I'm learning algebraic topology. My professor left me a question:
Prove that space pairs $(D^n,S^{n-1})$ and $(\mathbb R^n,\mathbb R^n\backslash\{0\})$ are NOT homotopy equivalent.
It surprises me very much because as we all know, $D^n$ and $\mathbb R^n$ are homotopy equivalent, and $S^{n-1}$ and $\mathbb R^n\backslash\{0\}$ are homotopy equivalent. Moreover, their relative singular homology groups are isomorphic: $$H_q(\mathbb R^n,\mathbb R^n\backslash\{0\})=\left\{\begin{array}{ll} \mathbb Z, & q=n, \\ 0,& q\neq n \end{array}\right.=H_q(D^n,S^{n-1}).$$ Usually, I don't ask questions without making any effort, but this time I really don't know how to start. The proof of this question may not be related to the knowledge of algebraic topology. Any help would be great appreciated!
The definition of a map $f$ between space pairs $(X,A)$ and $(Y,B)$ is that $f:X\rightarrow Y$ is continuous and satisfies $f(A)\subset B$.
The homotopy definition of a map between space pairs is as follows:
Let $f, g: (X,A)\rightarrow (Y,B)$ be two continuous maps. A homotopy between them is a continuous map $$H: (X\times [0,1],A\times [0,1])\rightarrow (Y,B)$$ which satisfies $$ H(\cdot,0)=f\quad \text{and}\quad H(\cdot, 1)=g\quad \text{and}\quad H(A\times [0,1])\subset B$$ Two spaces $(X,A)$ and $(Y,B)$ are said to be homotopy equivalent, iff by definition there exist $f: (X,A)\rightarrow (Y,B)$ and $g:(Y,B)\rightarrow (X,A)$ such that $g\circ f$ and $f\circ g$ are homotopic to ${\rm id}_{(X,A)} $ respectively ${\rm id}_{(Y,B)}$.
It turns out that there is no map $f:(\Bbb R^n, \Bbb R^n\setminus \{0\})\to(D^n, S^{n-1})$ that works nicely. You will find that whatever map you try to make, we have that the restriction of $f$ to $\Bbb R\setminus \{0\}$ must be nullhomotopic on $S^n$: Consider that since $f(\Bbb R^n\setminus \{0\})\subseteq S^{n-1}$, we must also have $f(\Bbb R^n)\subseteq S^{n-1}$, and this map is clearly nullhomotopic. It is not difficult to make this into the above claimed nullhomotopy.
So no matter what $g$ you pick, $f\circ g$ can't be homotopic to the identity, as that would make the identity on $S^{n-1}$ nullhomotopic.