Let me first state the question itself.
Let $R$ be a ring with $1$. Let $0\to M\to Q\to L\to 0$ and $0\to M\to Q'\to L'\to 0$ be short exact sequences of $R$-modules where $Q$ and $Q'$ are both injective. Prove that $Q\oplus L'=Q'\oplus L$.
(You can assume that pushout and pullback exist in the category of $R$-modules.)
I didn't learn category theory and my definitions may seem a bit old-fashion (because the book is old?). We say a chain $0\to A\to B\to C\to 0$ is short exact if ker$(g)$=image$(f)$, where $f:A\to B$ and $g:B\to C$, $f$ injective, $g$ surjective. We say a module $Q$ is an injective module if it is a submodule of some $R$-module $M$ and there is another submodule $K$ of $M$ such that $M=K+Q$ and $K\cap Q=\{0\}$.
I know a theorem that if we have $R$-module homomorphisms $\varphi: Q\to Q'$ and $\psi:L'\to L$, then $\varphi\oplus \psi: Q\oplus L'\to Q'\oplus L$ is also a $R$-module homomorphism. Therefore, if we can find two bijective $R$-module homomorphisms $\varphi: Q\to Q'$ and $\psi:L'\to L$, then we are done.
My specific questions are:
Q1: In the question, there is no an $R$-module containing $Q$, and so my definition of injective modules seems not working. Is there a way to solve this problem without using categorical language?
Q2: I don't know how to explicitly construct a map from $Q$ to $Q'$ because the two chains are "separated". There is no an intermediate stuff for us to build a "connection" between $Q$ and $Q'$. So, my second question is how to construct such an $R$-module isomorphism from $Q$ to $Q'$?
Any post involving categorical language is also welcome though I prefer non-categorical language. Just because I may not have enough time to learn the rigorous categorical language before my upcoming exam.
Thanks for help.