hom(Hom(-,X),-) commutes with all colimits in HOM(A^op,Ab), not just cokernels, by the Yoneda lemma and the fact that colimits in functor categories are pointwise. One way of expressing the problem with your argument is that the Yoneda embedding preserves almost no colimits, in particular, no cokernels. – Kevin Carlson Sep 10 '18 at 1:39
• @Kevin Carlson Thanks. – Ben Sep 11 '18 at 5:35