1
$\begingroup$

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?

$\endgroup$

1 Answer 1

1
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.