# Product of types in stable theories

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.

Unfortunately, this map is never continuous. The set $[x=y]$ is clopen in $S_{xy}(M)$, but its preimage in $S_x(M)\times S_y(M)$ is $\{(\mathrm{tp}(a/M),\mathrm{tp}(a/M))\mid a\in M\}$, which is not closed (for example, its closure contains the whole diagonal).
What is true is that the "nonforking extension" map is well-behaved when restricted to each fiber. That is, fix a type $q(y)\in S_y(M)$ and consider the map $i_q\colon S_x(M)\to S_{xy}(M)$ which maps a type $p(x)\in S_x(M)$ to $\mathrm{tp}(ab/M)$, where $b$ realizes $q(y)$ and $a$ realizes the unique nonforking extension of $p(x)$ to a type over $Mb$. Then $i_q$ is a homeomorphism of $S_x(M)$ onto its image, which is a closed subset of $S_{xy}(M)$.
The point is that $q$ is definable over $M$, so for every formula $\varphi(x;y)$, there is a formula $\delta(x)$ with parameters from $M$ such that $\varphi(c;y)\in q(y)$ if and only if $\delta(c)$. And $\varphi(x;y)\in i_q(p(x))$ if and only if $\delta(x)\in p(x)$. But the defining formula doesn't vary continuously in $q$ in the way which would be necessary to get a continuous map from the product space.