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.
Best Answer
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.