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Let $f: \mathcal{M} \rightarrow \mathcal{M}$ be a smooth diffeomorphism and $\phi: \mathcal{M} \rightarrow \mathbb{R}$ be a smooth function, where $\mathcal{M}$ is a $d$-dimensional manifold (which we assume to be $\mathcal{M}=\mathbb{R}^{d}$ ).

According to Takens' embedding theorem (1981) there exists an embedding of $\mathcal M$ via $$\Phi_{f,\phi}:\mathcal M \rightarrow \mathbb{R}^{2d+1}$$ with $$\Phi_{f,\phi}(x)=(\phi(x),\phi(f(x)),...,\phi(f^{2m}(x))^T.$$

In some references, I read that an embedding is a smooth one-to-one map with a smooth inverse. I would like to prove this result but I have no idea how to proceed rigorously.

For example, from the Chain Rule, all the entries of $\Phi$ are smooth by assumptions. But what to say for its inverse? Since $f$ is a diffeomorphism its inverse should be smooth. And then? References and suggestions are welcome.

EDIT Below you will find the changes I made to this post following the comments received.

This is Takens' theorem I was referring to.

enter image description here

It is the Theorem 1 of the paper:

Takens, Floris. "Detecting strange attractors in turbulence." Dynamical systems and turbulence, Warwick 1980. Springer, Berlin, Heidelberg, 1981. 366-381.

The following is taken instead from a paper by Casdagli et al

enter image description here

The paper is:

Casdagli, Martin, et al. "State space reconstruction in the presence of noise." Physica D: Nonlinear Phenomena 51.1-3 (1991): 52-98.

Below Eq. (4) I read: An embedding is a smooth, one-to-one coordinate transformation with a smooth inverse.

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    $\begingroup$ You must be quoting the theorem incorrectly. How can $\Phi_{f, \phi}$ be an embedding if $\phi$ is a constant function, i.e. $\phi=0$? Perhaps the statement has different quantifiers. Similarly, you can't do this if $f$ is the identity function. $\endgroup$
    – Ryan Budney
    Commented Jun 28, 2022 at 20:15
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    $\begingroup$ Takens' original theorem was about strange attractors. As noted by Ryan, you are misstating the result. $\endgroup$
    – Moishe Kohan
    Commented Jun 28, 2022 at 22:09
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    $\begingroup$ I believe the result says that for "generic" $\phi$, the map $\Phi_{f,\phi}$ above gives an embedding of $\mathcal M$ into $\mathbb R^{2d+1}$. As Ryan says, there should be conditions on $f$ also, such as $f$ having a dense orbit. $\endgroup$
    – Anthony Quas
    Commented Jun 28, 2022 at 22:33
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    $\begingroup$ For Theorem 1 to be correct, "genericness" needs to be taken in the space of pairs $(\phi, y)$, since it's false if $y$ is fixed, or $\phi$ fixed. To me this theorem looks like a souped-up differential topology denseness theorem. For example, due to Whitney, embeddings are dense in the space of smooth maps $N \to \mathbb R^{k}$ where $k > 2 dim(N)$. A similar argument implies Morse functions are dense in smooth maps $N \to \mathbb R$. This Takens theorem seems to be something of a combination of these kinds of ideas. Presumably the proof should be, as well. $\endgroup$
    – Ryan Budney
    Commented Jun 29, 2022 at 7:09
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    $\begingroup$ For a compact manifold $N$, a smooth one-to-one map $N \to M$ whose derivative is everywhere one-to-one is a smooth embedding. i.e. you get the inverse is smooth for free. There's two components to this, one is a compactness argument to get that it is a topological embedding, but local smoothness of the inverse comes from (essentially) the inverse function theorem, but in the context of maps from lower-to-higher dimensional spaces. It's a theorem that does not always get its own name. I sometimes call it the "mono-split lemma". $\endgroup$
    – Ryan Budney
    Commented Jun 29, 2022 at 17:25

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