product of path connected spaces is path connected

Proposition. Let $X$ and $Y$ be topological spaces. Then $X\times Y$ is path connected if and only if both $X$ and $Y$ are path connected.

Proof. ”$\Leftarrow$” Assume that $X$ and $Y$ are path connected and let $(x_1,y_1),(x_2,y_2)\in X\times Y$ be arbitrary points. Since $X$ is path connected, then there exists a continous map

$$\sigma :\mathrm{I}\to X$$

such that

$$\sigma (0)=x_1\mathit{\hspace{1em}}\mathrm{and}\mathit{\hspace{1em}}\sigma (1)=x_2.$$

Analogously there exists a continous map

$$\tau :\mathrm{I}\to Y$$

such that

$$\tau (0)=y_1\mathit{\hspace{1em}}\mathrm{and}\mathit{\hspace{1em}}\tau (1)=y_2.$$

Then we have an induced map

$$\sigma \times \tau :\mathrm{I}\to X\times Y$$

defined by the formula:

$$(\sigma \times \tau )(t)=(\sigma (t),\tau (t)),$$

which is continous path from $(x_1,y_1)$ to $(x_2,y_2)$.

”$\Rightarrow$” On the other hand assume that $X\times Y$ is path connected. Let $x_1,x_2\in X$ and $y_0\in Y$. Then there exists a path

$$\sigma :\mathrm{I}\to X\times Y$$

such that

$$\sigma (0)=(x_1,y_0)\mathit{\hspace{1em}}\mathrm{and}\mathit{\hspace{1em}}\sigma (1)=(x_1,y_0).$$

We also have the projection map $\pi :X\times Y\to X$ such that $\pi (x,y)=x$. Thus we have a map

$$\tau :\mathrm{I}\to X$$

defined by the formula

$$\tau (t)=\pi (\sigma (t)).$$

This is a continous path from $x_1$ to $x_2$, therefore $X$ is path connected. Analogously $Y$ is path connected. $\mathrm{\square }$

Title	product of path connected spaces is path connected
Canonical name	ProductOfPathConnectedSpacesIsPathConnected
Date of creation	2013-03-22 18:31:02
Last modified on	2013-03-22 18:31:02
Owner	joking (16130)
Last modified by	joking (16130)
Numerical id	4
Author	joking (16130)
Entry type	Theorem
Classification	msc 54D05