If f and g are composable diffeomorphisms, why does g∘ f keep that structure?

Its inverse is f^-1∘ g^-1, which is also smooth

Because distances always remain identical

Because it does not need to be bijective

Because every composition is commutative

Exercise · ZW-SMOOTH-DIFFEOMORPHIC-DEFORMATION-003 · ★★☆ Intermediate

If \(f\) and \(g\) are composable diffeomorphisms, why does \(g\circ f\) keep that structure?

Choose one option.

Show answer

The composition of diffeomorphisms is smooth and bijective, and its inverse \(f^{-1}\circ g^{-1}\) is also smooth.

A smooth diffeomorphic deformation \(\varphi\) is a bijective, smooth transformation whose inverse is also smooth.

\[\varphi^{-1}\text{ is smooth}\]