21.15 (2026)

*No, not necessarily: take $\Omega$ countably infinite, $B$ a subgroup of $S_\Omega$ which is not Borel, and $\phi : S_\Omega \ast_B S_\Omega \to S_\Omega$ an injective group homomorphism. Let $\phi_1$ be the restriction of $\phi$ to the first copy of $S_\Omega$ and $\phi_2$ that to the second. Each $\phi_i$ is continuous (A. S. Kechris, C. Rosendal, Proc. Lond. Math. Soc., 94, no. 2 (2007), 302–350), so $\text{im}(\phi_i)$ is Borel in $S_\Omega$ (as a continuous injective image of a Polish space). Now $B = \phi_1^{-1}(\phi(B)) = \phi_1^{-1}(\text{im}(\phi_1) \cap \text{im}(\phi_2))$ is Borel, a contradiction. (S. M. Corson, Letter of 26 January 2026; see also https://kourovkanotebookorg.wordpress.com/wp-content/uploads/2026/03/solution-of-21.15.pdf).