# Product space topology

Many times I've come across the expression:

Let $F:A\times B\to C$ be a continuous map.

I want to understand what it means for a function with a product space domain to be continuous. Assuming a topology on the codomain $\mathcal{O}_C$, (and also topologies on $A$ and $B$) what is the topology of $A\times B$?

I get the feeling that given $A\times B$ and the natural projections $\pi_A:A\times B\to A$, $\pi_B:A\times B\to B$, I need to induce a topology that "just" makes the projections continuous. Seperately, it should look probably something like $$\mathcal{O}_{A\times B}=\{\pi_A^{-1}(C) \ | \ C\in \mathcal{O}_C\}$$ and $$\mathcal{O}_{A\times B}=\{\pi_B^{-1}(\tilde{C}) \ | \ \tilde{C}\in \mathcal{O}_C\}.$$ My question is: how does one optimally choose something between the two above to induce a topology on $A\times B$?

(After I have defined a topology on $A\times B$ then I know what it means for $F$ to be continuous: preimages of open sets of the codomain should be open sets of the domain)

• The product topology is the initial topology with respect to the family of projections onto the factors. For $A\times B$, a basis of the product topology is $\{ U \times V : U \in \mathcal{O}_A,\, V \in \mathcal{O}_B\}$. – Daniel Fischer Dec 30 '17 at 10:55
• @DanielFischer I am sorry but I am a bit new to this. By saying initial, do you somehow refer to initial objects in category theory? If you have the time, please elaborate a bit. – EEEB Dec 30 '17 at 10:57
• No, I'm not referring to initial objects. And I don't see any obvious way to interpret it as such. If $X$ is a set, and $\langle Y_i : i \in I\rangle$ a family of topological spaces, and we have a family of maps $\langle f_i \colon X \to Y_i\rangle$, the initial topology on $X$ with respect to the $f_i$ is the coarsest topology on $X$ making all $f_i$ continuous. It has the universal property that for a topological space $Z$, a map $g \colon Z \to X$ is continuous if and only if $(f_i \circ g)$ is continuous for all $i\in I$. – Daniel Fischer Dec 30 '17 at 11:10
• Thus a subbasis of the initial topology is given by the family $$\{ f_i^{-1}(U) : i \in I, U \in \mathcal{O}_{Y_i}\}\,.$$ – Daniel Fischer Dec 30 '17 at 11:11
• Actually it is an initial object in the category of topologies making $\pi_A, \pi_B$ continuous (which is an ordered set and therefore a category - hence initial means smallest) . It's isomorphic to the opposite of a (small, not full) category of topological spaces – Maxime Ramzi Dec 30 '17 at 11:57

## 1 Answer

Let $T_A$ be the topology on $A$ and let $T_B$ be the topology on $B.$

Let $C=\{U\times V: U\in T_A\land V\in T_B\}.$ Let $W=\{\cup S: S\subset C\}.$

Then $W$ is a topology on $A\times B,$ and $C$ is a base (basis) for $W.$ The product topology on $A\times B$ is defined to be $W.$

$W$ is the weakest topology on $A\times B$ such that the projections $p_A:A\times B\to A$ and $p_B:A\times B\to B$ are continuous, where $p_A(a,b)=a$ and $p_B(a,b)=b.$

(I). Suppose $X$ is a topology on $A\times B$ such that $p_A$ and $p_B$ are continuous with respect to $X.$ For any $U\times V \in C$ (as defined above) we have $p_A^{-1}U=U\times B \in X$ and $p_B^{-1}V=A\times V\in X.$ So $U\times V=(U\times B)\cap (A\times V)\in X.$ Therefore $X\supset C,$ implying $X\supset W.$ That is, $X$ is stronger than or equal to $W.$

(II). If $U\in T_A$ then $p_A^{-1}U=U\times B\in C\subset W$ so $p_A$ is continuous with respect to $W.$ Similarly, $p_B$ is continuous with respect to $W.$

• Reference topic: Tychonoff product topology. – DanielWainfleet Dec 31 '17 at 4:21