# Continuity of cartesian product of functions between topological spaces

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.

• 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. Feb 26, 2016 at 12:18
• There are typos. Oops.
– Zzz
Feb 26, 2016 at 12:19
• 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 $X'\times Y' \rightarrow X'$.) Feb 26, 2016 at 12:21
• 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! :)
– Zzz
Feb 26, 2016 at 12:44
• Diagrams always help, glad I could help. Feb 26, 2016 at 12:46

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.
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.