The cartesian product of contractible spaces is contractible

Question

Let $X_i$, $i\in I$ be contractible spaces. Is the Cartesian product $\prod_iX_i$ contractible, too?

Answer 1

Since $X_i$ is contractible, there exists a continuous map $h_i\colon X_i\times [0,1]\to X_i$ such that $h_i(x,0)=x$ and $h_i(x,1)=x^0_i$ for some $x^0_i\in X_i$ for all $x\in X_i$. Define $$X = \prod_i X_i,\quad\text{and}\quad H\colon X\times [0,1]\to X,\ H\left((x_i)_i,t\right) = (h_i(x_i,t))_i.$$ This function is continuous. Moreover, $$H((x_i)_i,0)=(h_i(x_i,0))_i = (x_i)_i$$ and $$H((x_i)_i,1)=(h_i(x_i,1))_i = (x^0_i)_i,$$ therefore $H$ is a homotopy joining identity to a constant map.

Proof of the continuity For $i\in I$ let $\pi_i\colon X\to X_i$ be a projection. Then $$\pi_i\circ H = h_i\circ (\pi_i\times \mathrm{id}_{[0,1]}).$$ Therefore $\pi_i\circ H$ is continuous for all $i\in I$.