This may be more of a philosophical question. The question is thus: Let $X$ be a topological space, and $\mathcal{P_1}, \mathcal{P_2}$, two partitions of $X$. Now consider the identification spaces $X_{\mathcal{P_1}}$ and $X_{\mathcal{P_2}}$. If we can show $\mathcal{P_1} = \mathcal{P_2}$, my opinion is that we have $X_{\mathcal{P_1}} = X_{\mathcal{P_2}}$. But others say we don't have anything more than $X_{\mathcal{P_1}} \cong X_{\mathcal{P_2}}$. That is, these spaces are homeomorphic, but not equal.
If it helps, what follows is the real-world problem which gave rise to the debate.
We have a homeomorphism, $\varphi$, from $X+Z$, the disjoint union of $X$ and $Z$, to $X+Y$. We also have the identification spaces given by attaching maps $f$ and $g$, as shown in the diagram, and the projections onto these identification spaces, $\pi_1, \pi_2$. The map $\pi_2 \circ \varphi$ is the composition of identification maps, so it is an identification map. The statement is:
We wish to show $X \cup_g Z \; \cong X \cup_f Y$. But $X \cup_f Y \; \cong (X+Z)_{\pi_2 \circ \varphi}$. Thus, if we can show that the partitions induced on $X+Z$ by $g$ and by $\pi_2 \circ \varphi$ are equal, then we'll have $X \cup_g Z \; = (X+Z)_{\pi_2 \circ \varphi} \cong X \cup_f Y$.
So is asserting equality here correct? Is it correct but confusing? Is it wrong?