《代数拓扑》第一次作业

Suppose $H : X \times I \to Y$ is a homotopy from $h$ to $h'$, and $K : Y \times I \to Z$ is a homotopy from $k$ to $k'$. Define a map $F : X \times I \to Z$ by $$\begin{equation} F(x,t)=K(H(x,t),t). \end{equation}$$ It is clear that $F$ is continuous, and $$\begin{equation} F(x,0) = K(H(x,0),0) = K(h(x),0) = k(h(x)), \end{equation}$$ $$\begin{equation} F(x,1) = K(H(x,1),1) = K(h'(x),1) = k'(h'(x)). \end{equation}$$ Thus, $F$ is a homotopy from $k \circ h$ to $k' \circ h'$, as desired.

This problem is a special case of the next problem. So we only prove the next problem.

(a) We will prove that any star-shaped subset of $\mathbb{R}^n$ is contractible, which includes $I$ and $\mathbb{R}$ as special cases. Suppose $X \subseteq \mathbb{R}^n$ is star-shaped with respect to a point $x_0 \in X$. Define a homotopy $H : X \times I \to X$ by $$\begin{equation} H(x,t) = (1-t)x + t x_0. \end{equation}$$ It is clear that $H$ is continuous, and $$\begin{equation} H(x,0) = x, \end{equation}$$ $$\begin{equation} H(x,1) = x_0. \end{equation}$$ Thus, $H$ is a homotopy from the identity map on $X$ to the constant map at $x_0$, showing that $X$ is contractible.

(b) Let $X$ be a contractible space. For any two points $x_1, x_2 \in X$, since the identity map on $X$ is nullhomotopic, there exists a homotopy $H : X \times I \to X$ such that $$\begin{equation} H(x,0) = x, \end{equation}$$ $$\begin{equation} H(x,1) = x_0, \end{equation}$$ for some fixed point $x_0 \in X$. Define a path $\gamma : I \to X$ by $$\begin{equation} \gamma(t) = \begin{cases} H(x_1, 2t), & \text{if } 0 \leq t \leq \frac{1}{2}, \\ H(x_2, 2-2t), & \text{if } \frac{1}{2} \leq t \leq 1. \end{cases} \end{equation}$$ It is clear that $\gamma$ is continuous, and $\gamma(0) = x_1$, $\gamma(1) = x_2$. Thus, $X$ is path connected.

(c) Let $Y$ be a contractible space, and let $f, g : X \to Y$ be any two continuous maps. Since the identity map on $Y$ is nullhomotopic, there exists a homotopy $H : Y \times I \to Y$ such that $$\begin{equation} H(y,0) = y, \end{equation}$$ $$\begin{equation} H(y,1) = y_0, \end{equation}$$ for some fixed point $y_0 \in Y$. Define a homotopy $F : X \times I \to Y$ by $$\begin{equation} F(x,t) = \begin{cases} H(f(x), 2t), & \text{if } 0 \leq t \leq \frac{1}{2}, \\ H(g(x), 2-2t), & \text{if } \frac{1}{2} \leq t \leq 1. \end{cases} \end{equation}$$ It is clear that $F$ is continuous (use the pasting lemma [Theorem 18.3]), and $$\begin{equation} F(x,0) = f(x), \end{equation}$$ $$\begin{equation} F(x,1) = g(x). \end{equation}$$ Thus, $F$ is a homotopy from $f$ to $g$, showing that $[X, Y]$ has a single element.

(d) Let $X$ be a contractible space and $Y$ be a path connected space. Let $f, g : X \to Y$ be any two continuous maps. Since $X$ is contractible, there exists a homotopy $H : X \times I \to X$ such that $$\begin{equation} H(x,0) = x, H(x,1) = x_0, \end{equation}$$
for some fixed point $x_0 \in X$. Since $Y$ is path connected, there exists a path $\gamma : I \to Y$ such that $$\begin{equation} \gamma(0) = f(x_0), \gamma(1) = g(x_0). \end{equation}$$
Define a homotopy $F : X \times I \to Y$ by $$\begin{equation} F(x,t) = \begin{cases} f(H(x, 3t)), & \text{if } 0 ≤ t ≤ \frac{1}{3}, \\ \gamma(3t-1), & \text{if } \frac{1}{3} ≤ t ≤ \frac{2}{3}, \\ g(H(x, 3-3t)), & \text{if } \frac{2}{3} ≤ t ≤ 1. \end{cases} \end{equation}$$ It is clear that $F$ is continuous (use the pasting lemma [Theorem 18.3]), and $$\begin{equation} F(x,0) = F(H(x,0)) = f(x), \end{equation}$$ $$\begin{equation} F(x,1) = F(H(x,0)) = g(x). \end{equation}$$ Thus, $F$ is a homotopy from $f$ to $g$, showing that $[X, Y]$ has a single element.