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Show that Q is injective if and only if, whenever $$0\to A\stackrel{f}{\to} B\stackrel{g}{\to} C\to 0$$ is exact, then $$0\to \mathrm{Hom}_R(C,Q)\stackrel{g^*}{\to} \mathrm{Hom}_R(B,Q)\stackrel{f^*}{\to} \mathrm{Hom}_R(A,Q)\to 0$$ is exact.

I have no idea how to start on this problem. Please help. Thanks.

Jean Marie
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gws
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    $\mathrm{Hom}_R(-,Q)$ is always left exact, so this question is really about whether $Q$ is injective if and only if $f^*$ is onto. Think about what that means and how it relates to the definition of injective module. – Arturo Magidin Feb 20 '19 at 17:48
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    What is your definition of "injective"? This property is often taken as the definition itself. – Eric Wofsey Feb 20 '19 at 19:23
  • See https://math.stackexchange.com/questions/1168498/equivalence-of-definitions-of-injective-modules – Arnaud D. Feb 20 '19 at 19:32
  • Or alternatively https://math.stackexchange.com/questions/1155248/characterization-of-projective-and-injective-modules – Arnaud D. Feb 20 '19 at 19:48
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    I have taken the liberty to modify your title "Let Q be an R-injective R-module" which wasn't a title into a true one. Next time, pay attention to it. – Jean Marie Feb 20 '19 at 21:18
  • @JeanMarie Thanks for the clearer title. – gws Feb 20 '19 at 21:19
  • @ArturoMagidin I think I understand this reduction of the problem. But I don't see how to proceed from there. – gws Feb 20 '19 at 21:20
  • I had made a copy paste error on the new title. I just fixed it. – Jean Marie Feb 20 '19 at 21:20
  • @davidh: $f^$ being surjective is about how given a function from $A$ to $Q$, you need a function from $B$ to $Q$ that satisfies some condition. Being injective, by definition* is about how certain functions from $A$ to $Q$ will give some functions with some condition. Gee... I wonder if the two are related? – Arturo Magidin Feb 20 '19 at 22:10
  • @ArturoMagidin I see. Thanks a bunch. – gws Feb 21 '19 at 04:44

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