an $M$-definable function, then $f^*|_G\colon G \to C$ is definable.\
2) $\Delta$-definable context:\
i) If $f\colon G \to C$ is $\Delta$-definable, then it extends uniquely to a function $f^*\colon G^* \to C$ which is $\Delta$-definable over $M$. Moreover, $f^*$ is given by the formula $\{f^*(a)\}=\bigcap_{\varphi \in \operatorname{{tp}}_\Delta(a/M)} \operatorname{{cl}}(f[\varphi(M)])$.\
ii) Conversely, if a function $f^*\colon G^* \to C$ is a $\Delta$-definable over $M$, then $f^*|_G\colon G \to C$ is $\Delta$-definable.

\[rem: prolongation\] If $f\colon G \to C$ is $\Delta$-definable, then the unique function $f^*\colon G^* \to C$ which extends $f$ and is $\Delta$-definable over $M$ is also given by $\{f^*(a)\}=\bigcap_{\varphi \in \operatorname{{tp}}(a/M)} \operatorname{{cl}}(f[\varphi(M)])$.

This follows from the formula in point 2)(i) of Lemma \[prolongation\] and the observations that $\bigcap_{\varphi \in \operatorname{{tp}}(a/M)} \operatorname{{cl}}(f[\varphi(M)])$ is non-empty (which follows from the compactness of $C$) and contained in $\bigcap_{\varphi \in \operatorname{{tp}}_\Delta(a/M)} \operatorname{{cl}}(f[\varphi(M)])$.

The following remark follows easily from definitions.

\[remark: factorization\] A function $f\colon G^* \to C$ is $\Delta$-definable over $M$ if and only if there is a continuous function $h: S_{G,\Delta}(M) \to C$ such that $f=h\circ r$, where $r\colon G^* \to S_{G,\Delta}(M)$ is the obvious map $a \mapsto \operatorname{{tp}}_\Delta(a/M)$. In particular, a function $f\colon G^* \to C$ is $M$-definable if and only if there 