# 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 $N_2 \cong M_2$?

• It's not true in general that $M_2\subseteq N_2$, since they could very well be different modules that happen to be isomorphic. Everything up to that looks good, though. – Santana Afton Jul 31 '17 at 13:31
• For your second question, the answer is yes... But it's really just a reformulation of the Short Five Lemma (with $\alpha_1$ being an identity rather than an iso, but that doesn't make any difference)! – Arnaud D. Jul 31 '17 at 14:13
• Hmmm... I didn't use that the diagram is commutative... – Nicholas S Jul 31 '17 at 17:32
• @ArnaudD. With 2-year delay I can say that the answer is no. – Nicholas S Dec 2 '19 at 20:51

• 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 $A = B$.