Suppose I have a short exact sequence of modules $0 \to A \to B \to C \to 0$. Let $A' \subseteq A$ and $C' \subseteq C$ be submodules and suppose I have a short exact sequence $0 \to A'\to B'\to C'\to 0$. Then I have the following diagram:
$$
\require{AMScd}
\begin{CD}
0 @>>> A' @>>> B' @>>>C' @>>> 0 \\
@. @VVV @. @VVV @. \\
0 @>>> A @>>> B @>>>C @>>> 0
\end{CD}
$$
I know in general there is no map $B' \to B$ that would make the diagram commute, but are there sufficient conditions for such a map to exist?
Best Answer
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 $B$.