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Suppose I have a short exact sequence of modules $0\rightarrow A \rightarrow B \rightarrow C \rightarrow 0$. Let $A' \subseteq A$ and $C'\subseteq C$ be submodules and suppose I have a short exact sequence $0\rightarrow A'\rightarrow B'\rightarrow C'\rightarrow 0$. Then I have a diagram

$$ \newcommand{\ra}[1]{\!\!\!\!\!\!\!\!\!\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!\!\!\!} \newcommand{\da}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} % \begin{array}{lllllllll} 0 & \ra{} & A' & \ra{} & B' & \ra{} & C' & \ra{} & 0 \\ & & \da{} & & & & \da{} & & \\ 0 & \ra{} & A & \ra{} & B & \ra{} & C & \ra{} & 0 & \\ \end{array} $$ I know in general there is no map $B' \rightarrow B$ that would make the diagram commute, but are there sufficient conditions for such a map to exist?

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In general, given a diagram $$\begin{array}{cccccccccc} \varepsilon'\colon & 0 & \to & A' & \to & B' & \to & C' & \to & 0\\ &&& \phantom{\alpha}\downarrow\alpha&&&& \phantom{\gamma}\downarrow\gamma\\ \varepsilon\colon & 0 & \to & A & \to & B & \to & C & \to & 0 \end{array}$$ you can fill it in with a map $\beta\colon B'\to B$ if and only if the pushout $\alpha\varepsilon'$ equals the pullback $\varepsilon\gamma$ as extensions in $\mathrm{Ext}^1(C',A)$. Note that this doesn't required $\alpha$ and $\gamma$ to be monomorphisms.

Note that, in your situation, the pullback yields a submodule $B_2\leq B$ (the preimage of $C'$). Also, if there is such a map, then $B'$ would necessarily be (isomorphic to) a submodule of $B$.

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  • $\begingroup$ If $alpha$ is the identity map, then how is the consequence? $\endgroup$ – user29422 Mar 24 '20 at 18:47
  • $\begingroup$ If $\alpha$ is the identity, then this becomes: there exists $\beta$ making the diagram commute if and only if $\varepsilon’$ equals the pullback $\varepsilon\gamma$. $\endgroup$ – Andrew Hubery Mar 24 '20 at 22:22
  • $\begingroup$ Oh I see. Thank you very much! $\endgroup$ – user29422 Mar 25 '20 at 4:34

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