3
$\begingroup$

Note that if $\text{Hom}_R(D,-)$ functor takes short exact sequences to short exact sequences then it takes exact sequences of any length to exact sequences since any exact sequence can be broken up into a succession of short exact sequences.

Let $A \xrightarrow{\psi} B \xrightarrow{\phi} C$ be exact, then there is the short exact sequence $0 \to \ker \phi \to B \to \text{im} \ \phi \to 0$ but how does that imply $\text{Hom}_R(D,A) \xrightarrow{\psi}' \text{Hom}_R(D,B) \xrightarrow{\phi'} \text{Hom}_R(D,C)$ is exact?

$\endgroup$
  • $\begingroup$ Recall that the exactness of the first sequence means that $\ker(\phi) = \operatorname{im}(\psi)$. Can you take it from here? $\endgroup$ – A.P. Dec 25 '15 at 1:14
  • $\begingroup$ @A.P. Nope. I don't see what that gives us $\endgroup$ – EnjoysMath Dec 25 '15 at 1:20
  • $\begingroup$ I may be mistaken, but I think that all you need to prove is that the (right) exactness of $\operatorname{Hom}_R(D,-)$ implies $\operatorname{im}(f') = \operatorname{Hom}_R(D,\operatorname{im}(f))$ for every $f: G \to H$, for every $G,H$. Then the exactness of $$0 \to \operatorname{im}(\psi) \to B \to \operatorname{im}(\phi) \to 0$$ implies the exactness of $$0 \to \operatorname{Hom}_R(D,\operatorname{im}(\psi)) \to \operatorname{Hom}_R(D,B) \to \operatorname{Hom}_R(D,\operatorname{im}(\phi)) \to 0$$... $\endgroup$ – A.P. Dec 25 '15 at 1:35
  • $\begingroup$ ... and the above property gives $$0 \to \operatorname{im}(\psi') \to \operatorname{Hom}_R(D,B) \to \operatorname{im}(\phi') \to 0$ which means that $\operatorname{im}(\psi') = \ker(\phi')$, i.e. that your sequence in $\operatorname{Hom}$ is exact. $\endgroup$ – A.P. Dec 25 '15 at 1:37
  • $\begingroup$ Duplicate? math.stackexchange.com/questions/207551/… $\endgroup$ – Bruno Stonek Jan 1 '16 at 20:22
3
$\begingroup$

A sequence $A\stackrel\psi\to B\stackrel\phi\to C$ is exact iff there exist objects $D$, $E$, $F$, and $G$ and short exact sequences $$0\to D\to A\to E\to 0$$ $$0\to E\to B\to F\to 0$$ $$0\to F\to C\to G\to 0$$ such that the composition $A\to E\to B$ is $\psi$ and the composition $B\to F\to C$ is $\phi$. It follows that if you apply any functor $T$ which preserves short exact sequences, $T(A)\to T(B)\to T(C)$ is still exact (since you can just apply $T$ to the three short exact sequences above).

$\endgroup$
  • $\begingroup$ Sure, the epi-mono splitting was the basis for my comments above, but is there an easy way to prove this characterisation of exact sequences? (+1) $\endgroup$ – A.P. Dec 25 '15 at 1:47
  • 2
    $\begingroup$ If you're talking about modules, it should be very easy to verify the equivalence. In an abstract abelian category, you observe that $D\to A$ is a kernel of $\psi$ since $E\to B$ is monic (and dually $C\to G$ is a cokernel of $\phi$), and then the rest is straightforward. $\endgroup$ – Eric Wofsey Dec 25 '15 at 2:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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