Let $T$ be stable and $M \models T$. Given two types $p(x) \in S_x(M)$ and $q(y) \in S_y(M)$ there is a canonical way to get a type in $S_{xy}(M)$. Define $$p(x) \otimes q(y) = tp_{xy}(ab/M)$$ where $a$ realises $p$ and $b$ realises the unique non-forking extension of $q$ to $Ma$. This gives a canonical map $$S_x(M) \times S_y(M) \to S_{xy}(M)$$ I am wondering about the relationship of this map with the topologies on spaces of types. Does it satisfy some universal property? Is it even continuous?
I'll give some motivation for my question. We can view the structure $M$ in the multisorted language with sorts $M, M^2, M^3, ...$ Assume for simplicity that the language has relation symbols only. Say a binary relation is a relation on the sort $M^2$, etc. To establish connection between sorts we add the canonical functions $f_{ij} : M^i \times M^j \to M^{i+j}$ to the language and this is the only way that the sorts interact.
Now imagine we want to "compactify" our structure in some sense. Then for sorts we can take $S_1(M), S_2(M), S_3(M), ...$ We can identify $M$ with the dense subset of realised types in $S_1(M)$. And similarly for $M^2$ and $M^3$, etc. I imagine the above map $S_i(M) \times S_j(M) \to S_{i+j}(M)$ as being the appropriate interpretation of $f_{ij}$. But I feel there needs to be some universal property that justifies this.