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 $dim M=p=dim$ of $p$-sphere, so the corresponding frame are of $0$ dim, i.e. discrete points in $M$, so if we define $sgn(x)=1$ 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.