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If $\mathcal{R}'$ is a closed subsurface of a hyperbolic surface $\mathcal{R}$, then there is an inclusion homomorphism between the mapping class groups:

$$\text{Mod}(\mathcal{R}')\longrightarrow \text{Mod}(\mathcal{R})$$

I am concerned with the situation where $\mathcal{R}''$ is a general subsurface of $\mathcal{R}$. Such a surface has some handles, a number of boundaries, and a number of punctures. The presence of punctures makes the surface non-closed assuming that the boundary curves belong to the boundaries of $\mathcal{R}''$. It is possible some of the boundaries and/or punctures of $\mathcal{R}''$ are also boundaries and/or punctures of $\mathcal{R}$, i.e :

$$\partial\mathcal{R}''\cap \partial\mathcal{R}\ne \emptyset$$

Here a mapping-class fixes the boundary but can permute the punctures.

There are two questions:

  • Is there an inclusion homomorphism between the mapping-class groups in this case, i.e. does a homomorphism $\text{Mod}(\mathcal{R}'')\longrightarrow \text{Mod}(\mathcal{R})$ exist?
  • If yes, when is such a homomorphism injective? In particular, when $\text{Mod}(\mathcal{R}'')$ is a subgroup of $\text{Mod}(\mathcal{R})$?

A good reference is highly appreciated.

The case where the mapping-classes fix the punctures and all of the boundaries of $\mathcal{R}''$ belong to the interior of $\mathcal{R}$ is treated in Geometric Subgroups of Mapping Class Groups.

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I don't really understand the question, perhaps, but if the homeomorphism fixes the boundary, you can extend it by identity to the rest of the surface. This seems to be a homomorphism. Having it be injective is a lot to ask for, for example, the three-punctured sphere has rather complicated (planar) subsurfaces, and trivial mapping class group.

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  • $\begingroup$ Thank you for the answer. The case where the subsurface is closed is treated on page 83 of a primer on mapping-class group and the condition for injectivity is given. However, I am most interested in the case that the subsurface $\mathcal{R}''$ can be obtained by removing a number of pair of pants from the original surface $\mathcal{R}$. If $\mathcal{R}$ is a punctured surface, then by removing one or more pair(s) of pants we are left with a surface with boundaries and punctures. I am most interested in this case. $\endgroup$
    – QGravity
    Commented Oct 16, 2017 at 2:38
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    $\begingroup$ @QGravity, it sounds like you're interested in the case when every puncture of $\mathcal{R}''$ is a puncture of $\mathcal{R}$. In this case, the homomorphism is injective, for the simple reason that any homotopy can be extended across the boundary by the identity. $\endgroup$
    – HJRW
    Commented Nov 15, 2017 at 7:19

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