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Suppose $X$ is compact Hausdorff and $\mathcal P$ is a partition of $X$. Define the map $\pi:X \to \mathcal P$ taking each point to the unique partition element containing it. Give $\mathcal P$ the quotient topology under $\pi$. Assume the quotient topology is Hausdorff. This is equivalent to assuming $\pi$ is a closed map, and also equivalent to the condition that ($*$) every partition element is contained in an open subset of $X$ that is a union of partition elements.

Now take a closed set $Y \subset X$ which is a union of partition elements. There are two ways to put a topology on $\mathcal P_Y = \{P \in \mathcal P: P \subset Y\}$.

  1. Consider the subspace $\pi(Y) \subset \mathcal P$

  2. Give $\mathcal P_Y$ the quotient topology under the natural map $\Pi: Y \to \mathcal P_Y$.

Condition $(*)$ for $\mathcal P$ implies the same condition for $\mathcal P_Y$, and so the quotient topology (2) is Hausdorff. Since the two maps have the same fibers $-$ precisely the elements of $\mathcal P_Y$ $-$ it follows from the fact that surjections between compact Hausdorff spaces are quotient maps that the two topologies are natutally homeomorphic.

I suspect the above fails horribly in case the subspace $Y$ is not closed. However I am having trouble coming up with an example. Has anyone considered the above situation before?

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  • $\begingroup$ There is something wrong in your last statement in the first paragraph, because every partition element is always contained in $X$ which is an open subset of $X$ and the union of all partition elements. So the equivalence would mean that $\pi$ is always closed. $\endgroup$ Commented Nov 16 at 11:00
  • $\begingroup$ I'm not entirely sure, but if you just consider $Y$ as being the union of elements of the partition, you can prove that $\pi^{-1}(\pi(Y)) = Y$ and from there that the restriction $\pi\colon Y\to\pi(Y)$ is a quotient map. Now you have two quotient maps on $Y$ which have the same fibers. Each one induces a continuous map via the other, which (I guess) are inverse of each other. However, this is just a hint that comes to mind. $\endgroup$ Commented Nov 16 at 11:10

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The quotients you are considering are given by upper semi-continuous decompositions.

The subspace and quotient topology will coincide if the quotient map $\pi$ is closed.

See this answer of mine where I redirect you to theorem in Topology by Dugundji.

In particular, since your map $\pi:X\to \mathcal{P}$ is closed, the subspace topology on $\mathcal{P}_Y$ and quotient topology from the map $\pi\restriction_Y:Y\to\mathcal{P}_Y$ will coincide.

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  • $\begingroup$ This is exactly what I was looking for. You even corrected my faulty definition. Thanks a million! $\endgroup$ Commented 2 days ago

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