# Maps between short exact sequences

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


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 $$B$$.
• If $alpha$ is the identity map, then how is the consequence? – user29422 Mar 24 '20 at 18:47
• If $\alpha$ is the identity, then this becomes: there exists $\beta$ making the diagram commute if and only if $\varepsilon’$ equals the pullback $\varepsilon\gamma$. – Andrew Hubery Mar 24 '20 at 22:22