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:any 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 -1for x in this frame cobordism Pontryagin manifold due to its orientation given by the frame, we can conclude that frame cobordant Pontryain mfd 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.