Exercise 51.3

Proof of $(a)$. Since both $I$ and $ℝ$ are convex, we see that any two maps are homotopic by straight-line homotopy as on pp. 324–325 since then the straight lines are fully contained in convex set. In particular, $i_X\simeq f$ for any constant map $f$. □

Proof of $(b)$. If $X$ is our contractible space, and $F$ is our homotopy between $i_X$ and $f_0$ for $f_0$ a constant map with value $x_0\in X$, the map $\rho :I\to X$ where $\rho (t)=F(x,t)$ is a path connecting $x,x_0$ for any $x\in X$. □

Proof of $(c)$. Let $g_0:Y\to Y$ be the constant map with value $y_0\in Y$; we have $i_Y\simeq g_0$. Now define $f_0:X\to Y$ such that $f_0(x)=y_0$. Then, for any $f:X\to Y$, we have $f=i_Y\circ f\simeq g_0\circ f=f_0$ by Exercise 51.1, and so since $f$ was arbitrary, $[X,Y]=\{[f_0]\}$. □

Proof of $(d)$. Let $g_0:X\to X$ be the constant map with value $x_0\in X$; we have $i_X\simeq g_0$. Now define $y=f(x_0)$. Then, for any $f:X\to Y$, we have $f=f\circ i_X\simeq f\circ g_0$, which is the constant map at $y$. Fix $y_0\in Y$ with $f_0:X\to Y$ the constant map with value $y_0\in Y$, and let $\rho$ be a path connecting $y,y_0$. Define $H:X\times I\to Y$ such that $H(x,t)=\rho (t)$. Then, $H$ is a homotopy between $f\circ g_0$ and $f_0$. By transitivity of homotopy, we see $f\simeq f_0$, and so since $f$ was arbitrary, $[X,Y]=\{[f_0]\}$. □