Commutativity of diagram involving two arrows

Question

Hi suppose I have a diagram that looks like this:

but where we only have $fe = hf'$ and $ge = hf'$. What would I call the square? I can't say that it commutes yes? Is it true that in general given such a diagram, saying it commutes is the same as saying that all possible combinations of going around are always equal?

Edit: My two columns are actually part of two separate equalizers (namely the equalizers that express the sheaf properties of $\mathcal{F}$ and $\mathcal{G}$).

Answer 1

Yes, to say that a diagram commutes means that all possible paths of arrows from any given vertex to any other given vertex compose to the same morphism. The situation you describe can be stated as "the diagram, when neglecting the right vertical arrows, commutes".