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 $f$.


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 $f$.

  • $\begingroup$ 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

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