Contractibility is product-closed

This article gives the statement, and possibly proof, of a topological space property (i.e., contractible space) satisfying a topological space metaproperty (i.e., product-closed property of topological spaces)
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Statement

Property-theoretic statement

The property of topological spaces of being a contractible space, satisfies the metaproperty of topological spaces of being product-closed.

Statement with symbols

Let $X_{i}$ , $i \in I$ , be an indexed family of topological spaces. Then the product space, endowed with the product topology, is contractible.

Proof

Key idea (for two spaces)

Suppose $F : X \times I \to X$ and $G : Y \times I \to Y$ are contracting homotopies for $X$ and $Y$ . Then the map $F \times G$ defined as:

$(F \times G) (x, y, t) = (F (x, t), G (y, t))$

is a contracting homotopy for $X \times Y$ .

Thus $X \times Y$ is contractible.

Generic proof (for an arbitrary family)

Given: An indexing set $I$ , a collection ${X_{i}}_{i \in I}$ of contractible spaces. $X$ is the product of the $X_{i}$ s, endowed with the product topology

To prove: $X$ is a contractible space

Proof: Since each $X_{i}$ is contractible, we can choose, for each $X_{i}$ , a point $p_{i} \in X_{i}$ , and a contracting homotopy $F_{i} : X_{i} \times [0, 1] \to X_{i}$ , with the property that:

$F_{i} (a, 0) = a \forall a \in X_{i}, F_{i} (a, 1) = p_{i} \forall a \in X_{i}$

Now consider the point $p \in X$ whose $i^{t h}$ coordinate is $p_{i}$ for each $i \in I$ . We denote:

$x = (x_{i})_{i \in I}$

to be a point whose $i^{t h}$ coordinate is $x_{i}$ . Then, define a homotopy:

$F : X \times [0, 1] \to X$

given by:

$F (x, t) = (F_{i} (x_{i}, t))_{i \in I}$

In other words, the homotopy acts as $F_{i}$ in each coordinate. We observe that:

Since $F_{i} (x_{i}, 0) = x_{i}$ for each $i$ , $F (x, 0) = x$
Since $F_{i} (x_{i}, 1) = p_{i}$ for each $i$ , $F (x, 1) = p$
$F$ is a continuous map: Fill this in later

Thus, $F$ is a contracting homotopy on $X$ , so $X$ is contractible.