I am looking for an explicit presentation of the mapping class group of the annulus $\mathbb{A}^2$, after equipping it with $n$ interior punctures/marked points $\{x_1, \cdots, x_n\} \hookrightarrow \mathbb{A}^2$.
Here is how far I got:
The Birman-style exact sequence is (e.g. Massuyeau 2021, Thm. 3.13, p. 32)
$$ \cdots \to \mathrm{Homeos}^+(\mathbb{A}^2) \to \mathrm{Br}_n(\mathbb{A}^2) \longrightarrow \mathrm{MCG}(\mathbb{A}^2, \{x_1, \cdots, x_n\}) \longrightarrow \underset{\mbox{DT}}{\underbrace{\mathbb{Z}}} \to 1$$
(on the left: homeomorphisms preserving the orientation and fixing the boundary pointwise),
where the generator of $\mathbb{Z}$ on the right is given by Dehn-twisting (DT) the annulus (Farb & Margalit 2012 Prop. 2.4, p. 68): holding the inner boundary fixed while making a full rotation of the outer boundary.
On the other hand, the annulus is one of the exceptions to the rule that $\mathrm{Homeos}^+(\mathbb{A}^2)$ tends to be trivial (Farb & Margalit 2012, Thm. 1.14) which means that it is not immediate how to truncate to a short exact sequence on the left.
But, at least, by Theorem 1 of
- R. P. Kent & D. Pfeifer: "A Geometric and algebraic description of annular braid groups", Int. J. Algebra and Computation 12 01n02 (2002) 85-97 [doi:10.1142/S0218196702000997]
we have an explicit description of the annular braid group: it is the semidirect product
$$ \mathrm{Br}_n(\mathbb{A}^2) \,\simeq\, \mathrm{Br}_n^{\mathrm{aff}} \rtimes \underset{\mbox{Cyc}}{\underbrace{\mathbb{Z}}}$$
of the "affine braid group" (with a presentation just like Artin's, but for strands cyclically identified) with the free group generated by the operation "Cyc" of cyclically rotating all punctures (thought of as arranged equidistantly at equal radius in the annulus) "one step" in one direction.
If we assume that
$$ (\star) \;\;\;\;\; \mathrm{im}\big( \mathrm{Homeos}^+(\mathbb{A}^2) \to \mathrm{Br}_n(\mathbb{A}^2) \big) \overset{?}{\simeq} 1$$
were trivial, then
$$ 1 \overset{?}{\to} \big(\mathrm{Br}^{\mathrm{aff}}_n \rtimes \underset{\mbox{Cyc}}{\underbrace{\mathbb{Z}}}\big) \longrightarrow \mathrm{MCG}(\mathbb{A}^2, \{x_1, \cdots, x_n\}) \longrightarrow \underset{\mbox{DT}}{\underbrace{\mathbb{Z}}} \to 1 \,.$$
would imply that
$$ \mathrm{MCG}(\mathbb{A}^2, \{x_1, \cdots, x_n\}) \;\overset{?}{\simeq}\; \big(\mathrm{Br}^{\mathrm{aff}}_n \rtimes \underset{\mbox{Cyc}}{\underbrace{\mathbb{Z}}}\big) \rtimes \underset{\mbox{DT}}{\underbrace{\mathbb{Z}}} \,,$$
which is claimed in Remark 1.2 (p. 6) of:
- A. Gadbled, A.-L. Thiel & E. Wagner: "Categorical action of the extended braid group of affine type A", Comm. Contemporary Mathematics 19* 03 (2017) 1650024 [arXiv:1504.07596, doi;10.1142/S0219199716500243]
In fact, these authors also claim that the generators DT and Cyc commute --- which, I suppose, follows by noticing that both generators are given by radius-dependent rotations, and these all commute with each other.
But if, with assumption $(\star)$, we really have a short exact sequence as above, doesn't the Dehn twist DT actually conjugate the Artin generators to themselves, too (just shearing the form of the corresponding "whirl"-homeo), so that, under assumption $(\star)$, we'd actually have a direct product
$$ \mathrm{MCG}(\mathbb{A}^2, \{x_1, \cdots, x_n\}) \;\overset{?}{\simeq}\; \big(\mathrm{Br}^{\mathrm{aff}}_n \rtimes \mathbb{Z}\big) \times \mathbb{Z}$$
?
Finally: What's the image of the Dehn twist under the canonical map to the symmetric group? I gather the image is always the trivial permutation -- but I am not sure how to see this.