TITLE: Maps between short exact sequences
QUESTION [9 upvotes]: 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?

REPLY [2 votes]: 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 $B$.