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user50311
user50311
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Why are two diffeomorphims of $R^n$ are alsoalways homotopic (in the same category)?

Where can one find the proof of the following fact:

If there are two orientation-preserving diffeomorphisms $\phi_0$ and $\phi_1$ of $R^n$, then there exists a homotopy $\phi(t)$, such that $\phi(0)$ = $\phi_0$, $\phi(1)$ = $\phi_1$, and $\phi(t)$ is a diffeomorphism for any t$t$ (and $\phi$ is smooth for all variables).

This is easy to prove for $n=1$, and relatively easy for $n=2$, but for an arbitrary $n$ seems difficult.

This fact is used in the knot theory to justify the coincidence of two definitions of the knot equivalence.

Why are two diffeomorphims of $R^n$ are also homotopic (in the same category)?

Where can one find the proof of the following fact:

If there are two diffeomorphisms $\phi_0$ and $\phi_1$ of $R^n$, then there exists a homotopy $\phi(t)$, such that $\phi(0)$ = $\phi_0$, $\phi(1)$ = $\phi_1$, and $\phi(t)$ is a diffeomorphism for any t (and $\phi$ is smooth for all variables).

This is easy to prove for $n=1$, and relatively easy for $n=2$, but for an arbitrary $n$ seems difficult.

This fact is used in the knot theory to justify the coincidence of two definitions of the knot equivalence.

Why are two diffeomorphims of $R^n$ are always homotopic (in the same category)?

Where can one find the proof of the following fact:

If there are two orientation-preserving diffeomorphisms $\phi_0$ and $\phi_1$ of $R^n$, then there exists a homotopy $\phi(t)$, such that $\phi(0)$ = $\phi_0$, $\phi(1)$ = $\phi_1$, and $\phi(t)$ is a diffeomorphism for any $t$ (and $\phi$ is smooth for all variables).

This is easy to prove for $n=1$, and relatively easy for $n=2$, but for an arbitrary $n$ seems difficult.

This fact is used in the knot theory to justify the coincidence of two definitions of the knot equivalence.

Source Link
user50311
user50311
  • 305
  • 1
  • 7

Why are two diffeomorphims of $R^n$ are also homotopic (in the same category)?

Where can one find the proof of the following fact:

If there are two diffeomorphisms $\phi_0$ and $\phi_1$ of $R^n$, then there exists a homotopy $\phi(t)$, such that $\phi(0)$ = $\phi_0$, $\phi(1)$ = $\phi_1$, and $\phi(t)$ is a diffeomorphism for any t (and $\phi$ is smooth for all variables).

This is easy to prove for $n=1$, and relatively easy for $n=2$, but for an arbitrary $n$ seems difficult.

This fact is used in the knot theory to justify the coincidence of two definitions of the knot equivalence.