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I'm trying to brush up on some differential geometry, but there's a subtle point I don't understand. Suppose $h$ is a diffeomorphism. Then the lecture notes here suggest that it's derivative $df_x$ is an invertible linear map. Why precisely does the invertibility of $df_x$ follow from that of $f$?

Apologies if this is a trivial question - I'm a little out of practise with total derivatives!

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up vote 6 down vote accepted

A diffeomorphism is a smooth bijection with a smooth inverse. So if $f: M \longrightarrow N$ is a diffeomorphism, it is a smooth bijection and the inverse map $f^{-1}: N \longrightarrow M$ is smooth as well, so that $d(f^{-1})_y$ exists at all $y \in N$. $f^{-1} \circ f = \mathrm{Id}_M$ and $f \circ f^{-1} = \mathrm{Id}_N$, so we have that $$\mathrm{Id}_{T_x M} = d(f^{-1} \circ f)_x = d(f^{-1})_{f(x)} \circ df_x$$ and $$\mathrm{Id}_{T_{f(x)} N} = d(f \circ f^{-1})_{f(x)} = df_x \circ d(f^{-1})_{f(x)}$$ for any $x \in M$ (we applied the chain rule above), which implies that $df_x$ and $d(f^{-1})_{f(x)}$ are mutual inverses. Therefore $df_x$ is invertible and $(df_x)^{-1} = d(f^{-1})_{f(x)}$.

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Fantastic, clear answer. Many thanks! – Edward Hughes Oct 5 '12 at 11:25

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