Contractible Spaces and Homotopy Classes (9.51.3)

James Munkres Topology, chapter 9.51, exercise 3:

A space $X$ is said to be contractible if the identity map $i : X→X$ is nulhomotopic.
(a) Show that $I$ and $ℝ$ are contractible.
(b) Show that contractible spaces are path connected.
(c) Show that if $Y$ is contractible, then for any $X$, the set $[X,Y]$ has a single element.
(d) Show that if $X$ is contractible and $Y$ is path connected, then $[X,Y]$ has a single element.

Proof: (a) Since $I$ and $ℝ$ are both convex, the linear homotopies suffice, for $z : I→\mathcal{C}(ℝ,ℝ)$ (say) given by $(1-t)·f(x)+t·g(x)$ is a path in $\mathcal{C}(ℝ,ℝ)$ in the compact-open topology between arbitrary $f,g∈\mathcal{C}(ℝ,ℝ)$.

(b) Let $X$ be path connected, and let $x,y∈X$. If $G : X×I→X$ is a nulhomotopy to the constant map onto $e$, then $p_1 : I→X$ given by $p_1(t)=G(x,t)$ is a path from $p_1(0)=x$ to $p_1(1)=e$. Similarly there is a path from $y$ to $e$, so that $x$ and $y$ are connected by a path.

(c) First, let $Y$ simply be path connected, let $y,e∈Y$ be arbitrary, and let $p$ be a path from $y$ to $e$. Then $P : Y×I→Y$ given by $P(x,t)=p(t)$ is a homotopy from the constant map onto $y$ to the constant map onto $e$, so that all constant maps into a path connected space are homotopic. Hence, when $Y$ is contractible, it suffices to show that an arbitrary map $f : X→Y$ is homotopic to a constant map; and indeed, if $G$ is a nulhomotopy in $Y$ onto a constant map $x↦e$, then $F : X×I→Y$ given by $F(x,t)=G(f(x),t)$ is continuous such that $F(x,0)=G(f(x),0)=f(x)$ and $F(x,1)=G(f(x),1)=e$.

(d) As we saw, all constant maps $X→Y$ are homotopic, so it suffices to show an arbitrary map $f : X→Y$ is homotopic to a constant map. Let $G$ be a nulhomotopy in $X$ onto a constant map $x↦e$; then $F : X×I→Y$ given by $F(x,t)=f(G(x,t))$ is continuous such that $F(x,0)=f(G(x,0))=f(x)$ and $F(x,1)=f(G(x,1))=f(e)$ is constant.$~\square$