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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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How do these "others" define "identification space"? (I gather this is a synonym for quotient space) –  Niels Diepeveen Sep 17 '11 at 16:28
    
it is obvious that if $\mathcal P_1$ and $\mathcal P_2$ are the same partition of your space $X$, then the quotient spaces $X/\mathcal P_1$ and $X/\mathcal P_2$ are the same identification space. That is not your opinion! –  Mariano Suárez-Alvarez Sep 17 '11 at 18:17
    
@Niels: Yes, an identification space is a quotient space. I would say we all agree on that. –  Corey Harris Sep 17 '11 at 18:46
    
OK. I just wanted to check that there is not some weird definition in which $X/P$ is not uniquely defined for given $X$ and $P$. –  Niels Diepeveen Sep 17 '11 at 19:19
    
Well, on this point, there is not agreement. One opinion is that if the partition $\mathcal{P_1}$ is induced by a map $f$, and the partition $\mathcal{P_2}$ is induced by a map $g$ which is not equal to $f$, then $X_{\mathcal{P_1}}$ and $X_{\mathcal{P_2}}$ cannot be equal but may be homeomorphic. I disagree and say that once the partition has been defined, the map which induced it is no longer relevant. –  Corey Harris Sep 17 '11 at 19:42

1 Answer 1

up vote 2 down vote accepted

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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It might be helpful to make that last statement explicit: formally, points of $(X+Z)_{\pi_2\circ\varphi}$ are subsets of $X\cup Z$, while points of $X\cup_f Y$ are subsets of $X\cup Y$, so the two identification spaces cannot be identically equal unless $Z=Y$. –  Brian M. Scott Sep 17 '11 at 18:30
    
I just saw that I swapped the relations in the last statement (the one on which my questions depends!). I've fixed it now. It should have said $X \cup_g Z \; = (X+Z)_{\pi_2 \circ \varphi} \cong X \cup_f Y$. –  Corey Harris Sep 17 '11 at 18:34
    
@LostInMath: Your answer is helpful, but I don't want to accept it before you get a chance to respond to the errata. –  Corey Harris Sep 19 '11 at 3:41
    
@daswerth: I edited the answer. One more point regarding the formulation of the question: note that the attaching map $g$ is a map from a subset of $X$ to $Z$ so it does not induce a partition on the disjoint union $X+Z$. I think you mean the partition induced by the projection $\pi_1:X+Z\to X\cup_g Z$. –  LostInMath Sep 19 '11 at 10:50
    
@LostInMath: Oh, that makes sense. Thank you for your insight. –  Corey Harris Sep 20 '11 at 12:49

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