Given the following diagram
which is exact on all horizontal rows and the left and middle columns. I'm trying to show that it is exact at $A''$ which amount to showing that $\ker(\varphi_3) = 0$ that is $\varphi_3$ is injective. I'm having some issues with this. If we suppose that $\varphi_3(a'') = 0$, then as $\pi_1$ is surjective there exists $a \in A$ such that $a'' = \pi_1(a)$ so $$\varphi_3(a'')=\varphi_3(\pi_1(a)) = 0$$ now since this diagram commutes $$0 = \varphi_3(\pi_1(a)) = \pi_2(\varphi_2(a))$$ so $\varphi_2(a) \in \ker(\pi_2) = \operatorname{im}(i_2)$. We thus have that $$\varphi_2(a)= i_2(b')$$ for $b' \in B'$ and so $$0 = \pi_2(\varphi_2(a)) = \pi_2(i_2(b')).$$ This is where I'm stuck. I would want to have that $b' = 0$, but I cant get anywhere with this $\varphi_2(a) = i_2(b')$ expression as I cannot chase $b'$ any further. Any help with this?
From $\varphi_2(a) = i_2(b')$ you get $$0 = \psi_2(\varphi_2(a)) = \psi_2(i_2(b')) = i_3(\psi_1(b')) = 0.$$ Since $i_3$ is injective, $\psi_1(b') = 0$ and hence by exactness of the column at $B'$, $b' = \varphi_1(a')$ for some (unique) $a'$. Now $\varphi_2(a) = i_2(\varphi_1(a')) = \varphi_2(i_1(a'))$, and since $\varphi_2$ is injective, $a = i_1(a')$. So $a'' = \pi_1(i_1(a')) = 0$, as required.