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.