According to this other question, Why does there exist a unique quotient topology that makes a given surjective map a quotient map?

Munkres states the following:

If $X$ is a topological space and $Y$ is a set and if $p: X \to Y$ is a surjective map, then there exists exactly one topology on $Y$ relative to which $p$ is a quotient map.

I am trouble proving the existence part.

We know that every set $U$ is open if $f^{-1}(U)$ is open. But, I am having trouble showing that if $A$ and $B$ are open in $Y$ and $f^{-1}(A)$ and $f^{-1}(B)$ are open in $X$ then $f^{-1}(A \cap B)$ is open in $X$.

It seems obvious that $f^{-1}(A \cap B) \subseteq f^{-1}(A) \cap f^{-1}(B)$.

But, I'm not sure how to prove $f^{-1}(A \cap B) = f^{-1}(A) \cap f^{-1}(B)$ which we know to be open in $X$.

  • $\begingroup$ Erase, delete work around. $\endgroup$ – William Elliot Jan 21 '18 at 19:42
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    $\begingroup$ You're looking in the wrong direction. The quotient map need not be open. The quotient topology consists of those subsets whose preimage is open. $\endgroup$ – Daniel Fischer Jan 21 '18 at 19:42
  • $\begingroup$ I understand that, but for it to be a topology, every intersection of open sets in Y have to be also open. What I am trying to prove is that, if you start with all sets as open whose preimage is open, is the intersection of these sets also open, i.e. do we always have a valid topology. This would also imply that the intersection of the preimages for the two sets in Y should also be open, I believe. $\endgroup$ – Jeff Jan 21 '18 at 21:58
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    $\begingroup$ This follows from $f^{-1}[A\cap B] = f^{-1}[A] \cap f^{-1}[B]$ $\endgroup$ – Henno Brandsma Jan 21 '18 at 22:03

Henno's comment basically answered the question with the identity, $f^{-1}(A \cap B) = f^{-1}(A) \cap f^{-1}(B)$

Which is proved in other questions like how to prove $f^{-1}(B_1 \cap B_2) = f^{-1}(B_1) \cap f^{-1}(B_2)$


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