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Supose that $X\stackrel{f}{\to} Y \stackrel{g}{\to} Z$ are given, $f,g$ smooth and $X,Y,Z$ compact, oriented manifolds. Prove that $$\textrm{deg}(f\circ g) = \textrm{deg}(f)\textrm{deg}(g)$$ where $\textrm{deg}(f)=I(f,{z})$ for any $y\in Y$.

I say this on the answer by @Jared at The degree of antipodal map. and this is in fact an exercise of Guillemin-Pollack (chap.3 sec.3 number 10). I have no idea where start from. Any thoughts?

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rseallan
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1 Answers1

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Let $z\in Z$ a regular value of the composite. Then

$$\deg(g\circ f)=\sum_{x \in (g\circ f)^{-1}(z)}\pm 1$$

according whit the orientatión of the derivative $T_xX→T_zZ$. Let’s write

$$\deg(g\circ f)=\sum_{x \in (g\circ f)^{-1}(z)}\pm 1=\sum_{y \in g^{-1}(z)}\sum_{x \in f^{-1}(y)}\pm 1=\sum_{y \in g^{-1}(z)}\pm 1\sum_{x \in f^{-1}(y)}\pm 1=\deg(g)\deg(f).$$

jimbo
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