Why the rightmost square is a pullback?

Solution 1:

To prove that the rightmost square is a pullback, it is enough to prove that the canonical arrow $(f,g):X\to Y\times_{Y''}X''$ (which exists because the square commutes) is an isomorphism. To this end, look at the decomposition of your square into two commutative squares, one induced by the equality $\psi_2\circ (f,g)=g$ and the other one given by the actual pullback $Y\times_{Y''}X''\to X''$. Then taking kernels yields a commutative diagram $$\require{AMScd} \begin{CD}0 @>>> X' @>{\ker(g)}>> X @>{g}>> X'' @>>> 0 \\ & @V{u}VV @V{(f,g)}VV @VV{1_{X''}}V \\ 0 @>>> K @>{\ker(\psi_2)}>> Y\times_{Y''}X'' @>{\psi_2}>> X'' @>>> 0 \\& @V{v}VV @V{\psi_1}VV @VV{f''}V \\ 0 @>>> Y' @>>{\ker(h)}> Y @>>{h}> Y'' @>>> 0 \end{CD}$$ whose rows are exact ($\psi_2$ is an epi because $g$ is). Here $v$ is the factorisation of $\psi_1\circ \ker(\psi_2)$ through $\ker(h)$, which must exist since $$h\circ \psi_1\circ \ker(\psi_2)=f''\circ \psi_2 \circ \ker(\psi_2)=0;$$ and $u$ is the factorization of $(f,g)\circ \ker(g)$ through $\ker(\psi_2)$, which must exist since $$\psi_2\circ (f,g)\circ \ker(g)=g\circ \ker(g)=0.$$ Notice that $$\ker(h)\circ v\circ u= \psi_1\circ \ker(\psi_2)\circ u=\psi_1\circ (f,g)\circ \ker(g)=f\circ \ker(g)=\ker(h)\circ f',$$ and thus $v\circ u=f'$ since $\ker(h)$ is a monomorphism. Now the result in your previous question shows that $v$ must be an iso, so that if $f'$ is an iso, so is $u$. Then applying the Short Five Lemma to the upper part of the diagram shows that $(f,g)$ is an isomorphism.

Note : This answer was essentially adapted from my MO answer here.