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?


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 $\mathscr{P}(X)$.

  • $\begingroup$ 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
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.