1
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – DanielWainfleet Nov 21 '16 at 11:25
3
$\begingroup$

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.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.