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Suppose $f:X \to Y.$ Let $A$ be the assertion $f$ is continuous and maps saturated open sets to open sets, let $B$ be the assertion $f$ is a quotient map. I am struggling to prove $A \Leftrightarrow B,$ which was supposedly "proven" here.

My proof of $B \Rightarrow A$ is the same as the given proofs, and my proof of $A \Rightarrow B$ proceeds along the same lines until I get to the statement $f(f^{-1}(U)) = U.$ This is false because we do not know $f$ is surjective. All we know is that $f(f^{-1}(U)) \subseteq U.$ Did the book forget to mention surjectivity in the "is equivalent to..." discussion?

Alex Ortiz
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  • The entire discussion there is about surjective maps. That is stated explicitly in the definitions of quotient map and saturated with respect to a (surjective) map, and in that context it should be clear that the final sentence in the extract refers to a surjective map $p:X\to Y$. The author didn’t forget to mention it, but rather assumed that the reader would recognize that it was being assumed throughout that discussion. – Brian M. Scott Sep 21 '20 at 04:10

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Proof of $f(f^{-1}(U)) = U$ is the implication $1 \Rightarrow 2$ here. And then you get surjectivity from $f(f^{-1}(Y)) = Y,$ completing the proof of $f$ being a quotient map.

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The preamble of this discussion by Munkres (I assume that's what you're talking about, as you link here) assumes that the map $p: X \to Y$ is surjective. So that's assumed throughout this paragraph and "$p$ quotient" $\iff$ $p$ continuous and maps "saturated open sets to open sets" is under this assumption.

So yes, it's not explicitly mentioned in a theorem, but it is part of the discussion. We never talk about quotient maps and the like for non-surjective functions anyway.

Munkres in fact proved (implicitly, he doesn't give it a number as he doesn't need it later)

Let $p:X \to Y$ be a surjective function. Then TFAE:

  1. $p$ is quotient.
  2. $p$ is continuous and for every saturated open set $O$ in $X$, $p[O]$ is open in $Y$.
Henno Brandsma
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