Not a quotient map on K-topology

Question

Recall that $\Bbb R_K$ denotes the real line in the $\mathit{K}$-topology. Let $Y$ be the quotient space obtained from $\mathbb{R_K}$ by collapsing the set $\mathit{K}$ to a point; let $p: \Bbb{R_K} \to Y$ be the quotient map. Then show that $p \times p : \Bbb{R}_K \times \Bbb{R}_K \to Y \times Y$ is not a quotient map. I am not able to get this one. Please help me out.

Answer 1

In part $(a)$ of this question from Munkres you must have shown that $Y$ is $T_1$ (we collapse a closed set in a $T_1$ space ) but is not Hausdorff (as $0 \in Y$ (really the class of $0$) and $K \in Y$ (as a class, so a point of $Y$ (!)) do not have disjoint open neighbourhoods (recall that $\Bbb R_K$ is not regular because we do not have disjoint open neighbourhoods of $0$ and the closed set $K$). So for $(b)$ we can follow Munkres' helpful hint:

It follows from $Y$ not being Hausdorff that $\Delta_Y = \{(y,y) \mid y \in Y\}$ is not a closed subset of $Y \times Y$ (explicitly, $(0,K)$ is in its closure but not in it).

But $$(p \times p)^{-1}[\Delta_Y] = (K \times K) \cup \Delta_X$$

which is closed in the domain, where $\Delta = \{(x,x)\mid x \in \Bbb R_K\}$ is closed in $\Bbb R_K \times \Bbb R_K$ as the $\Bbb R_K$ is Hausdorff. So despite the inverse image under $p \times p$ being closed, $\Delta_Y$ is not closed in the $Y \times Y$, so this directly contradicts the definition of a quotient map.

Answer 2

Note that the map $p\times p$ is continuous, so the inverse images of open sets under $p\times p$ are open. Thus, we want to find a subset of $Y\times Y$ that is not open but that has an open inverse image under $p\times p$. And you can of course replace open with closed throughout. I see that Henno has posted the easiest example, using the diagonal in $Y\times Y$. Here’s a rather different example, using a non-open set in $Y\times Y$.

SKETCH: Choose positive reals $r_{m,n}$ for $m,n\in\Bbb Z^+$ so that $\langle r_{m,n}:n\ge 1\rangle$ is a strictly decreasing sequence with limit $0$ for each $m\in\Bbb Z^+$, and $\langle r_{m,n}:m\ge 1\rangle$ is a strictly decreasing sequence with limit $0$ for each $n\in\Bbb Z^+$. Choose them small enough that the sets

$$B_{m,n}=\left(\frac1m-r_{m,n},\frac1m+r_{m,n}\right)\times\left(\frac1n-r_{m,n},\frac1n+r_{m,n}\right)$$

are pairwise disjoint. Let $U=\bigcup\{B_{m,n}:m,n\in\Bbb Z^+\}$; $U$ is an open nbhd of $K\times K$ in $\Bbb R_K\times\Bbb R_K$.

Let $A$ be the image of $U$ under the map $p\times p$, and let $y_K$ be the point of $Y$ corresponding to $K$; clearly $\langle y_K,y_K\rangle\in A$. Any open nbhd of $y_K$ in $Y$ must contain a set of the form $$\{y_K\}\cup\bigcup_{n\in\Bbb Z^+}\left(\frac1n-s_n,\frac1n+s_n\right)\,,$$ where $0<s_n<\frac1{2n(n+1)}$. Thus, any open nbhd of $\langle y_K,y_K\rangle$ in $Y\times Y$ must contain the Cartesian product of two such sets. To complete the argument, show that $A$ does not contain the product of two such sets and is therefore not open in $Y\times Y$. Since its inverse image under $p\times p$ is $U$, which is open, $p\times p$ cannot be a quotient map.