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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 $X \cup_g Z \; = (X+Z)_{\pi_2 \circ \varphi} \cong X \cup_f Y$.

commutative diagram

So is asserting equality here correct? Is it correct but confusing? Is it wrong?

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    So basically they are saying that extensionality does not apply to partitions. That is an "unusual point of view".2011-09-17

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If $\mathcal{P}_1=\mathcal{P}_2$, then the equality $X_{\mathcal{P_1}}=X_{\mathcal{P_2}}$ is correct of course.

Let's say you have two maps $f:X\to Y$, $g:X\to Z$ inducing the partitions $\mathcal{P}_f$ and $\mathcal{P_g}$ on $X$. Now if $\mathcal{P}_f=\mathcal{P}_g$, then $X_{\mathcal{P}_f}=X_{\mathcal{P_g}}$. So in your specific situation we get the equality $(X+Z)_{\pi_1}=(X+Z)_{\pi_2\circ\varphi}$. But by definition $X\cup_g Z=(X+Z)_{\pi_1}$, so $X\cup_g Z=(X+Z)_{\pi_2\circ\varphi}$.

On the other hand the equality $(X+Z)_{\pi_2\circ\varphi}=X\cup_f Y$ is not correct in general, since $(X+Z)_{\pi_2\circ\varphi}$ is a identification space of $X+Z$ while $X\cup_f Y$ is a identification space of $X+Y$.

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    @LostInMath: Oh, that makes sense. Thank you for your insight.2011-09-20