In the following commutative diagram, with exact rows and columns, is $j_1$ surjective? I have the following commutative digram of abelian groups and homomorphisms between them. The rows and columns are exact:
\begin{array}
A  &  &&  && &  & & & 0 & \  \\
 &  &&  & &  & && &\downarrow{}\\
  &  &&  && A_1 & \stackrel{h_1}{\longrightarrow} & A_2 &\stackrel{p_1}{\longrightarrow} & A_3 &  \  \\
  &  &&  && \downarrow{i_1} & &\downarrow{i_2}& &\downarrow{i_3}\\
   &  &&  && B_1 & \stackrel{h_2}{\longrightarrow} & B_2 &\stackrel{p_2}{\longrightarrow} & B_3 & \  \\
  &  &&  && \downarrow{j_1} & &\downarrow{j_2}& &\downarrow{j_3}\\
 & 0 && \stackrel{}{\longrightarrow} &&C_1 & \stackrel{h_3}{\longrightarrow} &C_2 & \stackrel{p_3}{\longrightarrow} & C_3 & {\longrightarrow} & 0 & \\
  &  &&  && & && &\downarrow{}\\
&  &&  &&  &  & &  & 0  &   & \\
\end{array}
I need to show that:
$p_1$ and $j_2$ are surjective $\implies $ $p_2$ and $j_1$ are surjective
I've been able to use the four lemma to show that $p_2$ is surjective. But I don't know how to show $j_1$ is also surjective.
 A: A little diagram chase should suffice.

Start with $x\in C_1$. As $h_3(x)\in C_2$ and $j_2$ is surjective by assumption take $y\in B_2$ such that $j_2(y)=h_3(x)$. By commutativity of the diagram and by exactness we have
  $$(j_3\circ p_2)(y)=(p_3\circ j_2)(y)=(p_3\circ h_3)(x)=0$$
  Therefore $p_2(y)\in\ker j_3$ and again by exactness there is $z\in A_3$ such that $i_3(z)=p_2(y)$. As $p_1$ is surjective by assumption take $w\in A_2$ such that $p_1(w)=z.$ Now 
  $$(p_2\circ i_2)(w)=(i_3\circ p_1)(w)=i_3(z)=p_2(y)$$ 
  Let $k=y-i_2(w)$, then 
  $$p_2(k)=p_2(y-i_2(w))=p_2(y)-(p_2\circ i_2)(w)=0$$ 
  By exactness there is $u\in B_1$ with $h_2(u)=k$. Now, observe that 
  $$(h_3\circ j_1)(u)=(j_2\circ h_2)(u)=j_2(k)=j_2(y-i_2(w))=j_2(y)-(j_2\circ i_2)(w)=h_3(x)$$
  Finally, as $h_3$ is injective by assumption we obtain $j_1(u)=x$ and $j_1$ is surjective as desired.

I am almost entirely sure that the details are painful to read and I think there is a smarter way maybe involving the Four Lemma too. But I could not resist writing down the whole chase; my apologies.
