This is my first post. I am stuck on the following problem. Ive been at it for a few days and I cant seem to figure it out. I cant use the Yoneda lemma and it has to be general to any Abelian category.
Let $A \xrightarrow{f} B \xrightarrow{g} C$ be a sequence in an Abelian category. If the sequence $$Hom(X,A) \xrightarrow{Hom(f)} Hom(X,B) \xrightarrow{Hom(g)} Hom(X,C)$$ is exact for any $X$, then $A \xrightarrow{f} B \xrightarrow{g} C$ is exact.
Anything would help!
To get you started:
Let $X = \text{Ker}(g)$, and let $k\colon X\to B$ be the kernel map. Then $k\in \text{Hom}(X,B)$, and $\text{Hom}(g)(k) = g\circ k = 0$ by definition of the kernel. Thus $k\in \text{Ker}(\text{Hom}(g)) = \text{Im}(\text{Hom}(f))$ since the sequence of Hom groups is exact (in the category of abelian groups). That is, there is a map $k'\in \text{Hom}(X,A)$ such that $\text{Hom}(f)(k') = f\circ k' = k$.
At this point, if we were working in the category of abelian groups, we'd be done, since if $a\in \text{Ker}(g)\subseteq B$, then $a = k(a) = f(k'(a))\in \text{Im}(f)$, and the sequence is exact. I'll leave it to you to think about how to translate this argument to the general abelian categories case.