If I have a diffeomorphism $f:S_1 \rightarrow S_2$, where $S_1, S_2$ are regular surfaces, then is $(df)^{-1}=df^{-1}$?

How would I show this? I saw a hint that said to consider $f^{-1}\circ f=id$ at some point, but I dont know how the chain rule would reveal the answer.

## Best Answer

The idea is to take the differentiation rule for inverse functions: As

$$ f^{-1}\circ f = id $$ we have $$ E = d(f^{-1}\circ f) = (df^{-1})\circ f \cdot d f $$ As $df$ is regular we thus get $$ (df)^{-1} = (df^{-1})\circ f $$