I'd like to prove that if $f,g : \mathbb{S}^n \longmapsto \mathbb{S}^n$ are pointed maps with $deg(f) = deg(g)$, then $f,g$ are homotopic.
The degree is defined in the following way : given $f : \mathbb{S}^n \longmapsto \mathbb{S}^n$ continuos, this induces $f_* : \tilde{H}_n(\mathbb{S}^n) \longmapsto \tilde{H}_n(\mathbb{S}^n)$, so exists $d \in \mathbb{Z}$ such that $f_*(x) = dx$ (this is because $\tilde{H}_n(\mathbb{S}^n) \simeq \mathbb{Z}$), and we define $d$ as the degree of $f$.
We will use the Hurewicz theorem in the following form :
Theorem : There's an homomorphism $h_n : \pi_n(X,x_0) \longmapsto H_n(X)$ which is functorial. If $X$ is $(n-1)$-connected $h_n$ is an
isomorphism (or in the limit case where $n=1$ is isomorphic to the
quotient with respect to the commutator's subgroup).
I'd like to understand if my reasoning is correct, and in the case is not, improve it if possible :
If $f,g$ have same degree,let's say $d$, we have $f_*(x) = dx = g_*(x)$, so $f_* = g_*$, i.e they induce the same map in homology.
Now if $n>1$ take the diagram
$\require{AMScd}$
\begin{CD}
\pi_n(\mathbb{S}^n,x_0) @>{f}>> \pi_n(\mathbb{S}^n,x_0) \\
@VVV @VVV\\
\tilde{H_n}(\mathbb{S}^n) @>{f_*}>> \tilde{H_n}(\mathbb{S}^n)
\end{CD}
Where the vertical maps are given by $h_n$ and f is the induced map on $\pi_n(\mathbb{S}^n,x_0)$.
I don't the following, but if $h_n$ is natural in $X$, we have induced commuting diagram taking continuos maps $f : X \longmapsto Y$,
$\require{AMScd}$
\begin{CD}
\pi_n(\mathbb{S}^n,x_0) @>{f}>> \pi_n(\mathbb{S}^n,x_0) \\
@VVV @VVV\\
\tilde{H_n}(\mathbb{S}^n) @>{f_* = g_*}>> \tilde{H_n}(\mathbb{S}^n) \\
@VVV @VVV\\
\pi_n(\mathbb{S}^n,x_0) @>{g}>> \pi_n(\mathbb{S}^n,x_0)
\end{CD}
Hence the diagram above commutes and this should be sufficient to show that $[f] = [g] \in \pi_n(X,x_0)$, that implies that exists a pointed $H : \mathbb{S}^n \times I \longmapsto \mathbb{S}^n$ which is the desidered homotopy.
I am not convinced of the follownigs :
- Is $h_n$ natural in $X?$
- The statement is true only with pointed maps and so pointed homotopy? (Which would be the case of this proof)
- What about $n = 1?$
- If the proof is correct, what smoky part that can be improved?
Any help would be appreciated, thanks in advance.
Best Answer
Yes. The Hurewicz theorem says that $h_n$ is functorial.
In $\pi_n(X,x_0)$ we consider indeed pointed homotopy classes of pointed maps. But it is well-known (and easy to see) that for simply connected $X$ the canonical map $\pi_n(X,x_0) \to [S^n,X] =$ set of free homotopy classes of maps $S^n \to X$ is a bijection.
Since $\pi_1(S^1,x_0)$ is abelian, $h_1 : \pi_1(S^1,x_0) \to H_1(S^1)$ is an isomorphism. But it is well-known (and again easy to see) that $\pi_1(S^1,x_0) \to [S^1,S^1]$ is a bijection. You can for example use the covering map $\mathbb R \to S^1$ to prove it. Alternatively you can use the fact that $S^1$ is an H-space (it is even a topological group).
Your proof is correct.