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Suppose we have two smooth compact oriented surfaces $M_1$ and $M_2$ with boundary,both of them have genus $g$, and there are orientation preserving diffeomorphisms $\psi_1, \psi_2, \cdots, \psi_n$, which map each in-boundary of $M_1$ to an in-boundary of $M_2$, there are also orientation preserving diffeomorphisms $\phi_1, \phi_2, \cdots, \phi_m$, which map each out-boundary of $M_1$ to an out-boundary of $M_2$. My question is if there exist an orientation preserving diffeomorphism between $M_1$ and $M_2$ which extends all of the above boundary diffeomorphisms, i.e, $\phi_i$'s and $\psi_i$'s ? "Here, an in-boundary refers to a connected component of $\partial M_i$ with an orientation that, together with the inward normal to the boundary, forms the orientation of $M_i$. The same applies to the out-boundary."

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The way you phrased this makes it sound harder than it is. Your two surfaces are diffeomorphic, so we can identify them both with a single surface $M$. Do this in a way that reflects the identifications between the boundary components of the $M_i$ given by your maps. Orient the boundary components of $M$ in the usual way (so $M$ is to their left). Your various maps between boundary components then assemble into an orientation preserving diffeomorphism $f_0: \partial M \rightarrow \partial M$. Since we identified the $M_i$ with $M$ in a way that reflects the identifications of the boundary components given by your maps, the map $f_0$ takes each boundary component to itself. You want to extend $f_0$ to a diffeomorphism of the whole surface.

We now get to a special feature of this low-dimensional setting: the group of orientation-preserving diffeomorphisms of a circle is connected (even better, this group deformation retracts to the group of rotations; see below for a proof). This implies that we can find an isotopy $f_t$ from $f_0$ to the identity. Choosing a collar neighborhood $\partial M \times [0,1]$ of the boundary, the desired extension is the identity off the collar neighborhood and on the collar neighborhood takes $(p,t)$ to $f_t(p)$.

Here’s how to prove that the group $G$ of orientation-preserving diffeomorphisms of the circle deformation retracts to the group $SO(2)$ of rotations. Let $\pi: \mathbb{R} \rightarrow S^1$ be the universal cover and let $x = \pi(0)$. The map $G \rightarrow S^1$ taking $g$ to $gx$ is a fiber bundle. The inclusion of $SO(2)$ into $G$ is a section. It is thus enough to prove that the fiber $G_0$ over $x$ is contractible. This is the group of orientation-preserving diffeomorphisms fixing $x$. Using covering space theory, this can be identified with the space of strictly increasing smooth maps $F: \mathbb{R} \rightarrow \mathbb{R}$ that take each integer to itself. The desired deformation retraction of $G_0$ is then just the straight-line homotopy from such an $F$ to the identity.

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  • $\begingroup$ Thank you for your detailed explanation. However, I am concerned about the assumption that $f_0$ does not map one boundary component to a different boundary component, unless the initial diffeomorphism (your second sentences) preserves each boundary component separately. $\endgroup$
    – LDLSS
    Commented Jun 26 at 9:24
  • $\begingroup$ @LDLSS: When I said that you identify the two surfaces with $M$, I meant that you do so in a way that reflects the identifications of the boundary components of the $M_i$ given by your maps. $\endgroup$ Commented Jun 26 at 9:36
  • $\begingroup$ (I edited my answer to make this explicit) $\endgroup$ Commented Jun 26 at 9:53

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