Let $n$ be $\geq 2$, $$\mathbb{P}^n(\mathbb{R}) \supset S^{n-1}= \lbrace [1,x_1,...,x_n] | x_1^2+...+x_n^2=1 \rbrace$$ and $B(0,1)= \lbrace [1,x_1,...,x_n] | x_1^2+...x_n^2<1\rbrace $. Show that $S^{n-1}$ is not a deformation retract of $\mathbb{P}^n(\mathbb{R})/ B(0,1)$. Can this be done easily by using that the existence of a deformation retraction implies homotopy equivalence?


1 Answer 1


It seems that the homotopy equivalence way does not work for $n=2$, as in this case the two sets are in fact homotopy equivalent.

However, one can use the following approach. Think of $\mathbb{R}\mathbb{P}^n$ as the quotient $S^n/\sim$ where $p\sim q$ if $p,q$ are antipodal. Then the set $\mathbb{R}\mathbb{P}^n\setminus B(0,1)$ is the quotient $C/\sim$ where $C$ is obtained by cutting off two antipodal $n$-balls from $S^n$, and the copy of $S^{n-1}$ in question is the quotient $\partial C/\sim$.

Suppose $\partial C/\sim$ is a deformation retract of $C/\sim$ and let $H:C/\sim\times[0,1]\to C/\sim$ be the corresponding homotopy. Define a map $f:C\to \partial C$ as follows. Given $p\in C$, let $[p]\in C/\sim$ denote $p$'s equivalence class. The path $t\mapsto H([p],t)$ connects $[p]$ with some point in $\partial C/\sim$, and by lifting it to $C$ one obtains a path $\gamma:[0,1]\to C$ connecting $p$ with some point in $\partial C$. Define $f(p)=\gamma(1)$. It is clear that $f$ is continuous (see explanation below) and $f|_{\partial C}=id$, thus we have a retraction $C\to\partial C$, which is impossible as $C$ is connected and $\partial C$ is not.

Two words about the continuity of $f$. As continuity is a local property, we only need to verify that $f$ is continuous on a small neighborhood of a given $p\in C$. Let $U$ be an open neighborhood of $[p]$ which is homeomorphic to a neighborhood $\tilde{U}$ of $p$ (there is such, as $\pi:C\to C/\sim$ is a covering map), and let $G:U\times[0,1]\to C/\sim$ be the restriction of $H$. It follows from the homotopy lifting property (Hatcher, Proposition $1.30$) that $G$ can be lifted to $\tilde{G}:U\times[0,1]\to C$ satisfying $\tilde{G}_0=(\pi|_{\tilde{U}})^{-1}$. By construction we have $f|_{\tilde{U}}=\tilde{G}_1\circ\pi$, thus $f$ is continuous.

  • $\begingroup$ While trying to formalize the proof, I was not able to see why f is continuous. Of course we can lift H in such a way that $\gamma(0)=p$, but what about $\gamma(1)$? $\endgroup$
    – d. zeffiro
    Sep 10, 2014 at 17:29
  • 1
    $\begingroup$ @ d. zeffiro I added some explanation to the answer, please have a look. $\endgroup$ Sep 10, 2014 at 19:31

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