Question

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.

Answer 1

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 $f_0\simeq f_1$.

Answer 2

Suppose you have two maps $f$ and $g$ with common regular point p, and that each map has Brouwer degree k (and assume wlog $k>0$). Use $p_1,...,p_n$ to denote elements of $f^{-1}(p)$ and use $q_1,..., q_m$ to denote the elements of $g^{-1}(p)$.

Here's how to construct the framed cobordism between $f^{-1}(p)$ and $g^{-1}(p)$:

Choose k of the $p_i$ with positive orientation and choose $k$ of the $q_j$ with positive orientation. By reordering the $p_i$ and $q_j$, we may as well assume you've chosen $p_1,..,p_k$ and $q_1,...,q_k$. Now, connect $p_i$ to $q_i$ via a line segment (i.e, a path in M, which is possible because $M$ is connected).

The remaining $p_i$ can be paired off in such a way as each pair contains a point of positive orientation and another point of negative orientation. For each such pair, connect the two points with a curve (in $M$). Finally, doing the same to the remaining $q_j$ yields the desired framed cobordism.