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I want to prove the following theorem:

If $f:X\rightarrow X'$ and $g:Y\rightarrow Y'$ are continuous functions between topological spaces, then the mapping between product spaces $$f\times g:X\times Y\rightarrow X'\times Y', (x,y)\mapsto(f(x),g(y)) $$ is continuous.


I am using the theorem written below:

Theorem. Let $X, Y$ be topological spaces and $X\times Y$ their product space. If $Z$ is a topological space and $f:Z\rightarrow X\times Y$ a mapping, then $f$ is continuous iff $p\circ f, q\circ f$ are continous, where $p:X\times Y\rightarrow X, q:X\times Y\rightarrow Y$ are projections.


Assume that $f:X\rightarrow X'$ and $g:Y\rightarrow Y'$ are continuous functions between topological spaces.

Let $p':X'\times Y'\rightarrow X',q':X'\times Y'\rightarrow Y'$.

Then let's assume that $p'\circ f\times g, q'\circ f\times g$ are continous. Let $W\subseteq X'\times Y'$ be open. Then there exists open sets $U_{i}\subseteq X'$ and $V_{i}\subseteq Y'$ $(i\in I)$ such that $U_i$ is open in $X'$ and $V_{i}$ is open in $Y'$ by every $i\in I$ and also $W=\bigcup_{i\in I} U_i\times V_i$.

Because $$(f\times g)^{-1}(W)=(f\times g)^{-1}\bigg(\bigcup_{i\in I} U_i\times V_i\bigg)=\bigcup_{i\in I}(f\times g)^{-1}(U_{i}\times V_{i}), $$ it's enough to show that $(f\times g)^{-1}(U_{i}\times V_{i})$ is open by every $i\in I$.

Now, $U_{i}\times V_{i} = (U_{i}\times Y')\cap (X'\times V_{i})=p'^{-1}(U_i)\cap q'^{-1}(V_i)$.

Then, $$\begin{align*}(f\times g)^{-1}(U_{i}\times V_{i})&=(f\times g)^{-1}(p'^{-1}(U_i)\cap q'^{-1}(V_i))\\ &=(p'\circ f\times g)^{-1}(U_{i})\cap(q'\circ f\times g)^{-1}(V_i). \end{align*}$$ Now, there are also projections $p:X\times Y\rightarrow X, q:X\times Y\rightarrow Y$. Because functions $f,g$ are continuous, then $f\circ p, g\circ q$ are continuous.

And forward, because $f\times g\circ p' = f\circ p$ and $f\times g\circ q' = g\circ q$ we can continue $$(p'\circ f\times g)^{-1}(U_{i})\cap (q'\circ f\times g)^{-1}(V_i)=(f\circ p)^{-1}(U_i)\cap (g\circ q)^{-1}(V_{i})$$ which is open.


Fixed.

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  • $\begingroup$ Though, I'm not sure where you get in the third line $(p\circ f\times g)^{-1}[U] = f^{-1}(p^{-1}(U_i) \cap g^{-1}(p^{-1}(U_i) $ since $p^{-1}[U_i] \in X'\times Y'$ and so $g^{-1}[p^{-1}[U_i]]$ is not even defined. $\endgroup$
    – sqtrat
    Feb 26, 2016 at 12:18
  • $\begingroup$ There are typos. Oops. $\endgroup$
    – Zzz
    Feb 26, 2016 at 12:19
  • $\begingroup$ You should replace it by $(p\circ f\times g)^{-1}[U] = (f \circ p)^{-1}[U_i]$ using the hint in answer below, where the second $p$ is the projection from $X\times Y \rightarrow X$.( The first $p$ is the projection from $X'\times Y' \rightarrow X'$.) $\endgroup$
    – sqtrat
    Feb 26, 2016 at 12:21
  • $\begingroup$ Alright, I added new projections $p,q$ and replaced old $p,q$ by $p',q'$. I had to draw a diagram to figure your point! :) $\endgroup$
    – Zzz
    Feb 26, 2016 at 12:44
  • $\begingroup$ Diagrams always help, glad I could help. $\endgroup$
    – sqtrat
    Feb 26, 2016 at 12:46

2 Answers 2

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Hint: There is an easier way, if $\pi_i:X_1 \times X_2 \rightarrow X_i $ and $p_i:X'_1 \times X'_2 \rightarrow X'_i $ are the projections and $f_i: X_i \rightarrow X'_i$ are continuous for $i=1,2$, then simply show that $p'_i\circ (f\times g) = f_i \circ \pi_i$ which is continuous by assumption.

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This solution seems to simple to be correct (but I can't find a mistake):

Take a basic open set in $X'\times Y'$, call it $U\times V$. It's enough to show that $(f\times g)^{-1}(U\times V)$ is open. But $$(f\times g)^{-1}(U\times V)=\{(x,y)\mid (f(x),g(y))\in U\times V\}$$ $$= \{(x,y)\mid f(x)\in U\}\cap \{(x,y)\mid g(y)\in V\}$$ $$= f^{-1}(U)\times Y \bigcap X\times g^{-1}(V),$$ which is the intersection of two open sets and therefore open.

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  • $\begingroup$ @sqtrat Maybe you can help me check its correctness? $\endgroup$
    – JKEG
    Feb 21, 2019 at 14:43

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