Can anything be said about continuous functions whose domain is an infinite product of topological spaces?

Question

Based on my readings much can be said about functions that map into the product topology, but can anything be said about functions that map from the product topology? Do they preserve continuity?

Answer 1

Even before we ask about continuity, we have to ask about function formation in the first place. Here we run into trouble: there is in general no way to combine functions from a bunch of factor spaces into some common space into a single function from the product space into that same common space. And this is just a fact about sets:

There is no obvious way in general to combine a map $f: A_1\rightarrow B$ and a map $g: A_2\rightarrow B$ into a map from $A_1\times A_2$ to $B$.

(By contrast, the "into the product" version is easy: given $f:B\rightarrow A_1$ and $g:B\rightarrow A_2$, consider the map $B\rightarrow A_1\times A_2: b\mapsto (f(b), g(b))$.)

Instead, what we get is a map from the (disjoint) union of $A_1$ and $A_2$ to $B$, given by

$$h(x)=\begin{cases} f(x) & \mbox{ if }x\in A_1,\\ g(x) & \mbox{ if }x\in A_2.\\ \end{cases}$$

(If $A_1$ and $A_2$ aren't disjoint, we need to "disjointify" them - e.g. replace $A_1$ by $A_1\times\{1\}$ and $A_2$ by $A_2\times\{2\}$.)

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This may seem weirdly asymmetric, but in fact there is a symmetry here, it's just harder to see. The operations of disjoint union and Cartesian product are in a sense "dual" to each other:

• The key feature of the Cartesian product $S\times T$ of sets $S$ and $T$ is that there are maps $\pi_S:S\times T\rightarrow S$ and $\pi_T: S\times T\rightarrow T$ (the coordinate projection maps) such that for every set $U$ and every pair of maps $f_S:U\rightarrow S$ and $f_T: U\rightarrow T$, there is exactly one map $h: U\rightarrow S\times T$ such that $f_S=\pi_S\circ h$ and $f_T=\pi_T\circ h$. Specifically, $h$ is just $$u\mapsto (f_S(u), f_T(u)).$$ In a sense, the Cartesian product of two sets maps to both the factor sets in a "universal" way.

• The dual situation holds for disjoint unions! Specifically, the key feature of the disjoint union $S\sqcup T$ is that we have maps $\iota_S: S\rightarrow S\sqcup T$ and $\iota_T: T\rightarrow S\sqcup T$ (the inclusion maps) such that for any other set $U$ and any pair of maps $f_S: S\rightarrow U$ and $f_T: T\rightarrow U$ there is exactly one map $h: S\sqcup T\rightarrow U$ such that $f_S=h\circ \iota_S$ and $f_T=h\circ \iota_T$. Specifically, this is the map $$x\mapsto\begin{cases} f_S(x) & \mbox{ if } x\in S,\\ f_T(x) & \mbox{ if } x\in T.\\ \end{cases}$$

  Why is this "dual" to the picture for the Cartesian product? Well, first of all note the arrow reversal: the $\pi$-maps were from the product to the factors, while the $\iota$-maps are from the "base sets" to their union, and we're considering maps from those "base sets" rather than to those base sets - and correspondingly, we have "mirrored compositions" ($\pi_S\circ h$ and $\pi_T\circ h$ versus $h\circ\iota_S$ and $h\circ\iota_T$). Basically, when we think in terms of abstract behavior of functions, the disjoint union is a "flipped around" version of the Cartesian product - even though they are in many ways totally different (e.g. consider their respective cardinalities)!

This may feel weird and wishy-washy at first, but in fact it's a first step along the road to category theory. See here for some more details.

(And note that none of the above is about topological spaces specifically - already things are complicated just considering sets on their own!)