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Anton Petrunin
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I am looking for a reference to the following statement. (It should be known --- I saw it before, don't remember where; search by keywords did not help.)

Let $f\colon M\to N$ be a homeomorphism between two smooth $d$-dimensional manifolds. Suppose that $M$ and $N$ admit Riemannian metrics such that the $f$ is $e^{\mp\varepsilon}$-bi-Lipschitz for some $\varepsilon=\varepsilon(d)$. Then $M$ is diffeomorphic to $N$.

(It can be proved by taking a locally finite covering of manifolds by almost isometric charts and applying a smoothing in the charts recursively.)

I am looking for a reference to the following statement. (It should be known --- I saw it before, don't remember where; search by keywords did not help.)

Let $f\colon M\to N$ be a homeomorphism between two smooth $d$-dimensional manifolds. Suppose that $M$ and $N$ admit Riemannian metrics such that the $f$ is $e^{\mp\varepsilon}$-bi-Lipschitz for some $\varepsilon=\varepsilon(d)$. Then $M$ is diffeomorphic to $N$.

I am looking for a reference to the following statement. (It should be known --- I saw it before, don't remember where; search by keywords did not help.)

Let $f\colon M\to N$ be a homeomorphism between two smooth $d$-dimensional manifolds. Suppose that $M$ and $N$ admit Riemannian metrics such that the $f$ is $e^{\mp\varepsilon}$-bi-Lipschitz for some $\varepsilon=\varepsilon(d)$. Then $M$ is diffeomorphic to $N$.

(It can be proved by taking a locally finite covering of manifolds by almost isometric charts and applying a smoothing in the charts recursively.)

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Anton Petrunin
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Almost isometric manifolds are diffeomorphic

I am looking for a reference to the following statement. (It should be known --- I saw it before, don't remember where; search by keywords did not help.)

Let $f\colon M\to N$ be a homeomorphism between two smooth $d$-dimensional manifolds. Suppose that $M$ and $N$ admit Riemannian metrics such that the $f$ is $e^{\mp\varepsilon}$-bi-Lipschitz for some $\varepsilon=\varepsilon(d)$. Then $M$ is diffeomorphic to $N$.