Section 2.6 Products
The definition of a product of sets is engineered precisely so that a map into the product is determined by each of its components. More precisely, for any indexed family of sets \(\{X_a\}_{a\in A}\text{,}\) the product
is a set equipped with projection maps
one for every \(b\in A\text{,}\) which together determine the maps into \(\prod_{a\in A}X_a\text{.}\) That is, composition with the projections induce a bijection
If the sets \(X_a\) are all endowed with topologies, then we would like to topologise the product \(\prod_{a\in A}X_a\) so that the same thing will happen with continuous maps. That is, we would like to ensure that a map \(f\colon\fromto{Y}{\prod_{a\in A}X_a}\) is continuous if and only if, for every \(a\in A\text{,}\) the composite map \(\pi_a\circ f\colon\fromto{Y}{X_a}\) is continuous. But we know just how to arrange that:
Definition 2.6.1.
Let \(\{(X_a,\tau_a)\}_{a\in A}\) be a family of topological spaces. Then the product topology \(\prod_{a\in A}\tau_a\) on \(\prod_{a\in A}X_a\) is the initial topology with respect to the set \(\{\pi_a\}_{\alpha\in A}\text{.}\)
The definition of the product topology is engineered precisely so that a continuous map into the product is determined by each of its continuous components. More precisely, for any indexed family of topological spaces \(\{X_a\}_{a\in A}\text{,}\) the product
is a set equipped with continuous projection maps
one for every \(b\in A\text{,}\) which together determine the maps into \(\prod_{a\in A}X_a\text{.}\) That is, composition with the projections induce a bijection
Example 2.6.2.
In particular, for any set \(A\) and any topological space \(X\text{,}\) we may contemplate the product topology on the set \(\Map(A,X)\text{.}\) We write in particular
Proposition 2.6.3.
For any \(n\geq 0\text{,}\) the product topology on \(\RR^n\) coincides with the standard topology (from the Euclidean metric).
Proof.
The product topology on \(\RR^n\) is generated by sets of the form \(\RR\times\cdots\times \left]a,b\right[\times\RR\times\cdots\times\RR\text{.}\) Consequently, the sets of the form \(C(x,\varepsilon) \coloneq \prod_{i=1}^n\left]x_i-\varepsilon,x_i+\varepsilon\right[\) are a base for the product topology. Now for any \(x \in \RR^n\text{,}\) and for any \(\varepsilon>0\text{,}\) the exists a \(\delta>0\) such that
and
Since the subsets \(C(x, \varepsilon)\) generate the product topology, and the subsets \(B(x,\varepsilon)\) generate the standard topology, the two topologies coincide.
Warning 2.6.4.
Consider the product topological space
This topology has a base
Consequently, if \(U\) is an open subset of \(\RR^{\omega}\text{,}\) then for any point \(x \in U\text{,}\) there exists \(N \in \NN\) such that any point of the form
also lies in \(U\text{.}\) Conequently, the subset
is not open!
Example 2.6.5.
If \(S_1,\dots,S_n\) is a finite collection of discrete topological spaces, then the product
is also discrete. Indeed, for any \(i\) and any \(x_i \in S_i\text{,}\) the set \(S_1 \times \cdots \times S_{i-1} \times \{x_i\} \times S_{i+1} \times \cdots \times S_n\) is open, and since finite intersections of opens are open, it follows that any singleton \(\{(x_1, \dots, x_n)\}\) is open as well.
Warning 2.6.6.
If \(\{S_i\}_{i\in\NN}\) is a countable family of discrete finite sets of cardinality at least \(2\text{,}\) the product \(S \coloneq \prod_{i \in \NN} S_i\) is not discrete: if \((x_0, x_1, \dots) \in S\) is a point, then the open sets are unions of sets of the form
so that no singleton is open. In fact, it is always homeomorphic to our old pal the Cantor space \(C\) — irrespective of which finite sets \(S_i\) are chosen!
There is a relative form of the product as well.
Definition 2.6.7.
Let \(U\text{,}\) \(V\text{,}\) and \(X\) be three topological spaces, and let \(f\colon\fromto{U}{X}\) and \(g\colon\fromto{V}{X}\) two continuous maps. Then the fiber product
\(U\times_XV\) is the subspace
of \(U\times V\text{.}\) In this case, the square
is sometimes called a pullback square.
Example 2.6.8.
As a special case of this construction, if \(V\) is a subspace of \(X\) and if \(g\) is the inclusion map \(V \inclusion X\text{,}\) then the fiber product \(U \times_X V\) is homeomorphic to the subspace \(f^{-1}(V) \subseteq X\text{.}\) In particular, if \(V = \{x\}\) for some \(x\in X\text{,}\) then the fiber product \(U \times_X V\) is the fiber \(f^{-1}\{x\}\text{.}\)
Example 2.6.9.
Let \(f \colon X \to Y\) be a continuous map of topological spaces. Then the graph of \(f\) is the fiber product
The projection map \(\pr_1 \colon \Gamma(f) \to X\) is a homeomorphism.