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In Spivak's Calculus, when discussing manifolds with boundary, and trying to make a point that a point cannot be both on the boundary and not on the boundary on page 113 Spivak notes:

Since $\text{det} (h_2 \circ h_1^{-1})' \neq 0$, this contradicts Problem 2-36.

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I don't quite see how he got that $\text{det} (h_2 \circ h_1^{-1}) \neq 0$. I know that $h_1$ and $ h_2$ both are invertible, and hence their composition is invertible. But what makes the derivative of the composition invertible? In other words, why is $\text{det} (h_2 \circ h_1^{-1})' \neq 0$?

Snowball
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1 Answers1

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If $f$ is invertible and both $f$ and $f^{-1}$ are differentiable, then from the equation $f^{-1}\circ f=\text{id}$, it follows from the chain rule that for any $a$, \begin{align} D(f^{-1})_{f(a)}\circ Df_a &= D(\text{id})_a=\text{id}. \end{align} Hence, $Df_a$ is invertible (in fact the inverse is given by $D(f^{-1})_{f(a)}$); equivalently, the determinant is non-zero. Spivak is applying this line of reasoning to $f=h_2\circ h_1^{-1}$. The key point is that both the function and its inverse are differentiable, hence (chain rule) the derivatives are also invertible.

peek-a-boo
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