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 $i^{th}$ coordinate is $p_i$ for each $i \in I$. We denote:

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

to be a point whose $i^{th}$ 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.