2
$\begingroup$

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?

$\endgroup$
5
  • $\begingroup$ 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) $\endgroup$ – HallaSurvivor Mar 14 at 23:30
  • $\begingroup$ @HallaSurvivor: $\operatorname{Hom}(M,-)$ is not a right adjoint in an arbitrary abelian category. $\endgroup$ – Eric Wofsey Mar 15 at 0:18
  • $\begingroup$ @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? $\endgroup$ – HallaSurvivor Mar 15 at 0:19
  • 2
    $\begingroup$ @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. $\endgroup$ – Eric Wofsey Mar 15 at 0:21
  • $\begingroup$ Wow. That's embarrassingly simple. Thanks for the correction ^_^ $\endgroup$ – HallaSurvivor Mar 15 at 0:23
1
$\begingroup$

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 $f\in\operatorname{im}(i_*)$.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.