Proving homotopic equivalence between $^m$ and $R^{m+1}-\{0\}$

Question

I'm studying differential forms, integration on surfaces, and these concepts are defined in exercises. The first exercise defines this concept:

Let $M$ be a surface $C^{\infty}$. Two closed forms $\alpha, \beta$ of degree $r$ and $C^{\infty}$ are cohomologous when $\alpha-\beta$ is exact. Then the equivalence class $[\alpha]$ constitutes the vector space $H^r(M)$, called the cohomology group of degree $r$ in the surface $M$. Every map $f:M\to N$, $C^{\infty}$, induces the linear transformation $f^*:H^r(N)\to H^r(M)$ such that $(g\circ f)^* = f^* \circ g^*$, defined by $f^*[\alpha] = [f^*\alpha]$

Now, the exercise I'm supposed to answer is this one, which defines some concepts too:

We say that two surfaces $M, N$, $C^{\infty}$, have the same type of homotopy when there are maps $f:M\to N$ and $g:N\to M$, $C^{\infty}$, such that $g\circ f $ and $f\circ g$ are maps homotopics to the correspondent identities. In this case, $f$ is called a homotopic equivalence and $g$ its homotopic inverse. Prove that, in this case, $f^*:H^k(N)\to H^k(M)$ and $g^*:H^k(M)\to H^k(N)$ are, for $k=0,1, \cdots$ isomorphisms. Show that the inclusion $i: S^m\to \mathbb{R}^{m+1}-\{0\}$ and the radial projection $p: \mathbb{R}^{m+1}-\{0\}\to S^m$ (defined by $p(x) = x/|x|$) are homotopic equivalences between $S^m$ and $R^{m+1}-\{0\}$. Conclude that the cohomology groups $H^k(S^m)$ and $H^k(\mathbb{R}^{m+1}-\{0\})$ are isomorph for $k=0,1, \cdots$

I don't have much theory to solve this, I'm lost. For example, I know that $p$ maps from $\mathbb{R}^{m+1}-\{0\}$ to $S^m$. I know that $i \circ p$ is the identity on $\mathbb{R}^{m+1}-\{0\}$, and I can also make the 'reverse' identity. Is it trivial that these two are identities one of another? How to show that these composites are homotopic to the identity? How to show that $f^*$ and $g^*$ are isomorphisms?

I think this is a classical result and there are much content that I can read about, but I couldn t find anything specific to the sphere and $\mathbb{R}^m$.

Answer 1

Firstly, You are wrong, $i\circ p$ is not the identiy on $\mathbb{R}^n-\{0\}$.

Secondly, A little bit of definitions and facts that you can find in any book of differential forms.

Let $f,g; M\rightarrow N$ be $C^\infty$ maps between manifolds. We say that $f$ is homotopic to $g$ or $f \sim g$ if there is a $C^\infty$ map $F : M \times [0,1] \rightarrow N$ such that $F(x,0) = f(x)$ and $F(x,1) = g(x)$ for any $x \in M$. Such a map $F$ is called a homotpy between $f$ and $g$.

A main relation between the notion of homotopy and the maps induced in cohomology is the following remark

If $f \sim g$ then $f^* = g^* : H^*(N) \rightarrow H^*(M)$.

Now, we say that $f: M \rightarrow N$ is a homotopy equivalence with homotpy inverse $g: N \rightarrow M$ if there are homotpies

$$ g \circ f \sim id_M$$ and $$ f \circ g \sim id_N$$

Note that if $f$ is a homotpy equivalence with homotpy inverse $g$, by the above Remark we get in cohomology the equalities

$$f^*\circ g^* = (g \circ f)^* = (id_M)^* = id_{H^*(M)}$$

and

$$g^*\circ f^* = (f \circ g)^* = (id_N)^* = id_{H^*(N)}$$

This means, $f^*$ and $g^*$ are isomorphisms in Cohomology

Finally, going back to your original concern. where $i:S^m \rightarrow \mathbb{R}^{m+1}-\{0\}$ is the inclusion and $p: \mathbb{R}^{m+1}-\{0\} \rightarrow S^m$ is the radial projection. Notice

$$ p \circ i (x) = x = id_{S^m};$$

Therefore $(p \circ i)^* = id_{H^*(S^m)}$. On the other hand,

$$ i\circ p (x) = x/|x|$$

In this case, the composite is not the identiy on $\mathbb{R}^{m+1}-\{0\}$ but it is homotopic to $id_{\mathbb{R}^{m+1}-\{0\}}$

Question: Can you think on a map $F: \mathbb{R}^{m+1}-\{0\} \times [0,1] \rightarrow \mathbb{R}^{m+1}-\{0\}$ such that $F(x,0) = x$ and $F(x,1) = x/|x|$ for any $x \in \mathbb{R}^{m+1}-\{0\}$? Picturing everything in $\mathbb{R}^3$ would be very useful.

That is, $F$ is a homotpy between $i \circ p$ and $id_{\mathbb{R}^{m+1}-\{0\}}$ and conclude that $i$ is a homotpy equivalence with homotpy inverse $p$.