= Solution
First forget one puncture from the four-punctured sphere. Parts a and b give
$$
\operatorname{PMod}(S_{0,4})
\cong\pi_1(S_{0,3})\cong F_2.
$$
Now forget the fifth puncture of $\Sigma=S_{0,5}$. The <Birman exact sequence> gives
$$
1\longrightarrow K\longrightarrow\operatorname{PMod}(\Sigma)
\xrightarrow{\pi}F_2\longrightarrow1,
\qquad
K\cong\pi_1(S_{0,4})\cong F_3.
$$
Choose a minimal free generating pair $x,y$ of $F_2$ and lifts $\widetilde x,\widetilde y$ in $\operatorname{PMod}(\Sigma)$. Let $F=\langle\widetilde x,\widetilde y\rangle$. The restriction $\pi|_F:F\to F_2$ is surjective. Since $F$ is generated by two elements and $F_2$ has minimal generator number two, the stated Hopf-type fact makes $\pi|_F$ an isomorphism. Thus $F\cong F_2$ and $F\cap K=1$.
For any $g\in\operatorname{PMod}(\Sigma)$, choose $f\in F$ with $\pi(f)=\pi(g)$. Then $f^{-1}g\in K$, so $g\in FK$. The kernel is normal by exactness. We have proved the <semidirect-product decomposition of the pure mapping class group of the five-punctured sphere>
$$
\operatorname{PMod}(S_{0,5})=FK\cong F_3\rtimes F_2.
$$
The minimal generating-set sizes are therefore
$$
\boxed{d(F)=2,\qquad d(K)=3}.
$$
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