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 $Y_i$. $\mathrm{\square }$