I was doing Exercise A.36 in Lee's Introduction to smooth manifolds which states the following:

Let $q: X \rightarrow Y$ be an open quotient map. Then $Y$ is Hausdorff if and only if $R = \{(x_1, x_2) \mid q(x_1) = q(x_2) \}$ is closed in $X \times X$. I know this has been asked before and my answer agreed with the top answer in this post: $X/{\sim}$ is Hausdorff if and only if $\sim$ is closed in $X \times X$. However, I don't see where we use the fact that $q$ is actually a quotient map. It seems that the solution only uses that $q$ is open (so that $q \times q$ is open), continuous and surjective. Can someone please clarify?

On a related note, another answer to the same post also states that the product of two open quotient maps is an open quotient map. Is this really true though? I know that in general, the product of two quotient maps is not a quotient map but the counterexamples do not involve maps that are also open.

  • $\begingroup$ Okay, so I guess a quotient map $f:X\to Y$ is defined as a map between topological spaces which is surjective, and $U$ is open in $Y$ if and only if $f^{-1}(U)$ is open in $X$. So, I guess you don't need $q$ to be a quotient map, but you do need continuity and surjectivity, and these things are guaranteed by $q$ being a quotient map. $\endgroup$ – Ben W Jan 3 '19 at 20:46
  • $\begingroup$ Well, being a quotient map is not sufficient either. Quotient maps are not necessarily open and we do need openness. $\endgroup$ – Thomas Bakx Jan 3 '19 at 20:49
  • $\begingroup$ Right, you need it to be open, continuous, and surjective. The last two conditions are guaranteed by $q$ being a quotient map, and the first condition is given by hypothesis., $\endgroup$ – Ben W Jan 3 '19 at 20:50
  • $\begingroup$ All that is clear, this is not answering my question. $\endgroup$ – Thomas Bakx Jan 3 '19 at 21:25
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    $\begingroup$ @ThomasBakx What makes you think that a projection $\mathbb{R}^2\to\mathbb{R}$ is not a quotient map? Every surjective continuous open mapping is a quotient map. This is implied by $f(f^{-1}(V))=V$ which holds for any $V\subseteq Y$ if $f:X\to Y$ is surjective. $\endgroup$ – freakish Jan 3 '19 at 22:23

Let us agree that a map is a continuous function.

A quotient map $q : X \to Y$ is a surjective map such that $V \subset Y$ is open if and only $q^{-1}(V) \subset X$ is open.

A quotient map is in general not an open map, but is is well-known that any open surjective map is a quotient map (this follows from $q(q^{-1}(V)) = V$).

The assumption "Let $q : X \to Y$ be an open quotient map" is therefore the same as "Let $q : X \to Y$ be an open surjective map".

What is the relation to the post $X/{\sim}$ is Hausdorff if and only if $\sim$ is closed in $X \times X$?

In this post we do not start with a map $q : X \to Y$ between topological spaces, but with a space $X$ and an equivalence relation $\sim$ on $X$ and then define $Y = X / \sim$. The function $$q : X \to Y, q(x) = [x] ,$$ where $[x]$ denotes the equivalence class of $x$ with respect to $\sim$, is by definition a surjection. $X$ is a topological space, but $Y$ is defined as a set which does not yet have a topology. In this situation the set $Y$ is by default endowed with the quotient topology (which is the finest topology on $Y$ making $q$ continuous).

Thus, the phrase "If the quotient map is open" tells us two facts:

(1) The set $Y$ is endowed with the quotient topology.

(2) $q$ is an open map.

For the proof we only need (2). But without specifying the topology on $Y$ it does not make sense to say that $q : X \to Y$ is an open map, and that is the reason why the above phrase is used.

However, alternatively one could also say that $Y$ is endowed with a topology such that $q$ becomes an open map. In that case the topology on $Y$ must automatically be the quotient topology.

In your post we already have a topology on $Y$, and it is redundant to say that $q$ is an open quotient map.

Finally, the product of two open quotient maps is an open quotient map simply because the product of two open maps is an open map.

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