# homotopy with a contractible domain

Assume that $Y$ is an arbitrary topological space  and $X$ is a contractible  topological space. Then all maps $f:X\rightarrow Y$ 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 $H:I\times Y\rightarrow Y$ such that $H(0,y)=y_{1}$ and $H(1,y)=y_{2}$ for all $y\in Y$. Thus the map $\alpha:I\rightarrow Y$ 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  $H:I\times X\rightarrow X$ connecting the identity map and a constant map $c$. Let $f:X\rightarrow Y$ be an arbitrary map. Define a map $F:I\times X\rightarrow Y$ 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 $\alpha:I\rightarrow Y$ from $y_{1}$ to $y_{2}$. Therefore constant maps are homotopic via the homotopy $H(t,x)=\alpha(t)$.

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

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

which completes       the proof. $\square$

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 $f:X\rightarrow Y$ 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

 $\psi:[X,Y]\rightarrow\pi_{0}(Y)$
 $\psi([f])=Y_{i},$

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 $Y_{i}$. $\square$

Title homotopy with a contractible domain HomotopyWithAContractibleDomain 2013-03-22 18:02:12 2013-03-22 18:02:12 joking (16130) joking (16130) 26 joking (16130) Theorem msc 55P99 homotopy contractible