which is definable over $M$”, so we are in the definable context.

\[remark: existence of universal definable G-ambit\] There exists a unique (up to isomorphism) universal $\Delta$-definable $G$-ambit.

Uniqueness is clear. To show existence, consider a set $(G,X_i,x_i)_{i \in I}$ of all up to isomorphism $\Delta$-definable $G$-ambits (such a set exists, as there is a common bound on the cardinalities of all $G$-ambits). Now, let $X'=\prod_{i} X_i$, $x=(x_i)_i$, and let $X$ be the closure of the orbit of $x$ under the coordinatewise action of $G$. Then $(G,X')$ is $\Delta$-definable by Remark \[definability of products\], and so $(G,X,x)$ is a $\Delta$-definable $G$-ambit. From the construction, we see that $(G,X,x)$ maps on any $\Delta$-definable $G$-ambit, i.e. it is universal.

By the assumption on the form of the formulas in $\Delta$, we see that $G$ acts on $S_{G,\Delta}(M)$ by $$g \operatorname{{tp}}_{\Delta}(a/M)=\operatorname{{tp}}_{\Delta}(ga/M)$$ so that $(G,S_{G,\Delta}(M), \operatorname{{tp}}_{\Delta}(e/M))$ is a $G$-ambit.

Let $(G,{\mathcal U}, x)$ be the universal $\Delta$-definable $G$-ambit. Then, $f_x \colon G \to {\mathcal U}$ given by $f_x(g)=gx$ is $\Delta$-definable. Thus, by Fact \[prolongation\], it extends to the function $f^*_x \colon G^* \to {\mathcal U}$ which is $\Delta$-definable over $M$. Hence, by Remark \[remark: factorization\], there exists a continuous function $h\colon S_{G,\Delta}(M) \to {\mathcal 