Is $f \circ f^{-1}=id$ a sufficient criteria/check for diffeomorphism? Or what is?

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Is $f \circ f^{-1}=id$ a sufficient criteria for diffeomorphism?

I recall having seen some version which had three functions in the composition.

But I also think of reading that $f \circ f^{-1}=id$ implies that both $f$ and $f^{-1}$ must be smooth. But I don't know if this also proves bijectivity?

Particularly, http://www.maths.adelaide.edu.au/michael.murray/dg_hons/node7.html

If $ f$ is a diffeomorphism $ f \circ f^{-1} = 1_{ \mathbb{R}^n}$

But is this a $\iff$ or $\implies$?


Also confused as to whether $f \circ f^{-1}$ really produces $\mathbb{R}$, since $f:U \rightarrow \mathbb{R}^n$ and $U \subset \mathbb{R}^n$. If Lemma 1.2 requires that $\phi^{-1}$ would be onto $\mathbb{R}$.

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Invertibility is one requirement. Just as important is the requirement that both $f$ and $f^{-1}$ are $C^\infty$, and the definition of a function between manifolds being $C^\infty$ usually uses a certain composition of three functions.