$F^E$ is isomorphic to an $E$-indexed product space.

Question

In Jean Gallier "Aspects of harmonic Analysis and representation theory." on page 21:

In this section we study the space of functions $f : E → F$, where $E$ and $F$ are arbitrary topological spaces. We denote the set of all functions from $E$ to $F$ by $F^E$.

Our first goal is to make $F^E$ into a topological space in its own right. Surprisingly, one of the easiest ways to describe a topology on $F^E$ is to follow Tychonoff and observe that $$F^E\cong \prod_{x\in E}F_x, F_x=F.$$ Since $F^E$ is isomorphic to an $E$-indexed product space, we may give it a product topology as follows:...

Question 1: What does $F_x$ mean here? Does it mean projection? So $F_x : F×E\rightarrow F$. Does it correct?

Question 2: Also I don't understand what does he mean by $F^E\cong \prod_{x\in E}F_x, F_x=F.$ All $F_x$'s are projections onto $F$ space. How their product can be the $F^E$ space? Or even what is the meaning of such product? What do I not understand?

Question 3: I know what "isomorphic" means. But I don't understand what does he want to say by "Since $F^E$ is isomorphic to an $E$-indexed product space" preciesly. From Munkres, Topology, Chapter 15, "The Product Topology" I know something about this topology. But it does not help me here.

Please give me some clarifications.

Answer 1

Using the standard construction of Cartesian products, we even have $$F^E = \prod_{x\in E}F_x$$

Here $(F_x)_{x \in E}$ is the indexed collection of sets where all $F_x=F$. See for example here.

Given any indexed collection $(X_\alpha)_{\alpha \in J}$ of sets, one defines $$\prod_{\alpha \in J}X_\alpha = \{ \phi : J \to \bigcup_{\alpha \in J} X_\alpha \mid \phi(\alpha) \in X_\alpha \text{ for all } \alpha \in J \} .$$ See Munkres p. 113.

In our case we have $\bigcup_{x \in E} F_x = F$ and the condition $\phi(x) \in F_x = F$ is satisfied for all $\phi : E \to F$. Therefore $\prod_{x\in E}F_x$ is simply the set of function $F^E$.

Anyway, it does not matter whether $$F^E = \prod_{x\in E}F_x \tag{1}$$ or only $$F^E \cong \prod_{x\in E}F_x \tag{2}$$ where $\cong$ means that there exist a canonical bijection $b : F^E \to \prod_{x\in E}F_x$. The set $\prod_{x\in E}F_x$ can be given the product topology and thereby also $F^E$ gets a topology. This is trivial in case $(1)$. In case $(2)$ we use $b$ to induce a topology on $F^E$: Simply take all sets $b^{-1}(U)$, where $U$ is open in the product topology.

Let us finally try to understand the meaning of "canonical bijection" in $(2)$. There may be other constructions of Cartesian products, for example $\prod_{i \in \{1,\ldots,n\}} X_i = X_1 \times X_2 \times \ldots \times X_n = \{(x_1,x_2,\ldots,x_n) \mid x_i \in X_i \}$. Here $n$-tuples may be introduced as here and here. Working with alternative constructions prevents us from saying that $F^E = \prod_{x\in E}F_x$. However, all possible constructions are designed to give the following universal property in this abstract definition:

Given an indexed family of sets $\mathscr X = (X_\alpha)_{\alpha \in J}$, a Cartesian product of $\mathscr X$ is a system $(P,(p_\alpha)_{\alpha \in J})$ consisting of a set $P$ and functions $p_{\alpha} : P \to X_\alpha$, $\alpha \in J$ such that for any collection of functions $f_{\alpha} : Y \to X_\alpha$, $\alpha \in J$, having the same domain $Y$ there exists a unique function $f :Y \to P$ such that $f_{\alpha} = p_{\alpha} \circ f$ for all $\alpha \in J$.

It is now an easy exercise to show that if $(P',(p'_\alpha)_{\alpha \in J})$ is another Cartesian product of $\mathscr X$, then there exists a unique bijection $b : P \to P'$ such that $p_\alpha = p'_\alpha \circ b$ for all $\alpha \in J$.

Our above standard construction yields a Cartesian product in this sense. Simply define $p_\alpha(\phi) = \phi(\alpha)$. Now you should see what "canonical bijection" in $(2)$ means.

Answer 2

As is written, $F_x=F$.
Then, $\prod_{x\in E}F_x=\prod_{x\in E}F$ is the set of '$E$-indiced sequences' $(f_x)_{x\in E}$ of elements $f_x$ of $F$ .

In other words, these are just the functions $f:E\to F$ (arbitrary functions, not necessarily continuous).
The sequence $(f_x)_{x\in E}$ corresponds to the function $x\mapsto f_x$ and vice-versa.

So this way, as sets, $\prod_{x\in E}F_x\cong F^E$.

Now we can utilize the topology of $F$ to induce a topology on $\prod_{x\in E}F$, hence on $F^E$ by the above bijection.

Answer 3

If $(F_x)_{x \in E}$ is a family of sets, then $\prod_{x \in E} F_x$ is, by definition, the set of maps $f$ with domain $E$ such that for every $x \in E$, $f(x) \in F_x$.

Therefore, $\prod_{x\in E} F = F^E$.

So the open sets for the product topology are unions of intersections of sets of the form $V_{e,U} := \{f \in F^E \ \vert \ f(e) \in U\}$, with $e \in E$ and $U$ open in $F$.

Answer 4

I thought that $F^E$ is just notation without any meaning. I mean I thought we can write it as $F_E$ or $F(E)$. But this $E$ appears as power.

If $X=\{1, 2, 3\}$ and $Y=\{1, 2, 3\}$ then the number of functions from X to Y is equal to $3^3$ because there three element in $X$ and 3 elements in $Y$.

If we have two spaces $F$ and $E$ then a natural way to denote the space of all functions from $E$ to $F$ is $F^E$.

It is that I did not understand. Now everithing is ok. I will accept a detailed answer.

I wrote that I am familiar with the concept of the product space from Munkers.