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!
On
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.
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 $(df_x)^{-1} = d(f^{-1})_{f(x)}$.