Let $M$ be a multiple pointed space, i.e. $M$ is a topological space and there is a finite point set $M\supset P=\{p_1,...,p_k\}, k<\infty$. Such a $p_i$ is called a marked point. A map $$\varphi:M,P\to M,P$$ is called a $P$-relative map if $$\varphi|_P=id|_P.$$ Our question is that

For a disk $D^2$ (moreover we order that all maps are invariant on $S^1=\partial D^2$) or a sphere $S^2$ with a non-empty $P$, if a homeomorphism $h$ relative to $P$ (and $\partial D^2$ for $D^2$ case) is connected to identity via a homotopy $H$ relative to $P$ (and $\partial D^2$ for $D^2$ case), can we improve such homotopy $H$ to an isotopy $I$?

For example, if there is only one marked point, the answer to such question is "Yes" by Alexander, see CURVES ON 2-MANIFOLDS AND ISOTOPIES of Epstein, Section 5.

In generally, we want to study the homotopy classes and the isotopy classes of $P$-relative homeomorphisms for any two dimensional surfaces $S$ (may contain some boundary $\partial S\subset S$ and we order homeomorphisms on boundary are identity, or some puncture points $q\not \in S$). So the question above can be restated as

When the natural corresponding map $T$ from an isotopy class of homeomorphism of $S$ to a homotopy class of homeomorphism of $S$ $$T:[Homeo(S)]_{isotopy}\to[Homeo(S)]_{homotopy}$$ is an injection or an isomorphism (one-one)?

  • $\begingroup$ If there are no obstruction to the existence of a (non relative) isotopy connecting $h$ with the identity, then I guess that such an isotopy can be made into a relative to $P$ one. A relative isotopy is a special map from the cylinder $M\times [0,1]$ to $M$, sending each segment $\{p_i\}\times[0,1]$ to $p_i$, and I don't see why not a non relative one could not be suitably deformed to fulfill such a property. But this is just a guess. $\endgroup$ – Giovanni Moreno Oct 22 '15 at 11:09
  • $\begingroup$ Generally speaking, $[Homeo(S)]_{isotopy}$ with respect to some (bigger than 2) marked points is a non-tirvial group, see the definition of mapping class group in 'A primer on mapping class groups Version 5.0' by Benson Farb and Dan Margalit. $\endgroup$ – A.T.Saaki Oct 23 '15 at 1:41

The answer is No, a homotopy relative to $P$ cannot in general be improved to an isotopy. To see this, consider the fibration $$ {\rm HomEq}^+(M\ {\rm rel} \ P)\to {\rm HomEq}^+(M)\to {\rm Map}(P,M) $$ where ${\rm HomEq}^+$ denotes the space of orientation-preserving homotopy equivalences and ${\rm Map}(P,M)$ is the space of maps $P\to M$. Thus ${\rm Map}(P,M)$ is the product of $k$ copies of $M$, where $k=|P|$. Part of the long exact sequence of homotopy groups for this fibration is $$ \pi_1{\rm Map}(P,M)\to\pi_0{\rm HomEq}^+(M\ {\rm rel} \ P)\to\pi_0{\rm HomEq}^+(M) $$ In the special case $M=S^2$ the first and third terms of this three-term exact sequence are zero so the middle term must be zero as well. The natural map $$ \pi_0{\rm Homeo^+}(S^2\ {\rm rel} \ P)\to \pi_0{\rm HomEq^+}(S^2\ {\rm rel} \ P) $$ will then have a nontrivial kernel if the domain group is nonzero, which happens when $k\geq 4$ since $\pi_0{\rm Homeo^+}(S^2\ {\rm rel} \ P)$ is the (pure) mapping class group of a $k$-punctured sphere. Elements of this kernel give homeomorphisms that are homotopic to the identity rel $P$ but not isotopic to the identity rel $P$.

The case $M=D^2$ is similar, where homotopy equivalences and homeomorphisms are required to restrict to the identity on $\partial D^2$. In this case one only needs $k\geq 2$. For more complicated surfaces such as closed surfaces of positive genus the map $$ \pi_0{\rm Homeo^+}(M\ {\rm rel} \ P)\to \pi_0{\rm HomEq^+}(M\ {\rm rel} \ P) $$ again has a nontrivial kernel as long as $k$ is not too small, but the argument is a little more complicated since the first and third terms in the exact sequence above are no longer zero. The rough idea is that the kernel of the map $$ \pi_0{\rm Homeo^+}(M\ {\rm rel} \ P)\to \pi_0{\rm Homeo^+}(M) $$ is bigger than just the product of $k$ copies of $\pi_1M$, due to braiding phenomena in $M$. It shouldn't be too hard to make this precise.

For an explicit example in the case of $D^2$, say, consider the homeomorphism $f$ obtained by dragging the point $p_1$ around a small loop encircling $p_2$. By construction, $f$ is homotopic (and even isotopic) to the identity fixing $ p_2,\cdots,p_k$, and this homotopy can be deformed to fix $p_1$ as well by deforming the loop that $p_1$ traces out to the constant loop, crossing $p_2$ at some time during this deformation. This deformation can be achieved by an ambient homotopy of $M$ fixing $p_2,\cdots,p_k$ but not by an ambient isotopy fixing these points since $p_1$ crosses $p_2$ at some time during the deformation.

  • $\begingroup$ Thanks! Now I know that the number of marked points plays an important role to construct the non-trivial kernel. By Epstein's result, a homotopy of a surface with none or one marked point can be improved to an isotopy. So for $g$ genus surfaces with $l$-$S^1$ boudaries, there must be a maximal number $N_{g,l}$ such that any homotopy of it can be improved to isotopy? Is there some exact formula for such $N_{g,l}$? Moreover, how about the same question if we replace $P$ as a collection of disjiont closed disks? $\endgroup$ – A.T.Saaki Oct 26 '15 at 5:09

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