A problem of diagram chasing Consider the following diagram of functions between sets:

I know that the $4$ inner triangles (i.e. $\{X,X',Z\}$,$\{X',Y',Z\}$...) are all commutative diagrams  and moreover that $f_1$ and $f_3$ are bijective functions. 
Can I conclude that the outer square $\{X,X',Y',Y\}$ is a commutative diagram?
Thanks in advance.

I think that you can't. Take for example $X=X'=Z=\{a\}$, $Y=Y'=\{a,b\}$, $f_1,f_3,a,b$ the identity maps. Then set $f_2\colon a\mapsto a$ and $f_4\colon a\mapsto b$. Then every triangle commute but the big square doesn't.
More generally, you can take $X=X'$, $Y=Y'$, $f_1,f_3$ the identity maps. Then you pick two different maps $f_2,f_4\colon X\to Y$ and you set $Z$ as the equalizer of those maps and $X\to Z$ the map that you get composing $f_2$ with the equalizer.
