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The pure mapping class group $\mathrm{PMod}(S)$ consists of orientation-preserving mapping classes fixing every puncture individually. If $S^{\ast }$ adds a distinguished puncture and $\gamma$ is a loop in $S$, the point-pushing map moves that puncture once around $\gamma$ while leaving the old punctures fixed. For a simple loop, a thin annular neighborhood of $\gamma$ has boundary curves ${\gamma }_{+}$ and ${\gamma }_{-}$, and with consistent twist conventionsThe Birman exact sequence isfor the finite-type negative-Euler-characteristic surfaces under consideration.

First forget one puncture from the four-punctured sphere. Parts a and b giveNow forget the fifth puncture of $\Sigma =S_{0,5}$. The Birman exact sequence gives

Choose a minimal free generating pair $x,y$ of $F_2$ and lifts $\tilde{x},\tilde{y}$ in $\mathrm{PMod}(\Sigma )$. Let $F=⟨\tilde{x},\tilde{y}⟩$. 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 $g\in \mathrm{PMod}(\Sigma )$, 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 sphereThe minimal generating-set sizes are therefore

The Birman exact sequence and $\mathrm{PMod}(S_{0,3})=1$ identify $\mathrm{PMod}(S_{0,4})$ with ${\pi }_1(S_{0,3})$, the free group of rank two.

Forgetting the fifth puncture gives a split Birman exact sequenceLifts of two free generators generate a copy of $F_2$ meeting the point-pushing kernel $F_3$ trivially.