Equality of (sub)modules with a specific exact sequence of modules When reading these notes of Andy McLennan  I came across Lemma A4.3 and it's proof. 
$\textbf{Lemma 4.3}.$ If $0\to L\xrightarrow{f}M\xrightarrow{g} N\to 0$ is a short exact sequence of $R$-modules, $M'$ is a submodule of $M$, $f(L)\subset M'$, and $g(M')=N$, then $M'=M$. 
Proof. We regard $L$ as a submodule of $M$ and identify $N$ with $M/L$. Then $M/L=M'/L$, so any $m\in M$ is $m'+\ell$ for some $m'\in M'$ and $\ell\in L$, and consequently $m\in M'+L=M'$. Thus $M'=M$. $\Box$
I am not completely convinced by the above proof. First of all why can we regard $L$ as a submodule of $M$ and identify $N$ with $M/L$? Sure, by the definition of the exact sequence we know that $L$ is isomorphic to the submodule $f(L)$ of $M$ and that $N\cong M/f(L)$. But doesn't the proof now only covers a specific case of the stament in the Lemma?
I tried to prove the statement of the Lemma without making the above assumptions. However I only managed to get so far. By considering the following diagram
$$\require{AMScd}
\begin{CD}
0 @>>> f(L) @>\text{inclusion}>> M' @>g>> N @>>> 0 \\
@. @VV{f^{-1}}V @VV\text{inclusion}V @VV{id_N}V  \\
0 @>>> L @>f>> M @>g>> N @>>> 0
\end{CD}
$$
(one can prove that it's commutative and that the upper row is exact) and noting that the downward maps $f^{-1}$ and $id_N$ are isomorphims one can show that we have $M'\cong M$. From here I couldn't get any further in proving that $M'=M$. I know that $M'\subset M$ but as far as I know together with $M'\cong M$ it doesn't follow that $M'=M$. Here my intuition is based upon the example that $2\mathbb{Z}\subset\mathbb{Z}$ but $2\mathbb{Z}\cong\mathbb{Z}$ (as abelian groups).
So in short, why don't we loose information by considering $L$ as a submodule of $M$ and how can we complete the proof for the general statement (without assuming that $L$ is a submodule of $M$) as I tried to do.
 A: The proof is correct, but unnecessarily complicated.
Saying that $g(M')=N$ is the same as saying that $M'+\ker g=M$: let $x\in M$ and consider $x'\in M'$ such that $g(x)=g(x')$; then $x-x'\in\ker g$, so $x=x'+(x-x')\in M'+\ker g$. Since $\ker g=f(L)$ by exactness, you have
$$
M=M'+f(L)
$$
However, $f(L)\subset M'$, so $M'+f(L)=M'$.
Also your proof is correct: the final remark is that the inclusion map is an isomorphism, so it is surjective, hence $M'=M$.
A: Hint:
Use  the snake lemma  with the  above diagram. There results that, since the leftmost and rightmost vertical maps are bijective, the middle vertical map is surjective. A surjective inclusion is an equality.
A: Let me just rephrase what he's doing without the "reduction" and using elements throughout. It's a little less slick but it does show that you can just follow your nose.
Take any $m \in M$. Since $g(M') = N$ you know that there exists an $m' \in M'$ such that $g(m) = g(m')$ and hence $g(m-m') = 0$, i.e., $m - m' \in \ker g = \operatorname{im} f$. Thus there exists an $\ell \in L$ such that $f(\ell) = m - m'$. But $f(L) \subseteq M'$ so in particular $f(\ell) \in M'$. From here you should be able to conclude.
In your proof I think you're trying to use some version of the five lemma. This implies more than merely the fact that $M'$ is isomorphic to $M$ in some way over which you have no control: it says that the middle vertical arrow you have is such an isomorphism.
To make his reduction: let $L' = f(L)$ and consider the commutative diagram
\begin{CD}
0 @>>> L @>>> M @>>> N @>>> 0 \\
@. @VVV @| @VVV  \\
0 @>>> L' @>>> M @>>> M/L' @>>> 0
\end{CD}
where the vertical arrows are all isomorphisms. The situation is that $M'$ contains $L'$ and $M'$ maps surjectively onto $M/L'$, and now you can apply his proof.
