The famous hopf theorem says that a smooth map from a oriented closed dimension $p$ manifold to $S^{p}$ is homotopic if and only if $f$ and $g$ have the same brower degree. To prove the theorem Milnor suggested us three theorems in the book 'topology from the differential view point':


* Theorem A: any two such homotopic smooth mapping induce the framed cobordant Pontryagin manifold .
* Theorem B: If two Pontryagin manifold induced by $f$ and $g$ are frame cobordant, the $f$ and $g$ are homotopic (smooth).
* Theorem C: any frame cobordism Pontryagin manifold are induced by some smooth mapping $f$.


First, it is well-know that if $f$ and $g$ is smooth homotopic, then they have the same brower degree.


Second, we need to prove that if $f$ and $g$ have the same degree, then they are homotopic. From above three theorems, we only have to prove that $f$ and g have the frame cobordant Pontryagin manifold. Since $\operatorname{dim} M=p=$ dimension of $p$-sphere, so the corresponding frame are of $0$ dim, i.e. discrete points in $M$, so if we define $\operatorname{sgn}(x)=1$ or $-1$ for $x$ in this frame cobordism Pontryagin manifold due to its orientation given by the frame, we can conclude that frame cobordant Pontryain manifold have the same degree(=sum sgn(x)), but i don't know how to prove that if they have the same degree, they are frame cobordant in the particular case of dim 0 ?


Note : we have the notations and definitions as in Milnor's book.



a small questions about hopf theorem
I realize that this doesn't answer your question, but there is also an approach using the methods of homotopy theory and CW complexes. If $M$ is a closed smooth orientable $p$-manifold, then $M$ is homeomorphic to a finite CW complex with cells of dimension $\leq p$, and $H^p(M)=\mathbb{Z}$. 


We may construct a $K(\mathbb{Z},p)$ as a CW complex by taking $S^p$ and attaching cells of dimension $\geq p+2$ to kill the higher homotopy groups (for example, $K(\mathbb{Z},2)= \mathbb{CP}^{\infty}$ has a cell decomposition with one cell in every even dimension, and $2$-skeleton $S^2$). Let $i:S^p\hookrightarrow K(\mathbb{Z},p)$ be the induced inclusion, representing the fundamental class $[S^p]\in H^p(S^p)$. 


If we have maps $f\_i: M \to S^p$, $i=0,1$, such that the induced maps $H^p(f\_i): H^p(S^p)\to H^p(M)$ are equal, then this implies that $H^p(f\_0)([S^p])=H^p(f\_1)([S^p])$, and therefore that the maps $i\circ f\_i: M\to K(\mathbb{Z},p)$ are homotopic (by Brown Representability, Thm. 4E.1 Hatcher), and thus are realized by a map $F: M\times [0,1] \to K(\mathbb{Z},p)$, $F\_{| M\times i}=f\_i$. By the Cellular Approximation Theorem (Theorem 4.8 Hatcher), the map $F$ may be homotoped rel $M\times \{ 0, 1 \}$ to a map $F': M\times [0,1] \to K(\mathbb{Z},p)^{(p+1)}=S^p$, the $p+1$ skeleton of $K(\mathbb{Z},p)$, and thus $f\_0\simeq f\_1$.


