TITLE: Mapping class group and pure mapping class group
QUESTION [5 upvotes]: "A Primer on Mapping Class Groups" wrote

Let $\mathrm{Homeo}_+(S, \partial S)$ denote the group of orientation-preserving
homeomorphisms of $S$ that restrict to the identity on $\partial S$. $\mathrm{Mod}(S)$ is the
group of isotopy classes of elements of $\mathrm{Homeo}_+(S, \partial S)$, where isotopies
are required to fix the boundary pointwise.

and

Let $\mathrm{PMod}(S_{g,n})$ denote the pure mapping class group of $S_{g,n}$, which is
defined to be the subgroup of $\mathrm{Mod}(S_{g,n})$ consisting of elements that
fix each puncture individually.

What is the difference between the two definitions? Doesn't "restrict to the identity on $\partial S$" mean for for all $x\in \partial S$ we have $\phi(x)=\mathrm{id}(x)=x$, which implies $\phi$ fixes each puncture individually?

REPLY [12 votes]: Just to give an explicit description of the difference: if one takes a loop "around a boundary component," the Dehn twist around this loop is not isotopic to the identity. On the other hand, if one takes a loop around a puncture, it is isotopic to the identity, by "sliding the loop" to the puncture.
In fact this is the only difference between the two groups if I remember correctly. If $(S,\partial S)$ is the surface with boundary obtained by deleting open discs around each puncture of $S_{g,n}$, the inclusion $(S, \partial S)\to S_{g,n}$ induces a map $\text{Mod}(S,\partial S)\to \text{PMod}(S_{g,n})$. This map is surjective with kernel the free abelian group generated by the Dehn twists I describe above.