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?

  • $\begingroup$ 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 $\endgroup$ – AbstractAlgebraLearner 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

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).

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  • $\begingroup$ 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
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    $\begingroup$ 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

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