Building diffeomorphism

Question

I am trying to build a diffeomorphism between $S^1 = \{x^2 + y^2 = 1; x,y \in \mathbb{R}\}$ with subspace topology and $\mathbb{R P}^1 = \{[x,y]: x,y \in \mathbb{R}; x \vee y \not = 0 \}$ with quotient topology and I am struggling a little bit.

I have shown that both are smooth manifolds, and I used stereographic projection for $S^1$, but now I am runing into trouble when I give the homeomorphism between $S^1$ and $\mathbb{RP}^1$ as the map that takes a line in $\mathbb{RP}^1$ to the point in $S^1$ that you get when letting the parallel line go through the respective pole used in the stereographic projection.

Answer 1

I think your method is correct. Fix $\varepsilon < \frac{1}{M}$.

First we show that $f := f_{\varepsilon}$ is a bijection. So let $y \in \mathbb{R}^n$ be arbitrary. We must show that there exists a unique $x \in \mathbb{R}^n$ such that $x + \varepsilon h(x) = y$. Following the hint, we will show that $g \colon \mathbb{R}^n \to \mathbb{R}^n$ defined by $g(x) := y - \varepsilon h(x)$ has a unique fixed point using the contraction mapping theorem. We have $||Dg(x)|| = \varepsilon||Dh(x)|| \leq \varepsilon M < 1$. Thus $g$ is a contraction. Thus there exists a unique $x \in \mathbb{R}^n$ such that $g(x) = x$. Thus $f$ is a bijection.

It remains to show that $f$ is a local diffeomorphism at every $x \in \mathbb{R}^n$. It is not too difficult, though takes some effort, to show that if $A$ is invertible, then $A + H$ is invertible for any $H$ with $||H|| < \frac{1}{||A^{-1}||}$. See here for example: Prove if $\lim_{k \rightarrow \infty} A_k = A$, then $\lim_{k \rightarrow \infty} A_k^{-1} = A^{-1}$. We have $Df(x) = I + \varepsilon Dh(x)$. Since $||\varepsilon Dh(x)|| < 1$ and $||I|| = 1$, it follows that $Df(x)$ is invertible, so by the inverse function theorem, $f$ is a local $C^1$ diffeomorphism at $x$.

Answer 2

Let $x,u\neq 0\in \mathbb{R}^n$ suppose that $df_\epsilon(x).h=0$, this is equivalent to saying that $u+\epsilon dh(x).u=0$, we deduce that $\|u\|=\epsilon\|dh(x).u\|\leq \epsilon\|dh(x)\|\|u\|\leq \epsilon M\|u\|<\|u\|$. Contradiction, this implies that $df_\epsilon(x)$ is bijective an $f_\epsilon$ is a local homeomorphism.

Let $y$ be a point of $\mathbb{R}^n$ distinct of $x$, define $l(t)= h (x+t(y-x)), t\in [0,1]$ $\|l(1)-l(0)\|=\|h(y)-h(x)\|\leq sup_{t\in[0,1]}\|dl'\|\|x-y\|<M\|x-y\|$. Suppose that $f_\epsilon(x)=f_\epsilon(y)$, this implies that $x+\epsilon h(x)=y+\epsilon h(y)$, we deduce that $\|x-y\|=\epsilon\|h(x)-h(y)\|<\epsilon M\|x-y\|<\|x-y\|$. Contradiction. We deduce that $f_\epsilon$ is injective.

The image of $f_\epsilon$ is open since it is a local homeomorphism. Let $x_n$ a sequence of points such that $lim_nf_\epsilon(x_n)=y$, we deducee that $f_\epsilon(x_n)$ is a Cauchy sequence: For every $c>0$, there exists $N_c$ such that $n,m>N_c$ implies that $\|f_\epsilon(x_n)-f_\epsilon(x_m)\|<c$. This is equivalent to saying that:

$\|x_n+\epsilon h(x_n)-x_m-\epsilon h(x_m)\|<c$.

We have $\|x_n+\epsilon h(x_n)-x_m-\epsilon h(x_m)\|\geq |\|x_n-x_m\| -\epsilon\|h(x_n)-h(x_m)\||\geq |\|x_n-x_m\|-\epsilon M\|x_n-x_m\||=(1-\epsilon M)\|x_n-x_m\|$. (see the argument above about $l$). This implies that $(x_n)$ is a Cauchy sequence and converges towards $x$, we have $f_\epsilon(x)=y$. We deduce that $f_\epsilon$ is surjective since its image is non empty, open and closed and $f_\epsilon$ is an homeomorphism since it is a local homeomorphism and is bijective.