Exercise 51.2

Question

Given spaces $X$ and $Y$, let $[X,Y]$ denote the set of homotopy classes of maps of $X$ into $Y$.
\begin{enumerate}[label=(\alph*)]
\item Let $I = [0,1]$. Show that for any $X$, the set $[X,I]$ has a single element.
\item Show that if $Y$ is path connected, the set $[I,Y]$ has a single element.
\end{enumerate}

Amit Awasthi · Accepted Answer

\begin{proof}[Proof of $(a)$]
  Fix $f_0\colon X 	o I$. For arbitrary $f\colon X 	o I$, $f \simeq f_0$ by straight-line homotopy since any straight line in $I$ is contained in $I$. Thus, since $f$ was arbitrary, $[X,I] = \{[f_0]\}$.
\end{proof}
\begin{proof}[Proof of $(b)$]
  Let $f\colon I 	o Y$ and $y = f(0)$. Let $g\colon I 	o Y$ be the constant
  map with value $y \in Y$. Define $F\colon I 	imes I 	o Y$ by $F(s,t) =
  f(s(1-t))$. Since $F(s,0) = f(s)$ and $F(s,1) = g(s)$, we see $F$ is a
  homotopy $f \simeq g$. Now fix $y_0 \in Y$, and let $ho$ be a path
  connecting $y,y_0$. Define $H\colon X 	imes I 	o Y$ such that $H(s,t) = ho(t)$. Then, if $f_0$ is the constant map with value $y_0 \in Y$, $H$ is a homotopy $g \simeq f_0$. By transitivity of homotopy, we see $f \simeq f_0$, and so since $f$ was arbitrary, $[X,Y] = \{[f_0]\}$.
\end{proof}

Exercise 51.2

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Exercise 51.2

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