# homotopy with a contractible domain

Theorem. Assume that $Y$ is an arbitrary topological space^{} and $X$ is a contractible^{} topological space. Then all maps $f:X\to Y$ are homotopic^{} if and only if $Y$ is path connected.

Proof: Assume that all maps are homotopic. In particular constant maps are homotopic, so if ${y}_{1},{y}_{2}\in Y$, then there exists a continous map $H:I\times Y\to Y$ such that $H(0,y)={y}_{1}$ and $H(1,y)={y}_{2}$ for all $y\in Y$. Thus the map $\alpha :I\to Y$ defined by the formula^{} $\alpha (t)=H(t,{y}_{0})$ for a fixed ${y}_{0}\in Y$ is the wanted path.

On the other hand assume that $Y$ is path connected. Since $X$ is contractible, then for any $c\in X$ there exists a continous homotopy^{} $H:I\times X\to X$ connecting the identity map and a constant map $c$. Let $f:X\to Y$ be an arbitrary map. Define a map $F:I\times X\to Y$ by the formula: $F(t,x)=f(H(t,x))$. This map is a homotopy from $f$ to a constant map $f(c)$. Thus every map is homotopic to some constant map.

The space $Y$ is path connected, so for all ${y}_{1},{y}_{2}\in Y$ there exists a path $\alpha :I\to Y$ from ${y}_{1}$ to ${y}_{2}$. Therefore constant maps are homotopic via the homotopy $H(t,x)=\alpha (t)$.

Finaly for any continous maps $f,g:X\to Y$ and any point $c\in X$ we get:

$$f\simeq f(c)\simeq g(c)\simeq g,$$ |

which completes^{} the proof. $\mathrm{\square}$

Corollary. If $X$ is a contractible space, then for any topological space $Y$ there exists a bijection between the set $[X,Y]$ of homotopy classes of maps from $X$ to $Y$ and the set ${\pi}_{0}(Y)$ of path components of $Y$.

Proof: Assume that $Y=\bigcup {Y}_{i}$, where ${Y}_{i}$ are path components of $Y$. It is well known that contractible spaces are path connected, thus the image of any continous map $f:X\to Y$ is contained in ${Y}_{i}$ for some $i$. It follows from the theorem that two maps from $X$ to $Y$ are homotopic if and only if their images are contained in the same ${Y}_{i}$. Thus we have a well defined, injective map

$$\psi :[X,Y]\to {\pi}_{0}(Y)$$ |

$$\psi ([f])={Y}_{i},$$ |

where $i$ is such that $f(X)\subseteq {Y}_{i}$. This map is also surjective^{}, since for any $i$ there exists $y\in {Y}_{i}$, so the class of the constant map $f(x)=y$ is mapped into ${Y}_{i}$. $\mathrm{\square}$

Title | homotopy with a contractible domain |
---|---|

Canonical name | HomotopyWithAContractibleDomain |

Date of creation | 2013-03-22 18:02:12 |

Last modified on | 2013-03-22 18:02:12 |

Owner | joking (16130) |

Last modified by | joking (16130) |

Numerical id | 26 |

Author | joking (16130) |

Entry type | Theorem |

Classification | msc 55P99 |

Related topic | homotopy |

Related topic | contractible |