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Let $M$ $ $ be a differential manifold, and $f$ a diffeomorphism on $M$ which is isotopic to $id$. Assuming that $x\in M$ is a fixed point of $f$ and the orbit of $x$ under the isotopy is a trivial loop(trivial in $\pi_1(M)$). How to prove that there is an isotopy from $id$ to $f$ relative to $x$? (i.e. there is an isotopy $f_t$ satisfied that $f_0=id$, $f_1=f$ and $f_t(x)=x$.

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up vote 12 down vote accepted

The quick argument is to use the fibration $Diff(M,x)\to Diff(M) \to M$ whose total space is the diffeomorphism group of $M$ and whose fiber is the subgroup fixing the point $x$. The projection $Diff(M)\to M$ is given by evaluation at $x$. Since this is a fibration, we have $\pi_1(Diff(M),Diff(M,x)) \cong \pi_1(M)$, this isomorphism being induced by the projection map of the fibration, evaluation at $x$. This isomorphism gives exactly what you want: An isotopy from the identity to a diffeomorphism fixing $x$ can be deformed to fix $x$ at all times, provided that the loop traced out by $x$ during the isotopy is trivial in $\pi_1(M)$.

It is possible to trace back through this argument to see exactly what is happening. The key point is the homotopy lifting property for the fibration. This can be proved in the case needed here by an argument using suitably constructed vector fields.

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