Contractibility is product-closed

This article gives the statement, and possibly proof, of a topological space property satisfying a topological space metaproperty
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Statement

For two spaces

Let $X$ and $Y$ be contractible spaces. Then the product space $X\times Y$ is contractible.

For an arbitrary family of spaces

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. $P$ is the product of the $X_i$s, endowed with the product topology

To prove: $P$ 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:

$F_i(a,0)=a\mathrm{\forall }a\in X_i,F_i(a,1)=p_i\mathrm{\forall }a\in X_i$

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.