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Let $∼$ be an equivalence relation on a topological space X. $\ Y = X/∼ $ equipped with the quotient topology. How to show that if X is Hausdorff and the set $ \big\{ (x, y) : x ∼ y \big\} ⊆ X × X$ is closed then the quotient map is open.

Could someone possibly provide some hints in proving this statement.

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  • $\begingroup$ That set is not necessarily the diagonal of $X\times X.$ In particular, it won't be unless $\sim$ is the equality relation on $X.$ The diagonal is automatically a closed subset of $X\times X$ (as this is equivalent to $X$ being Hausdorff). $\endgroup$ – Cameron Buie May 16 '16 at 12:39
  • $\begingroup$ @CameronBuie yes that is true, but by diagonal in the title what I really meant was diagonal in the sense of equivalence. $\endgroup$ – Topology guy May 16 '16 at 12:42
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This is false, I think: take $X = \mathbb{R}$, $A = \mathbb{Z}$ and let $\sim$ be the relation that identifies $A$ to a point. Then the relation as a subset of $X \times X$ is $(A \times A) \cup \Delta_X$, so this is closed, and $X$ is clearly Hausdorff.

The space $X/\negthinspace\negthinspace\sim$ is not first countable at the point that corresponds to $A$, and if $q$ were open, its image would even be second countable.

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