# About degree of a smooth map between Manifolds

So from chapter 5 of Milnor's Topology from the Differentiable Viewpoint, there are two theorems about the degree of a smooth map where the degree is defined as follows:
$$f: M \to N$$ is a smooth map between manifolds of same dimension, M is compact without boundary and N is connected.

$$deg(f;y) = \sum_{x \in f^{-1}(y)} sign(df_x)$$ , where $$y$$ varies over the set of regular values of $$f$$.

From the fact that $$\#f^{-1}y$$ is a locally constant function and determinant is a smooth map, we can say that $$deg(f;y)$$ is locally constant.

Now the theorems are as follows:

Theorem A: The integer $$deg(f; y)$$ does not depend on the choice of regular value y.

Theorem B: If $$f$$ is smoothly homotopic to $$g$$, then $$deg$$ $$f$$ = $$deg$$ $$g$$.

For the proof of theorem A, I thought about using the fact that since $$deg(f;y)$$ is a locally constant function and $$N$$ being a connected manifold, we should have that $$deg(f;y)$$ is constant but then $$y$$ varies over all regular values of f which may not be even connected. I think I have to use Sard's theorem but I have got no clue.

For the proof of theorem B, Milnor states that "The proof will be essentially the same as that in §4" which states that $$deg$$ $$mod (2)$$ of smoothly homotopic maps are equal. But then again I don't understand how to use that.

I have been thinking about these two for over a day now. A sketch of the proofs or even some hint will be appreciated.

• Right after stating these two theorems, Milnor proceeds to prove them. What part of his proof are you struggling with? Oct 17, 2019 at 3:55

For theorem 2 (assuming theorem 1): let $$y$$ be a regular value of $$f$$ and assume $$f_n: M \rightarrow N$$ be converging in $$\mathscr{C}^1$$ to $$f$$.

Consider small coordinate patches $$U_i$$ around the points $$x_i$$ and a coordinate patch $$V$$ around $$y$$. We can choose the $$U_i$$ so that they are disjoint and such that the sign of $$d_uf$$ (as seen in the patches $$U_i$$ and $$V$$) does depend on $$U$$: in particular $$f: U_i \rightarrow V$$ is a local diffeomorphism. Now consider smaller $$U’_i \subset U_i$$ containing the point of $$f^{-1}(y)$$ such that the closure in $$M$$ of $$U’_i$$ is contained in $$U_i$$.

Now, assume that $$f_n \rightarrow f$$ in $$\mathscr{C}^1$$ topology.

Then, if $$n$$ is large enough, for each $$i$$, $$f_n(\overline{U’_i}) \subset V$$, and the sign of $$(df_n)_{|U’_i}$$ is constant equal to that of $$df_{|U_i}$$.

Also, we have some $$\delta > 0$$ such that $$d(f(x),y) \geq \delta$$ for $$x \notin U’_i$$. So for $$n$$ large enough, $$x \notin U’_i$$, $$d(f_n(x),y) > \delta/2$$, hence $$f_n^{-1}(y)$$ is contained in the reunion of the $$U_i$$.

Note now that if $$f_n(a_n)=y$$, then $$a_n$$ can be partitioned in finitely many subsequences that each converge to some point in $$f^{-1}(y)$$.

So all it remains to do is to show that with these assumptions, for all $$i$$ and all $$n$$ large enough there is some open $$V$$ containing $$f^{-1}(y) \cap U’_i$$ such that $$f_n^{-1}(y)$$ has exactly one element in $$V$$.

Uniqueness: assume $$f_n(a_n)=f_n(b_n)$$ in the same $$U’_i$$ for infinitely many $$n$$. We can assume (in coordinate patches) that $$\frac{a_n-b_n}{|a_n-b_n|}=\delta_n$$ converges to some unit vector $$u$$. Since $$0=\int_0^1{f_n’(a_n+t(b_n-a_n)) \cdot \delta_n}$$, $$0=f’(p) \cdot u$$, where $$p$$ is the point in $$U’_i \cap f^{-1}(y)$$ and we get a contradiction since $$y$$ is regular.

Existence: with the appropriate coordinate patches or subpatches, we may assume $$U’_i=V=B(0,1)$$ and $$f=Id$$ and $$f_n$$ converge uniformly on $$U’_i$$ to $$f$$ (that’s mostly to simplify, it works in a more general setting). Let $$x_n \in \overline{B}(0,1/2)$$ a minimum of $$|f_n|$$. Then $$|f_n(x_n)| \leq |f_n(0)| \rightarrow 0$$. Let $$x_{k_n}$$ be a convergent subsequence to $$t$$, then $$f(t)=0$$, so $$t=0$$, so if $$n$$ is large enough $$x_n$$ is an actual local minimum (on $$B(0,1/2)$$) of $$|f_n|$$. Now, $$0=|(\nabla |f_n|^2)(x_n)|=|f_n(x_n)||(\nabla f_n)(x_n)|$$. Since $$|\nabla f_n| \geq 1/2$$ on $$B(0,1/2)$$ for large enough $$n$$, it follows $$f_n(x_n)=0$$ and we are done.

• Looking at the proof we can say that for $f_n$ converging uniformly to f we have that they have same degrees but the question was for smoothly homotopic maps f and g. Does the proof follow from here which I'm not seeing or is it something else? Oct 16, 2019 at 16:35
• It means that if $(f_t)$ is a smooth homotopy, then $\deg{f_t}$ is locally constant, thus constant. Oct 16, 2019 at 17:45

For a detailed answer, It would help to refer to section 3.2, "Differential Topology" by Guillemin and Pollack. Especially the following part : Given X and Y to be two manifolds with boundary, Let $$\alpha$$ and $$\beta$$ to be bases of $${T}_{x}$$X and $${T}_{y}$$Y res. Then their product space acquires orientation given by sign($$\alpha$$ $$\times$$ 0 , 0 $$\times$$ $$\beta$$) = sign($$\alpha$$)sign($$\beta$$) . Then it follows that, given a homotopy f : X $$\times$$ I $$\rightarrow$$ Y, The orientation of $$\partial$$(X $$\times$$ I) is $${X}_{1}$$ $$\cup$$ -$${X}_{0}$$ .

Now going back to Milnor's book , because of homogeneity lemma , there exists a diffeomorphism from N to itself taking regular value x to another regular value y,composing that diffeomorphism with f we will get map g homotopic to f composed with identity map ( N is given to be connected and such a homotopy exists due to proof similar to homogeneity lemma). Let that homotopy be F : M $$\times$$ I $$\rightarrow$$ N to the boundary will have degree deg(g;x) -deg(f,x) where g=F(1) and f=F(0) which is proved to be zero by first lemma Note that deg(g;y) = deg(f;y) .Now we can take the homotopy map F to be anything i.e. any two homotopic maps then same procedure proves both theorems A and B.