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.
Not a quotient map on K-topology
general-topologyquotient-groupquotient-spaces
Best Answer
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.