Proving Short Five Lemma without the Five Lemma We have the following (commutative) diagram:
$$\begin{matrix}
0 & \to & M_1 & \xrightarrow{\phi_1} & M_2 & \xrightarrow{\phi_2} & M_3 & \to & 0 \\
\ & \   & \downarrow^{\alpha_1} & \ & \downarrow^{\alpha_2} \ & \ & \downarrow^{\alpha_3} \\
0 & \to & N_1 & \xrightarrow{\tau_1} & N_2 & \xrightarrow{\tau_2} & N_3 & \to & 0
\end{matrix}$$
$M_i,N_i$ are modules and $\phi_i,\tau_i$ are homomorphisms with exact rows.
We have injections $\phi_1:M_1 \to M_2$ and $\tau_1:N_1 \to N_2$ and surjections $\phi_2, \tau_2$. $\alpha_1,\alpha_3$ are isomorphisms, so that $\tau_1 \circ \alpha_1$ is an injection and $\alpha_3 \circ \phi_2$ is a surjection. 
Therefore, we have two exact sequences:
$$0 \to M_1 \to M_2 \to N_3 \to 0$$
$$0 \to M_1 \to N_2 \to N_3 \to 0$$
Considering the first one we have $M_2/M_1 \cong N_3$ and from the second one we obtain $N_2/M_1 \cong N_3$ and finally
$$N_2/M_1 \cong M_2/M_1$$
<<(without loss of generality we can assume that $M_1 \subset M_2 \subset N_2$).>>

My questions:
1. Is this part of proof correct?
2. Does $N_2/M_1 \cong M_2/M_1$ imply $N_2 \cong M_2$? 

Thanks in advance.

I think your identifications more obfuscate what’s going on than simplify the problem.


*

*If $α_2$ is to make both the left and the right square commute, in the reduced case it must be an inclusion – as all other arrows of the left square are inclusions as and the vertical arrow in the right square is assumed to be the identity as well.

*Thus, the question becomes: Does $N_2 / M_1 = M_2 / M_1$ imply $N_2 = M_2$?
And yes, it does. Let me rephrase it as: Does $A / C = B / C$ imply $A = B$? The premise says that for any element $x$ of either $A$ or $B$, there is some $y$ in either $B$ or $A$ respectively such that $x + C = y + C$, so in particular $x ∈ B + C$ and $y ∈ A + C$. Since $C ⊆ A$ and $C ⊆ B$, it follows that $A = B$.
But, I think it’s less complicated to do the diagram chase and conclude this fact above as an easy corollary.
