# If $\mathcal{A}$ is an abelian category then $\text{Hom}_{\mathcal{A}}(M,-)$ is left exact functor [duplicate]

Let $$0\to A\xrightarrow{i} B \xrightarrow{j}C$$ be an exact sequence in an abelian category $$\mathcal{A}$$. For an object $$M\in \mathcal{A}$$, I want to show that the sequence $$0\to \text{Hom}(M,A)\xrightarrow{i_*} \text{Hom}(M,B) \xrightarrow{j_*}\text{Hom}(M,C)$$ is exact. I know the proof of this when $$\mathcal{A}$$ is the category of $$R$$-modules for some ring $$R$$, and the same proof applies to show that $$i_*$$ is injective and $$j_*i_*=0$$, but I can't see how to show $$\ker(j_*)\subset \text{image}(i_*)$$. In the case $$\mathcal{A}=R$$-mod, given $$f\in \ker(j_*)$$ I may take $$i^{-1}f\in \text{Hom}(M,A)$$ to get $$i_*(i^{-1}f)=f$$. How can we show this in a general abelian category?

• How much abstract nonsense are you comfortable using? In particular $\text{Hom}(M,-)$ is a right adjoint. This automatically makes it left exact (cf. here, for instance) – HallaSurvivor Mar 14 at 23:30
• @HallaSurvivor: $\operatorname{Hom}(M,-)$ is not a right adjoint in an arbitrary abelian category. – Eric Wofsey Mar 15 at 0:18
• @EricWofsey: wait really? well there's a misconception I didn't know I had. Do you happen to have a counterexample on hand, or a link to one? – HallaSurvivor Mar 15 at 0:19
• @HallaSurvivor: Take $M$ to be any nonzero object in any small abelian category. The left adjoint would have to send $\mathbb{Z}$ to $M$ and thus arbitrary coproducts of copies of $M$ would have to exist, which is impossible by smallness. – Eric Wofsey Mar 15 at 0:21
• Wow. That's embarrassingly simple. Thanks for the correction ^_^ – HallaSurvivor Mar 15 at 0:23

By exactness of the original sequence, $$i$$ is a kernel of $$j$$. So by definition of a kernel, a morphism $$f:M\to B$$ factors through $$i$$ iff $$jf=0$$. This says exactly that if $$f\in\ker(j_*)$$ then $$f\in\operatorname{im}(i_*)$$.