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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 ${\rm id}_{(Y,B)}$.

HeroZhang001
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1 Answers1

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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.

Arthur
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  • Thank you so much! I will try to write a complete and detailed proof based on the ideas you provided :) – HeroZhang001 May 15 '23 at 08:23
  • I have proven that $\forall f:(\Bbb R^n, \Bbb R^n\setminus {0})\to(D^n, S^{n-1})$, and $\forall g:(D^n, S^{n-1})\to(\Bbb R^n, \Bbb R^n\setminus {0})$, $f\circ g:(D^n, S^{n-1})\to(D^n, S^{n-1})$ is nullhomotopic. But I don't know what the relationship between "$f\circ g:(D^n, S^{n-1})\to(D^n, S^{n-1})$ can't be homotopic to the identity" and "the identity on $S^{n−1}$ nullhomotopic". In other words, why ${\rm id}{(D^n, S^{n-1})}\simeq {\rm const}:(D^n, S^{n-1})\to(D^n, S^{n-1})$ will lead to ${\rm id}{S^{n-1}}\simeq {\rm const}$? – HeroZhang001 May 15 '23 at 11:35
  • Take a (hypothetical) homotopy $H$ between $f\circ g$ and the identity on $(D^n, S^{n-1})$. Restrict it to $S^{n-1}$. This is then a homotopy between the identity on $S^{n-1}$ and a map that is nullhomotpoic. Which shows that the identity is nullhomotopic. Which can't happen, so $H$ can't exist. – Arthur May 15 '23 at 12:12
  • Thanks! This question has been completely resolved. – HeroZhang001 May 15 '23 at 12:47