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For a topological space $X$ and a equivalence relation $\mathcal{R}$ on $X$, and $p:X\to X/\mathcal{R},\ x\mapsto[x]$, my notes has the proposition:

A subset $\tilde{V}$ in $X/\mathcal{R}$ endowed with the quotient topology is closed if and only if $p^{−1} (\tilde{V} )$ is closed in $X$.

With proof following from the fact that

$$p^{-1}\big((X/\mathcal{R})\setminus\tilde{V}\big)=X\setminus p^{-1}(\tilde{V}).$$

I can see that, as $p$ is continuous, $\tilde{V}$ closed in $X/\mathcal{R}$ $\Leftrightarrow$ $(X/\mathcal{R})\setminus\tilde{V}$ open in $X/\mathcal{R}$ $\Rightarrow$ $p^{-1}\big((X/\mathcal{R})\setminus\tilde{V}\big)=X\setminus p^{-1}(\tilde{V})$ open in $X$ $\Leftrightarrow$ $p^{-1}(\tilde{V})$ closed in $X$.

I can't see the other direction though. It seems to fail as $p^{-1}(S)$ open $\not\Rightarrow\ S$ open.

What am I missing to complete the proof?

Is $p^{-1}$ continuous?

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This is the definition of the topology on the quotient space : a subset of the quotient is open by definition if and only if its inverse image by $p$ is open, and if we note $Y$ the quotient space, we have $p^{-1}(Y\backslash U) = X\backslash p^{-1} (U)$, which shows that a subset of $Y$ is closed if and only if its inverse image by $p$ is closed, as being closed is by definition being the complementary of an open set.

By the way $p^{-1}$ is not an application between a priori topological spaces, so asking about its continuity is wrong. The application $p^{-1}$ is an application from $\mathscr{P}(Y)$ to $\mathscr{P}(X)$.

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  • $\begingroup$ Thanks, I'd missed that that was defintion. My course defined it with open sets, and then had this to prove it for closed sets. $\endgroup$ – Ori Feb 18 '15 at 20:50
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    $\begingroup$ You're welcome. If you're ok with the answer, don't hesitate to validate it. ;-) $\endgroup$ – ujsgeyrr1f0d0d0r0h1h0j0j_juj Feb 18 '15 at 20:52
  • $\begingroup$ I've been hovering over the tick. It needs another minute before I can accept. $\endgroup$ – Ori Feb 18 '15 at 20:53

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