Why is the mapping class group of a surface with nonempty boundary torsion-free? On page 201 of Farb and Margalit's Primer on Mapping Class Groups, they explain why the mapping class group $\mathrm{Mod}(S)$ is torsion-free when $\partial S \neq \varnothing$. Here is my understanding of the argument:
Let $S$ be a surface with a hyperbolic metric and let $\phi\colon S \to S$ be an isometry fixing $\partial S$ pointwise. Then $\phi$ fixes a frame (a basis for the tangent space) at every point of $\partial S$. Since isometries of surfaces are determined by their action on a frame, we must have that $\phi = \mathrm{id}$.
The Nielsen realization theorem (Theorem 7.1) states that for every order $k < \infty$ element $f \in \mathrm{Mod}(S)$, there is an isometry $\phi \in \mathrm{Homeo}^+(S)$   of order $k$ representing $f$. However, there is no guarantee that $\phi$ will fix the boundary, we only know that $\phi$ is in the free homotopy class $f$. Up to here I understand all the points that have been made.
What I don't understand is how they go from this, and the fact that Dehn twists about boundary components have infinite order, to conclude that $\mathrm{Mod}(S)$ is torsion-free whenever $\partial S \neq 0$. Is the point that if $f$ is a torsion element, then the isometry representative $\phi$ given by the NIelsen realization theorem must fix the boundary pointwise and therefore be the identity? I don't know how you would show this. I also don't understand why Dehn twists entered the argument.
 A: I think the reason that Dehn twists enter is that we can take the differential of  a (orientation-preserving) diffeomorphism $f$ of $S_g$ that fixes a chosen basepoint $\ast \in S_g$ at this point, and will get a map $d \colon \text{Diff}^+(S_g,\ast) \to \text{GL}_2^+(\mathbb R)$. The fiber of this fibration is then $\text{Diff}_{\partial}(S_{g,1})$. When you study the effect of the resulting fiber sequence $\text{Diff}_{\partial}(S_{g,1}) \to \text{Diff}^+(S_g,\ast) \to \text{GL}_2^+(\mathbb R)$ on homotopy groups, you obtain a short exact sequence $1 \to \mathbf{Z} \to \Gamma_{g,1} \to \Gamma_g^1 \to 1$ where $\Gamma_g^1$ stands for the mapping class group of a once-punctured surface of genus $g$, and the map $\mathbf{Z} \to \Gamma_{g,1}$ is given by Dehn twisting around the boundary curve.
Ok, so far so good, but now let us prove that $\Gamma_{g,1}$ is torsion-free, by contradiction: suppose $f \in \Gamma_{g,1}$ has finite order $k > 1$. Then (as you remarked), we can find $\phi \in \text{Diff}(S_g,\ast)$ of order $k$ representing $f$. We can also find a compatible metric on $S_g$. Then $d(f)$ actually lands in $SO(2) \subset \text{GL}_2^+(\mathbb R)$ and is given by a rotation, say around angle $\theta$. Then $k\theta = \ell2\pi$ for some non-zero integer $\ell$, and you can convince yourself that $f^k$ represents a non-trivial Dehn twist in $\Gamma_{g,1}$, contradicting $f^k = 1$.
