quotient map in topological space

Question

Excuse me i have a question :: Let $$f:X \to Y$$ be a surjective , continuous map of a topological space . Show that if f is an open map then it is quotient map

Here i dealt with it like the following : Let $$V \subset Y$$ s.t V is open in Y then since f continuous this implies to $$f^{-1} ({V})$$ open in X . and on the other hand since f is open suppose $$f ^{-1} (V)$$ is open in X then $$f(f^{-1}(V)) = V$$ since f is surjective is open in Y , This shows that Y is the strongest topology determined by T and f >> so it is a quotient map >>

But my Dr. told me that i must use the saturated and i could not fully understand him << can you help me please

Answer 1

Your proof is essentially correct: the definition of $f: X \to Y$ being a quotient map is

for all $U \subseteq Y$ the set $U$ open in $Y$ iff $f^{-1}[U]$ open in $X$.

As you say the left to right implication (of the iff) follows from the continuity of $f$. The right to left one from the observation that $f[f^{-1}[U]] = U$ as $f$ is surjective and then $f$ open implies that when $f^{-1}[U]$ is open, so is its image $U$.

What your teacher uses is an alternative characterisation of quotient maps (a theorem you might have covered, or maybe your text book/notes uses it as the definition?):

A continuous map $f: X \to Y$ is quotient iff the image of a saturated open $V \subseteq X$ is open in $Y$.

And the condition about saturated open sets is obvious for open maps $f$: images of open sets are open, saturated or not. So in this light, a quotient map is a sort of weakened version of open map.

In the proof above, the set $V=f^{-1}[U]$ is a typical example of a saturated set (wrt $f$). So there the connection is quite clear: we apply openness of $f$ only to a saturated open set.