Solution
= Solution
The monodromy around both $0$ and $1$ must be a product of two disjoint transpositions, hence a $(2,2)$ cycle type in $S_4$. The product relation for loops around the three punctures says $σ_0σ_1σ_∞=1$. By the stated fact, $σ_0σ_1$ is either the identity or another $(2,2)$ permutation. Consequently $σ_∞=(σ_0σ_1)^{-1}$ cannot contain a $3$-cycle. But ramification index $3$ over infinity would force exactly such a cycle, a contradiction.