U}$ such that $f^*_x=h \circ r$, where $r \colon G^* \to S_{G,\Delta}(M)$ is the obvious map (namely $a \mapsto \operatorname{{tp}}_\Delta(a/M)$). The function $h$ is uniquely determined by $f^*_x$, and we see that $h(\operatorname{{tp}}_{\Delta}(e/M))=x$.

\[remark: homomorphism of ambits\] The above map $h$ yields a homomorphism from the $G$-ambit $(G,S_{G,\Delta}(M), \operatorname{{tp}}_{\Delta}(e/M))$ to $(G,{\mathcal U}, x)$.

It remains to show that $h(gp)=gh(p)$ for any $g \in G$ and $p \in S_{G,\Delta}(M)$. By the explicit formula for $f^*_x$, we have $$\{gh(p)\}=g\bigcap_{\varphi \in p} \operatorname{{cl}}(f_x[\varphi(M)])=\bigcap_{\varphi \in p} \operatorname{{cl}}(f_x[g\varphi(M)]) = \bigcap_{\psi \in gp} \operatorname{{cl}}(f_x[\psi(M)])=h(gp).$$

If $(G,S_{G,\Delta}(M), \operatorname{{tp}}_{\Delta}(e/M))$ was a $\Delta$-definable $G$-ambit, we could proceed with the development of the theory exactly as in the well-understood externally definable context. However, in general, this ambit is not necessarily $\Delta$-definable (see Example \[example: E non-trivial\]). Recall that if all types in $S_G(M)$ are definable, then $(G,S_{G}(M), \operatorname{{tp}}(e/M))$ is definable and coincides with $(G,S_{G,\operatorname{{ext}}}(M),\operatorname{{tp}}_{\operatorname{{ext}}}(e/M))$, and the categories of definable and externally definable $G$-flows coincide. Our goal is to start to develop a theory of definable $G$-flows without the definability of types assumption.

Let $E_\Delta$ be the equivalence relation on $S_{G,\Delta}(M)$ given by $$E_\Delta(p,q) \iff h(p)=h(q).$$ In the definable context (i.e. when $\Delta$ consists of all formulas of the appropriate form), 