I am trying to solve the following problem.
Question: Consider the commutative diagram of abelian groups given below, where the rows in the diagram are exact.
Prove or disprove: if $p,q,s$ and $t$ are zero homomorphisms, then so is $r.$
My attempt: At first I thought this was false, so I tried to find a counterexample. I thought I had one with the sequence $0\longrightarrow Z/2Z\longrightarrow Z/4Z \longrightarrow Z/2Z \longrightarrow 0$ in both rows, but the diagram ended up being not commutative. After several failed attempts to find a counterexample, I tried to prove the statement, which also failed. Could someone please help?
Thank you!
This is false. As someone mentioned in the comments (I noticed a lot of people are writing answers as comments lately, am I missing something?), you can use as both rows $0 \to \mathbb{Z}/2\mathbb{Z} \to \mathbb{Z}/4\mathbb{Z} \to \mathbb{Z}/2\mathbb{Z} \to 0$ (the exactness of this sequence is just the fact that $(\mathbb{Z}/4\mathbb{Z})/(\mathbb{Z}/2\mathbb{Z}) \cong \mathbb{Z}/2\mathbb{Z}$). Use as vertical maps multiplication by 2. Then the middle vertical map is not zero while the other obviously are zero.