Help with product maps in topological spaces

Question

Let $A, B, X, Y$ be topological spaces.

Given two functions $f : A \to B$ and $g : X \to Y $, let $f \times g : A × X \to B \times Y$, $ (f \times g)(a, x) = (f(a), g(x))$.

can you help me to show that if f and g is cont. then fxg is too

and if the sets A,B is non-empty, then it holds the other way around too.

I'm pretty stuck.. can you please help me?

My solution

To show the statement, I will use Theorem 4.1. If I can show that $f\times g(\overline{S\times T})\subseteq \overline{f\times g(S\times T)}$ for a subset $S\times T\subseteq A\times X$.

Take a $(b,y)\in f\times g(\overline{S\times T})$. Then there exists a $s\in\overline S$ and $t\in\overline T$ such that $f(s)=b$ and $g(t)=y$. We know that both $f$ and $g$ is continuous, so from 4.1 we get $f(s)\in\overline B$ and $g(t)\in\overline Y$, and then $(b,y)\in \overline{f\times g (S\times T)}$. And from 4.1 we get that $f\times g$ is continuous.

hi Mees. I have just added my solution to (a). Could u maybe give your opinion on that? My theoreme 4.1 says that f is continuous iff. f(closure(A)) is in Closure(f(A)) — Esteban Cambiasso, Dec 03 '18 at 12:49
I just saw that i miss a bar over f×g(SxT) in the last part — Esteban Cambiasso, Dec 03 '18 at 13:39
For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to. To help you get started, I have retyped the text from your picture. — Martin Sleziak, Dec 03 '18 at 16:21
You might also have a look at some related questions: Continuity of cartesian product of functions between topological spaces, Characterization of the Product Topology as the finest in terms of a continuous function into the product.. — Martin Sleziak, Dec 03 '18 at 16:28
There are several problems with your attempt. You are using result which characterizes continuity using the condition $f\times g(\overline C)\subseteq \overline{f\times g(C)}$ for any subset $C\subseteq A\times X$. You are working with sets of the form $C=S\times T$ instead. Also you wrote that you are using $f(s)\in\overline B$ and $g(t)\in\overline Y$. Since $\overline B=B$ and $\overline Y=Y$ this does not give you any new information, so I assume you wanted to write something else there. — Martin Sleziak, Dec 03 '18 at 16:32
why is \overline(B)=B and \overline(Y)=Y? My goal is to show that if, (b,y) is in f(\overline(SxT)) then (b,y) is in \overline(f(SxT)) — Esteban Cambiasso, Dec 03 '18 at 16:47
@MarcHansen If your most recent was addressed to me, you should use @username so that I get the notification. For any topological space you have $\overline X=X$ (and $\overline\emptyset=\emptyset$. So this is true also for topological space $B$ and $Y$. (Maybe you wanted to use $f(S)$ instead of $B$ and $g(T)$ instead of $Y$?) — Martin Sleziak, Dec 03 '18 at 16:50
@MartinSleziak ah okay, i see. So if i say f(s)$\in$$\overline{f(S)}$ and g(t)$\in$$\overline{g(T)}$? then my answer is better? — Esteban Cambiasso, Dec 03 '18 at 17:53
In that way you would get $(b,y)\in\overline{f(S)}\times\overline{g(T)}$. Together with the fact hat $\overline{M\times N} =\overline M \times \overline N$ this seems to go in the right direction. — Martin Sleziak, Dec 03 '18 at 18:06
The closure approach is too messy IMHO. The categorical approach via the universal property seems cleaner to me. — Henno Brandsma, Dec 03 '18 at 22:05

Henno Brandsma · Answer 1 · 2018-12-03T22:04:16.733

By the universal property of products, a function : $F: Z \to X \times Y$ is continuous iff $\pi_X \circ F$ and $\pi_Y \circ F$ (the compositions with the continuous projections) are all (both) continuous.

This allows us to quickly solve the problem, applied to the product $B \times Y$ and $Z = A \times X$: $$\pi_B \circ (f \times g) = f\circ \pi_A$$ where the right hand side is continuous as a composition of continuous maps. The identity can be seen by evaluating both sides on an arbitary point $(a,x)$ and both expressions become $f(a)$.

Also $$\pi_Y \circ (f \times g) = g\circ \pi_X$$ and so both compositions with projections of $f \times g$ are continuous and so $f \times g$ is continuous.

On the other hand suppose that $f \times g$ is continuous, and fix $x_0 \in X$ and define $i: A \to A \times X$ by $i(a) = (a,x_0)$. As $\pi_A \circ i$ is the identity on $A$ (hence continuous) and $\pi_X \circ i$ is a constant map with the value $x_0$ (also continuous) the trusted universal property tells us that $i$ is continuous, and then note that $f = \pi_B \circ (f \times g) \circ i$ and then $f$ can be written as the continuous composition of continuous maps.

That $g$ is also continuous can be seen with the same idea and another embedding (now using a fixed $a_0 \in A$) from $X$ into $A \times X$.

okay thanks for your time. Have you looked at (b)? For me, that was the hardest one — Esteban Cambiasso, Dec 04 '18 at 08:59
@EstebanCambiasso it’s already part of the solution after “ on the other hand”. — Henno Brandsma, Dec 04 '18 at 09:13

Help with product maps in topological spaces

1 Answers1