# Proving $f$ is continuous if $f_1$ and $f_2$ are continuous and the reciprocal

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.

Questions:

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 $$f_2=p_2\circ f$$.

• $X_1 \times a_2$ is sloppy notation. 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 $$i(x)=(a,x)$$.