"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.


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?

  • $\begingroup$ Your last sentence sounds right but I don't think the reverse implication holds. $\endgroup$
    – Will Sawin
    Mar 1, 2022 at 16:05
  • $\begingroup$ I think $S_g$ here means the closed surface of genus $g$, i.e., $\partial S_g=\varnothing$. $\endgroup$ Mar 1, 2022 at 18:01

2 Answers 2


I do not recall the conventions adopted in the Primer, but there is a wide difference between boundary components (which will be embedded circles or lines) and punctures (which are “missing points” and are typically given by specifying a “punctured disk”). Under most conventions punctures are not “part” of the boundary (whatever that might mean…)

I greatly prefer, when possible, to avoid discussions of punctures and instead use the formalism of “marked points”. So, suppose that $S$ is a connected compact surface. Suppose that $Z \subset S$ is a finite set disjoint from the boundary. Then a mapping class of $(S, Z)$ is any homeomorphism of $S$ that permutes the points of $Z$. A pure mapping class of the pair $(S,Z)$ is required to fix each component of $\partial S$ setwise and also fix $Z$ pointwise.

There are further variations (if $Z$ is allowed to meet the boundary, if we fix the boundary pointwise, and so on). It really depends on what you want to do.

  • 3
    $\begingroup$ While boundary components are not the same thing as punctures/marked points, you can relate them. Given a mapping class fixing a boundary component, you can collapse the component to a single (marked) point and get a mapping class fixing a marked point, giving a homomorphism of mapping class groups, but there is no inverse map and it's not an isomorphism. I suspect this may be important to what the OP is trying to figure out, but I don't know. $\endgroup$
    – Will Sawin
    Mar 1, 2022 at 17:22
  • 2
    $\begingroup$ The last sentence of the original post claims that punctures are part of the boundary. That suggests to me that the poster is a bit confused (which is good, as that is how we learn, etc). $\endgroup$
    – Sam Nead
    Mar 1, 2022 at 17:31

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.


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