# Why is the total derivative of a diffeomorphism invertible?

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!

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 $(df_x)^{-1} = d(f^{-1})_{f(x)}$.
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.
• 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? Commented Jul 3, 2021 at 17:29
• "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. Commented Sep 26, 2021 at 8:42
• @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. Commented Sep 26, 2021 at 13:33