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)?** 



Homotopy classes of homeomorphisms of a multiple pointed space
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.


