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If

$0\rightarrow{A'}\rightarrow{A}\rightarrow{A''}\rightarrow{0}$

is an exact sequence of modules, then there exists an exact secuence

$0\rightarrow{}Hom(A'',B)\rightarrow{}Hom(A,B)\rightarrow{}Hom(A',B)\xrightarrow \partial{Ext}^1(A'',B)\rightarrow ...$

Suppose $A'\subseteq A$ and $f:A'\to B$. Prove that $f$ can be extended to $A$ if and only if $\partial f = 0$.

Any hint? Thanks!

user73564
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1 Answers1

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Let $i: A' \to A$ be the inclusion. If $f$ can be extended, then it's in the image of what map? What do you know about the composition of two consecutive maps in a complex?

Conversely, say $\partial f$=0. Then $f$ is in the kernel of $\partial$, but that sequence is exact, so...

John Myers
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  • We have a map $G:Hom(A,B)\to Hom(A',B)$ such that $Im(G)=Ker(\partial)$. If $f:A'\to B$ can be extended to $\overline{f}:A\to B$, then $\partial G(\overline{f})= 0$.

    But who is that map $G$?

    Conversely, if $\partial f =0$, then $f\in Im(G)$, then there exists $\overline{f}:A\to B$ such that $f= G(\overline{f})$... Again, who is $G$?

     – user73564 May 28 '13 at 02:25
  • The reason I gave the inclusion $A' \to A$ a name is because the induced map (after Hom-ing) is usually denoted $i^$ (for contravariant Hom). So your $G$ is my $i^$. Perhaps by "who is that map" you mean "what is that map?" If that's true, then $G$ is composition with $i$, i.e., $G(f) = fi$. So you have shown $\partial G(\tilde{f}) = 0$. But $G(\tilde{f})=f$, so you have $\partial f = 0$. The converse is similar. – John Myers May 28 '13 at 02:33
  • I see it now, thank you. By the way, I'm sorry for my english. – user73564 May 28 '13 at 02:40
  • No problem! By the way, I see you have asked a few questions here with none accepted. Usually when your question is answered, you should click the little check on the answer to "accept" it. – John Myers May 28 '13 at 03:24
  • Oh, I didn't know that. Thanks. – user73564 May 28 '13 at 09:59