# product of path connected spaces is path connected

. 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

 $\sigma(0)=x_{1}\ \ \mathrm{and}\ \ \sigma(1)=x_{2}.$

Analogously there exists a continous map

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

such that

 $\tau(0)=y_{1}\ \ \mathrm{and}\ \ \tau(1)=y_{2}.$

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

 $\sigma(0)=(x_{1},y_{0})\ \ \mathrm{and}\ \ \sigma(1)=(x_{1},y_{0}).$

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. $\square$

Title product of path connected spaces is path connected ProductOfPathConnectedSpacesIsPathConnected 2013-03-22 18:31:02 2013-03-22 18:31:02 joking (16130) joking (16130) 4 joking (16130) Theorem msc 54D05