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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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2 Answers 2

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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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Let me give an explanation based on intuition (not a rigorous proof).

For every tangential vector $v$ at $x$, the tangential vector $df_x \cdot v$ at $f(x)$ has the following significance: if you stand at $x$ and move in the direction $v$, then your image at $f(x)$ moves in direction $df_x \cdot v$.

Suppose that $df_x$ is not invertible. Then there exists $v$ such that $df_x \cdot v = 0$. In other words, moving in direction $v$ does not change your image $f(x)$, which intuitively contradicts the idea that $f$ is a diffeomorphism.

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  • $\begingroup$ I don’t quite yet see the point here - couldn‘t it be that f just has a saddle point in the direction of v, which would imply $df_x \cdot v = 0$? How would this contradict f being an isomorphism? $\endgroup$ Commented Jul 3, 2021 at 17:29
  • $\begingroup$ "Suppose that $df_x$ is not invertible. Then there exists $v$ such that $df_x \cdot v = 0$." Sorry but is this because of the inverse function theorem? I know the theorem states that if $f$ is continuously differentiable and has non zero derivative then $f$ is invertible. $\endgroup$
    – RopuToran
    Commented Sep 26, 2021 at 8:42
  • $\begingroup$ @RopuToran: no, this does not invoke the inverse function theorem. It is only a heuristic argument anyways, so the abstract result gets more intutitive. $\endgroup$
    – shuhalo
    Commented Sep 26, 2021 at 13:31
  • $\begingroup$ @returntrue: as indicated, it is not a rigorous proof. if it is an extremal point, it can't be a diffeomorphism, but yes it's more subtle when it's a saddle point. The example $x \mapsto x^2$ shows that the mapping is still bijective but the inverse won't be differentiable; I don't know a good intuitive heuristic argument for that. $\endgroup$
    – shuhalo
    Commented Sep 26, 2021 at 13:33

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