A question on diagram of groups

Question

Question: Assume that the rows of the following diagram of abelian groups are exact.

Prove or disprove the following statements:

1. If $p,q,s,t$ are zero homomorphisms, so is $r.$
2. If $p,q,s,t$ are surjective homomorphisms, so is $r.$

My attempt:

1. By taking the same exact sequence for both rows, namely, $0\longrightarrow Z/2Z\longrightarrow Z/4Z \longrightarrow Z/2Z \longrightarrow 0$ and $r$ to be the homomorphism that multiplies by $2,$ I was able to disprove this statement. Am I correct about this one?

2. I am stuck with this one and I don't know how to proceed. Could someone please help?

Thank you in advance!

Answer 1

For the first one, take $$\begin{array}{rcccccccl} 0&\longrightarrow &0&\longrightarrow&\mathbb{Z}&\stackrel{f}{\longrightarrow}&\mathbb{Z}&\longrightarrow &0\\ {\scriptstyle p}\downarrow &&\!\!\!\!\!{\scriptstyle q}\downarrow&&\!\!\!\!\!{\scriptstyle r}\downarrow&& \!\!\!\!\!{\scriptstyle s}\downarrow&&\!\!\!\!\!{\scriptstyle t}\downarrow\\ 0 & \longrightarrow & \mathbb{Z} &\stackrel{g}{\longrightarrow} & \mathbb{Z} & {\longrightarrow} & 0 & \longrightarrow & 0 \end{array}$$ with $p,q,s,t$ the zero maps, and $f,g,r$ all equal to the identity. The rows are exact (the maps $f$ and $g$ are isomorphisms); the diagram commutes: first and last squares clearly commute; the second square commutes since $g(q(0)) = g(0) = 0 = r(0)$; the third square commutes since $s(f(a))=0$ and the image of $r(a)$ is zero. Here $p,q,s,t$ are the zero maps, but $r$ is not the zero map.

For the second one, take $$\begin{array}{rcccccccl} 0&\longrightarrow &0&\longrightarrow&\mathbb{Z}&\stackrel{f}{\longrightarrow}&\mathbb{Z}\oplus\mathbb{Z}&\longrightarrow &\mathbb{Z}\\ {\scriptstyle p}\downarrow &&\!\!\!\!\!{\scriptstyle q}\downarrow&&\!\!\!\!\!{\scriptstyle r}\downarrow&& \!\!\!\!\!{\scriptstyle s}\downarrow&&\!\!\!\!\!{\scriptstyle t}\downarrow\\ 0 & \longrightarrow & 0 & \longrightarrow & \mathbb{Z}\oplus\mathbb{Z} & \stackrel{g}{\longrightarrow} & \mathbb{Z}\oplus \mathbb{Z} & \longrightarrow & 0 \end{array}$$ where $f(a)=r(a)= (a,0)$, and $s$ and $g$ both the identity map.

All of $p$, $q$, $s$, and $t$ are surjective. The rows are exact, but $r$ is not surjective.

Cf. the Five Lemma.

(Obviously you can essentially replace $\mathbb{Z}$ with any nonzero abelian group)