In this link it is proved that in an abelian category $\mathcal C$ we have that $f:A\rightarrow B$ is mono iff the sequence $0\rightarrow A\rightarrow B$ is exact, where the arrow from $A$ to $B$ is $f$, then they put as a corollary that an object $I$ is injective iff the hom-functor $\operatorname{Hom}_{\mathcal C}(-,I):\mathcal C^{\rm op}\rightarrow \mathsf{Set}$ is exact; an exact functor is a functor which presereves exact sequences. They show no proof of this, but I simply can't see it, then, why is this true?


  • $\begingroup$ Hint: the $Hom_C(\cdot, I)$ is always right exact, regardless of whether $I$ is injective. $\endgroup$ – xyzzyz Jun 10 '13 at 17:56
  • $\begingroup$ what is right exact? $\endgroup$ – Родион Раскольников Jun 10 '13 at 17:57
  • 1
    $\begingroup$ @xyzzyz You mean left exact. $\endgroup$ – Zhen Lin Jun 10 '13 at 18:20
  • $\begingroup$ @ZhenLin yes, of course. $\endgroup$ – xyzzyz Jun 10 '13 at 21:29

As stated in the comments, $\operatorname{Hom}_C(\cdot,I)$ is always left exact, so we wish to show that $I$ is injective if and only if $\operatorname{Hom}_C(\cdot,I)$ is right exact.

First, let $0\to B\to C$ be exact, and consider the sequence given by applying $\operatorname{Hom}(\cdot,I)$: $\operatorname{Hom}(C,I)\to\operatorname{Hom}(B,I)\to0$. This sequence is exact if and only if for every morphism $B\to I$ there exists a morphism $C\to I$ such that $B\to C\to I=B\to I$. This is exactly the requirement that $I$ be an injective object.


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