# 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 |