Let $S\_g^n$ be a surface of genus g with n boundaries and let $Mod(S\_g^n)$ be its mapping class group.
We will also denote by $S\_{g,m}^n$ a surface of genus g with n boundaries and m punctures.


The classical Birman exact sequence is


$$ 1 \to \pi\_1(UT(S\_g^{n-1}) \to Mod(S\_g^n) \to Mod(S\_g^{n-1}) \to 1$$


where $UT(S\_g^{n-1})$ is the unitary tangent bundle over $S\_g^{n-1}$ and the last map corresponds to capping one of the boundaries by a disk.


I am wondering what is the kernel of the map $f \colon Mod(S\_g^n) \to Mod(S\_g^{n-k})$ obtained by capping $k$ boundaries by $k$ disks.


It is a known fact, proven by Birman, that the kernel of the map $Mod(S\_{g,m}) \to Mod(S\_{g,m-k})$ obtained by forgetting $k$ punctures can be identified with $B\_k(S\_{g,m})$, the braid group on $k$ strands over $S\_{g,m}$.


Therefore, if $PB\_k(S\_{g,m})$ is the pure braid group on $k$ strands over $S\_{g,m}$, my best guess is that the kernel of $f$ is isomorphic to $PB\_k(S\_{g,m}) \oplus \mathbb{Z}^k$, where $\mathbb{Z}^k$ corresponds to the free abelian group with generators the Dehn twists around the boundaries we capped. Indeed, we can factor $f$ into $f = f\_1f\_2$ where $f\_2$ caps the boundaries by punctures disks and $f\_1$ forgets the punctures.


Is this correct? If so, is there another expression for this kernel? (Similarly to how $\pi\_1(S\_g^{n-1}) \oplus \mathbb{Z}$ can be identified with $\pi\_1(UT(S\_g^{n-1})$).



Generalized Birman exact sequence for surfaces with boundaries
It's almost right, your guess.


The easiest description is geometric, generalizing Birman's description $\pi\_1(UT(S\_g^{n-1}))$. You look instead at the configuration space parametrizing $k$ distinct points in $S\_g^{n-k}$, together with unit tangent vectors at each of these $k$ points. So this is an open subspace of the cartesian product $UT(S\_g^{n-k})^k$. The fundamental group of this space is the group you care about. It is sometimes called the *framed* braid group.


The framed braid group is an extension of the pure braid group $PB\_k(S^m\_g)$ by $\mathbf Z^k$, as one sees geometrically by fibering the framed configuration space over the usual configuration space with fiber $(S^1)^k$. For an open surface the tangent bundle is trivial, which implies that this $(S^1)^k$-bundle is trivial, and thus the extension is trivial, too. In this case your guess is correct. But for closed surfaces you get a nontrivial central extension.


