For $i=1,2$ let $(X_i,\tau_i)$ and $(X'_i,\tau_i)$ be topological spaces, and $f_i: X_i\to X'_i$ functions. Consider the function $f:X_1\times X_2\to X'_1\times X'_2$ defined by $f(x_1,x_2)=(f_1(x_1),f_2(x_2))$. Show that if $f_1,f_2$ are continuous then $f$ is continuous for the product topologies. Prove the reciprocal.

I tried to prove the statement in the following way:

$\Rightarrow$ Let $\mathscr{U}$ be an open set such that by definition of product topology $\mathscr{U}=A\times B$. As $f_1$ and $f_2$ are continuous functions, then $f_1^{-1}(A)\in\tau'_1$ and $f_2^{-1}(B)\in\tau'_2$.

So $f^{-1}(\mathscr{U})=f^{-1}(A\times B)=f_1^{-1}(A)\times f_2^{-1}(B)$so that by definition of open sets in the product topology are open hence $f_1^{-1}(A)\times f_2^{-1}(B)\in\tau'_i$.

$\Leftarrow$ Let's define the following functions $I_1: (X'_i,\tau'_i)\to (X_1\times a_2)$ and $I_2: (X'_i,\tau'_i)\to (a_2\times X_2)$ such that $(a_1,a_2)\in X'_i $ By known theorem $I_1$ and $I_2$ are homeomorphisms. Using the $p_1$ and $p_2$ as the projection functions. $f_1=p_1\circ f\circ I_1$ and $f_2=p_2\circ f\circ I_2$. As the composition of continuous functions is continuous hence $f_1$ and $f_2$ are continuous.


1) Is my proof right? Can I give the following step $f^{-1}(A\times B)=f_1^{-1}(A)\times f_2^{-1}(B)$?

2) I was advised to define $I_1$ and $I_2$. Why are the projections not enough? I mean $f_1=p_1\circ f$ and $f_2=p_2\circ f$.

Thanks in advance!

  • $\begingroup$ You need a small argument for the step in 1, plus some extra reasoning. See below. $\endgroup$ Jan 20 '19 at 13:18
  • $\begingroup$ $X_1 \times a_2$ is sloppy notation. $\endgroup$ Jan 20 '19 at 13:19

Firstly, an open set in the product topology is not always of the form $U \times V$, it is a union of such open sets, as these form a base for the product topology. So if $O$ is product open and so $O= \bigcup_{i \in I} (U_i \times V_i)$ we indeed have that $$f^{-1}[U_i \times V_i] = \{(x,x'): f(x,x') \in U_i \times V_i\} =\ \{(x,x'): (f_1(x), f_2(x') \in U_i \times V_i\} = \{(x,x'): f_1(x) \in U_i \text{ and } f_2(x') \in V_i\} =\ \{(x,x'): x\in f_1^{-1}[U_i] \text{ and } x' \in f_2^{-1}[V_i]\} =f_1^{-1}[U_i] \times f_2^{-1}[V_i]$$ which is basic product open, and so $f^{-1}[O] = \bigcup_i f^{-1}[U_i \times V_i]$ is a union of open sets and hence open.

The reverse needs the embeddings, as $\pi_1 \circ f$ is a map defined on $X_1 \times X_2$ and not on $X_1$, so is not equal to $f_1$. In order to use $f$ we need two arguments, so fix $a \in X_1$ and $a' \in X_2$ and define $j: X_1 \to X_1 \times X_2 $ by $j(x)=(x,a')$. It's continuous because the composition of $j$ with the two projections are either the identity on $X_1$ or a constant map, so continuous always. And then $(\pi_1 \circ f \circ j)(x)=\pi_1(f(x,a'))=\pi_1(f_1(x), f_2(a'))= f_1(x)$ for all $x\in X_1$ so $f_1$ is then a composition of three continuous maps, hence continuous.

$f_2$ is similarly continuous using $i: X_1 \to X_1 \times X_2$ defined by $i(x)=(a,x)$.


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.