# Is the identity map a diffeomorphism?

Unfortunately, googling this question leads to conflicting answers. According to this source, the identity map on any smooth manifold is a diffeomorphism, but it's not according to this. I appreciate it if someone gave a definitive answer.

-
The statement that the questioner is referring to begins at the bottom of page 38, and continues onto page 39, ending with "Thus the identity map is not a diffeomorphism." – MJD Oct 15 '12 at 4:08
@Smith: the second link you found refers to a map which is the identity on underlying sets, but uses a different smooth structure in the source and the target. Personally, I think this is a bad abuse of notation. It does not refer to the identity map from a smooth manifold to itself considering only a single smooth structure, which is a diffeomorphism. – Qiaochu Yuan Oct 15 '12 at 4:55

## 3 Answers

Let us fix a smooth manifold $M$.

• is the identity map $i:M\to M$ smooth?

• is it bijective?

• what is the inverse function?

• is the inverse function bijective?

Can you answer these questions?

As for your reference: the book does not say that the identity map of a smooth manifold is not a diffeo: it gives an example to show that if $M$ and $M'$ and two smooth manifolds on the same topological space, then the identity function $M\to M'$ is not necessarily smooth. This is a claim rather rather different to «the identity map of a smooth manifold is not smooth».

-
You are free to provide more pertinent answers, MJD. – Mariano Suárez-Alvarez Oct 15 '12 at 4:20
I didn't know the answer. I would have tried to provide it if I had known. – MJD Oct 15 '12 at 4:22

Yes, the identity map is a diffeomorphism, and the derivative at any point $p$ is just the identity on $T_pM$. Maybe it is best to see this in terms of directional derivatives. Write $I$ for the identity map. Fix a curve $\phi(t): \mathbb{R} \to M$ with $\phi'(t) = X$ for some $X \in T_pM$ and then compose with the identity map. Then $D_X I = (I\circ \phi)'(t) = \phi'(t) = X$.

Regarding your second reference, the author there is giving an example of two $C^{\infty}$ structures on $\mathbb{R}$ that are different. The issue you are having is that the identity map'' there really takes $\mathbb{R}$ with one smooth structure to $\mathbb{R}$ with another smooth structure. So it doesn't have to be smooth!

But if you consider the identity map on a manifold $M$ (with fixed smooth structure- if someone utters the words "smooth manifold" then they mean a topological manifold together with a smooth structure so that is embedded in the definition) then the identity map is always a diffeomorphism.

-

I think it does not necessarily a diffeomorphism if you think identity map as a map from a set to itself. However if you consider identity map from a smooth manifold to itself (i.e to same set with same smooth structure) then it is a diffeomorphism. Here the point is this, differentiability related with your smooth structure and you can put different smooth structures on the same set.

-