# (path) connectness as a homotopy invariant

Theorem. Let $X$ and $Y$ be arbitrary topological spaces^{} with $Y$ (path) connected^{}. If there are maps $f:X\to Y$ and $g:Y\to X$ such that $g\circ f:X\to X$ is homotopic^{} to the identity map, then $X$ is (path) connected.

Proof: Let $f:X\to Y$ and $g:Y\to X$ be maps satisfying theorem’s assumption. Furthermore let $X=\bigcup {X}_{i}$ be a decomposition of $X$ into (path) connected components^{}. Since $Y$ is (path) connected, then $g(Y)\subseteq {X}_{i}$ for some $i$. Thus $(g\circ f)(X)\subseteq {X}_{i}$. Now let $H:I\times X\to X$ be the homotopy^{} from $g\circ f$ to the identity map. Let ${\alpha}_{x}:I\to X$ be a path defined by the formula: ${\alpha}_{x}(t)=H(t,x)$. Since for all $x\in X$ we have ${\alpha}_{x}(0)\in {X}_{i}$ and $I$ is path connected, then ${\alpha}_{x}(I)\subseteq {X}_{i}$. Therefore $H(I\times X)\subseteq {X}_{i}$, but $H(\{1\}\times X)=X$ which implies that ${X}_{i}=X$, so $X$ is (path) connected. $\mathrm{\square}$

Straightforward application of this theorem is following:

Corollary. Let $X$ and $Y$ be homotopy equivalent spaces. Then $X$ is (path) connected if and only if $Y$ is (path) connected.

Title | (path) connectness as a homotopy invariant |
---|---|

Canonical name | pathConnectnessAsAHomotopyInvariant |

Date of creation | 2013-03-22 18:02:15 |

Last modified on | 2013-03-22 18:02:15 |

Owner | joking (16130) |

Last modified by | joking (16130) |

Numerical id | 8 |

Author | joking (16130) |

Entry type | Theorem |

Classification | msc 55P10 |

Related topic | Homotopy |

Related topic | homotopyequivalence |

Related topic | path |

Related topic | connectedspace |