we will write $E$ instead of $E_\Delta$.

\[cor: definable ambit is a quotient\] $(G,S_{G,\Delta}(M)/E_\Delta,\operatorname{{tp}}_{\Delta}(e/M)/E_\Delta)$ is the universal $\Delta$-definable $G$-ambit.

By Remark \[remark: homomorphism of ambits\], $E_\Delta$ is closed and invariant under the action of $G$. This implies that $(G,S_{G,\Delta}(M)/E_\Delta,\operatorname{{tp}}_{\Delta}(e/M)/E_\Delta)$ naturally becomes a $G$-ambit. From the very definition of $E_\Delta$, we get that $h$ factors through the quotient map $S_{G,\Delta}(M) \to S_{G,\Delta}(M)/E_\Delta$ yielding an isomorphism from the $G$-ambit $(G,S_{G,\Delta}(M)/E_\Delta,\operatorname{{tp}}_{\Delta}(e/M)/E_\Delta)$ to $(G,{\mathcal U}, x)$. This completes the proof as $(G,{\mathcal U}, x)$ was chosen to be the universal $\Delta$-definable $G$-ambit.

\[remark: uniqueness of E\] i) $E_\Delta$ is the unique equivalence relation $F$ on $S_{G,\Delta}(M)$ for which $(G,S_{G,\Delta}(M)/F,\operatorname{{tp}}_{\Delta}(e/M)/F)$, with the action of $G$ defined by $g(p/F):=(gp)/F$, is the universal $\Delta$-definable $G$-ambit.\
ii) In the definable context (i.e. when $\Delta$ consists of all formulas of the appropriate form), if all types in $S_G(M)$ are definable, then $E$ is trivial (i.e. it is the equality).

i\) Suppose $F_1$ and $F_2$ are two such relations. It is enough to show that $F_1 \subseteq F_2$. By the universality of $(G,S_{G,\Delta}(M)/F_1,\operatorname{{tp}}_{\Delta}(e/M)/F_1)$, there is $f\colon S_{G,\Delta}(M)/F_1 \to S_{G,\Delta}(M)/F_2$ which is a homomorphism of $G$-ambits. Let $f'\colon S_{G,\Delta}(M) \to S_{G,\Delta}(M)/F_2$ be the composition of $f$ with the quotient map $S_{G,\Delta}(M) 