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I'm trying to understand why the two definitions of a quotient map are equivalent. Suppose $p:X\to Y$ is a quotient map in the first definition. Then certainly $p$ is continuous and maps all open sets to open sets (in particular it maps saturated open sets to open sets). But the reverse implication is not clear. Suppose $p$ is continuous and maps saturated open sets to open sets. There are two things to prove:

1) $p$ is surjective.

2) if $U\subset X$ is an arbitrary open set (not necessarily saturated), then $p(U)\subset Y$ is open.

How do I show this?


Use the following easily proved facts: a set $V\subseteq X$ is saturated with respect to $f$ if and only if there is a $U\subseteq Y$ such that $f^{-1}(U)=V,$ and then $V=f^{-1}(f(V))$

The claim is that if $f:X\to Y$ is continuous and surjective, then $f$ is a quotient map if and only if it takes saturated open (or closed) sets to open (closed) sets. Let's prove the "open" case.

$(\Rightarrow )$ If $S\subseteq X$ is open and saturated, then $f(S)$ is open because $f^{-1}(f(S))=S$ is open and $f$ is a quotient map.

$(\Leftarrow )$ Suppose $f$ is continuous and takes saturated open sets to open sets. Then, if $U\subseteq Y$ is open, so is $f^{-1}(U)$. On the other hand, if $f^{-1}(U)$ is open in $X$, then $f^{-1}(U)$ is saturated and so by assumption $f(f^{-1}(U))=U$ is open in $Y$ so $f$ is a quotient map.

  • $\begingroup$ Why is $f^{-1}(U)$ saturated if $f^{-1}(U)$ is open? $\endgroup$ – user531587 Sep 17 '18 at 2:01
  • $\begingroup$ I guess you don't use openness here, you just use that $f^{-1}(U)$ is saturated, regardless of whether or not $U$ is open. $\endgroup$ – user531587 Sep 17 '18 at 2:09
  • $\begingroup$ Yes. Indeed, any set of the form $f^{-1}(U)$ is saturated $\endgroup$ – Matematleta Sep 17 '18 at 11:05

Suppose $f$ is quotient. (So for all $U \subseteq Y$, $f^{-1}[U]$ open iff $U$ is open)

Let's check it satisfies: $f$ is continuous and maps open saturated sets to open sets.

$f$ continuous is clear: if $U \subseteq Y$ is open, so is $f^{-1}[U]$, by the right to left implication in the definition of quotient map.

Suppose that $S$ is saturated and open. $S$ saturated means that $S = f^{-1}[C]$ for some $C \subseteq Y$. So now we know $S = f^{-1}[C]$ is open and the other implication of the definition of quotient map gives us that $C$ is open and as $f[S] = f[f^{-1}[C]] = C$ (last equality by surjectivity of $f$) we know that f[S]$ is indeed open, as required.

Suppose now that $f$ is continuous and maps saturated open sets to open sets.

To see that $f$ is quotient we need to show $U \subseteq Y$ open in $Y$ iff $f^{-1}[U]$ open in $X$. Now, if $U$ is open in $Y$, $f^{-1}[U]$ is open in $X$ by continuity of $f$. And if $f^{-1}[U]$ is open in $X$ we note that $f^{-1}[U]$ is saturated (and open) so by assumption $f[f^{-1}[U]] = U$ is open. This shows that $f$ is quotient.

The saturated closed case is exactly similar, using the alternative definition of quotient maps in terms of closed sets.

  • $\begingroup$ In the last but one paragraph I guess you also need to assume $f$ is surjective, otherwise it's not a quotient map. $\endgroup$ – user531587 Sep 18 '18 at 2:38
  • $\begingroup$ @user531587 It's common to globally assume $f$ is surjective in such discussions. This characterisation does need it, IIRC. $\endgroup$ – Henno Brandsma Sep 18 '18 at 3:46

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