3
$\begingroup$

Given smooth maps $f:M\to N$ and $g:N\to W$ (between closed, compact, oriented, $n$-manifolds), I want to show that $\text{deg}(g\circ f)=\text{deg}(g)\text{deg}(f)$ (Brouwer degree).

I know, given, a regular point $x\in M$, that $sign\ d(g\circ f)_x= (sign\ dg_{f(x)})(sign\ df_x)$, but I'm not sure how I can extend that fact, or if that is even the right direction.

The Brouwer degree is defined as follows (Milnor):

Let $f:M\to N$ be a smooth map (between closed, compact, oriented n-manifolds) and $p\in M$ a regular point of $f$ such that $df_x:T_xM\to T_{f(x)}N$ is a linear isomorphism. Define $sign\ df_x$ to be $\pm 1$, depending on whether $df_x$ preserves or reverses orientation. Then, for any regular value $q\in N$ $\text{deg}(f,q)=\sum_{p\in f^{-1}(q)} sign\ df_x$. Since, in fact, $\text{deg}(f,q)=\text{deg}(f,r)$ for any two regular values $q,r\in N$, $\text{deg}(f,q)$ is just called the (Brouwer) degree of $f$.

$\endgroup$
3
$\begingroup$

Let $w\in W$ be a regular value for $g\circ f$. Note that, in particular, if $n\in N$ with $g(n)=w$, then $g(n)$ is also a regular value of $f$.

Then \begin{align} \sum_{x\in (f\circ g)^{-1}(w)} \operatorname{sign} d(g\circ f)_x &= \sum_{x\in (f\circ g)^{-1}(w)} \operatorname{sign} dg_{f(x)} \cdot \operatorname{sign} df_x \\ &= \sum_{n\in g^{-1}(w)} \, \, \sum_{x\in f^{-1}(n)} \operatorname{sign} dg_n \cdot \operatorname{sign} df_x \\ &= \sum_{n\in g^{-1}(w)} \operatorname{sign} dg_n \cdot \sum_{x\in f^{-1}(n)} \operatorname{sign} df_{x} \\ &= \sum_{n\in g^{-1}(w)}\operatorname{sign}dg_n \cdot \deg(f,n) \\ &= \deg(f) \cdot \sum_{n\in g^{-1}(w)} dg_n \\ &= \deg(f) \cdot \deg(g,w)\\ &=\deg(f) \cdot \deg(g).\end{align}

The first line is the fact you quoted in your post. Moving to the second is just the set theoretic fact that $(f\circ g)^{-1}(w) = \bigcup_{n\in g^{-1}(w)}f^{-1}(n)$. The third line uses the fact that $x$ varying does not affect $\operatorname{sign}dg_n$. The fourth is the definition of degree. The fifth uses the fact that $\deg(f,n_1) = \deg(f, n_2)$ for any regular values $n_i$ of $f$.

$\endgroup$
  • $\begingroup$ Why do you have your first assumption? In other words why does $g(n)$ necessarily have to be a regular value of $f$? $\endgroup$ – Bajo Fondo Jun 28 '17 at 12:39
  • $\begingroup$ @Bajo: Because the manifolds all have the same dimension, being a regular value is the same as saying the induced map on tangent space is an ismorphism. Now, use the following linear algebraic fact: If $U,V,W$ are vector spaces of the same dimension, $T:U\rightarrow V$, $S:V\rightarrow W$ are linear, then $ST$ is an isomorphism iff both $S$ and $T$ are. (This strongly depends on all three vector spaces have the same dimension - there are easy counterexamples if $\dim V$ is larger than the other two). $\endgroup$ – Jason DeVito Jun 28 '17 at 14:00
  • $\begingroup$ Thank you, skipped over the oriented n-manifold part of the question. So the proposition is not necessarily true if the manifolds are of different dimensions? Can you provide a counter-example? $\endgroup$ – Bajo Fondo Jun 29 '17 at 0:41
  • $\begingroup$ @Bajo: Rereading, my first line can't be right: $g(n)$ can't be a regular value of $f$ since the range of $f$ doesn't hit $g(n)\in W$. I meant to write that $n$ is a regular value of $f$. Anyway, here is a counterexample: Let $M = W = \mathbb{R}$, $N = \mathbb{R}^2$ with $f(x) = (x,0)$ and $g(x,y) = x$. Then $g\circ f$ is the identity map, so every element of $\mathbb{R}$ is a regular value of $g\circ f$. However, the preimage $(x,0)$ of $x\in \mathbb{R}$ is not a regular value of $f$. $\endgroup$ – Jason DeVito Jun 29 '17 at 1:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.