# Closed, continuous and surjective map from $X\times Y$ to $X$?

We know that for any two topological spaces $X(\ne\emptyset)$ and $Y(\ne\emptyset)$, there exists an open, continuous and surjective map $\pi:X\times Y\to X$ defined by $\pi(x,y)=x$ for all $(x,y)\in X\times Y$ where the topology on $X\times Y$ is the product topology.

My question is,

Let $X(\ne\emptyset)$ and $Y(\ne\emptyset)$ be two topological spaces and $X\times Y$ is given the the product topology. Does there always exist a closed, continuous and surjective map $\Psi:X\times Y\to X$? If so then can anyone give an explicit example?

• This may be nit-picking but you should specify that $Y$ is not empty unless $X$ also is, both for the open and the closed mapping. – DanielWainfleet Nov 21 '16 at 11:25

I have not been seen a proof, but in the abstracts for the 2006 International Conference on Topology and its Applications Mikhail A. Patrakeev announced the theorem that there is no closed, continuous map of $\Bbb S^n$ onto $\Bbb S$ for $n>1$, where $\Bbb S$ is the Sorgenfrey line. In particular there is none from $\Bbb S\times\Bbb S$ to $\Bbb S$.

Added: And here is a simple counterexample. Let $X=\{0\}\cup\left\{\frac1n:n\in\Bbb Z^+\right\}$, and suppose that $f:X\times\Bbb R\to X$ is a continuous surjection. Let $x_n=\frac1n$ for $n\in\Bbb Z^+$. For each $k\in\Bbb Z^+$ there are $n(k)\in\Bbb N$ and $r_k\in\Bbb R$ such that $f(\langle x_{n(k)},r_k\rangle)=x_k$. Each of the sets $\{x_k\}\times\Bbb R$ is connected, so $f$ is constant on each of these sets, and

$$f\big[\{x_{n(k)}\}\times\Bbb R\big]=\{x_k\}$$

for each $k\in\Bbb N$. Let

$$F=\{\langle x_{n(k)},k\rangle:k\in\Bbb Z^+\text{ and }n(k)\ne 0\}\;;$$

then $F$ is closed in $X\times\Bbb R$, but $0\in\big(\operatorname{cl}f[F]\big)\setminus f[F]$, so $f[F]$ is not closed in $X$, and the map $f$ is not closed.

• I'm sorry but I don't understand your hint. Was it intended to give an answer to my first question? – user 170039 Nov 21 '16 at 6:13
• @user170039: Never mind: I misread the question. – Brian M. Scott Nov 21 '16 at 6:14