Contractibility is product-closed

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.

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:

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