Fact 2.4. Assume $\mathcal M$ is t-minimal, and consider real tuples a and b, parameter sets A and B, and type-definable sets X and Y.

1. (1) $\dim (M^n)=n$ for all n.

2. (2) If $X\subseteq Y$ , then $\dim (X)\leq \dim (Y)$ .

3. (3) $\dim (X\times Y)=\dim (X)+\dim (Y)$ .

4. (4) If $f:X\rightarrow Y$ is relatively definable and surjective with finite fibres, then $\dim (X)=\dim (Y)$ .

5. (5) If $A\subseteq B$ , then $\dim (a/B)\leq \dim (a/A)$ .

6. (6) $\dim (ab/A)\leq \dim (b/A)+\dim (a/Ab)$ .

7. (7) If $\mathrm {acl}(Aa)=\mathrm {acl}(Ab)$ , then $\dim (a/A)=\dim (b/A)$ .

8. (8) $\dim (a/A)$ is the smallest dimension of an A-definable set containing A.

Assumption 2.5. From now until the end of Section 6, assume $\mathcal M$ is t-minimal.

Proof. The case when a is independent (that is, $\dim (a/A)=n$ ) is, essentially, due to Johnson. More precisely, in this case, the lemma just says that if Y is A-definable and contains a, then Y contains a neighbourhood of a. But if not, then a belongs to $Y-\mathrm {Int}(Y)$ , and thus $\dim (Y-\mathrm {Int}(Y))=n$ . By Theorem 1.7(3) of Johnson’s paper [Reference Johnson11], this implies that $Y-\mathrm {Int}(Y)$ has non-empty interior, which is absurd.

Now in the general case, let X be any minimal basic witness. So X is a minimal a-witness for some basis b, witnessed by the projection $\pi :M^n\rightarrow M^d$ with $\pi (a)=b$ .

Let $Y\subseteq X$ be any other A-definable set containing a. Since $Y\subseteq X$ , Y is also an a-witness for b over A (also witnessed by $\pi $ ). Since the fibre size above b in X is minimal among all such sets, this means X and Y agree at all points in the fibre above b. Now let $W\subseteq M^d$ be the set of points $u\in \pi (X)\cap \pi (Y)$ such that $\pi ^{-1}(u)\cap X=\pi ^{-1}(u)\cap Y$ . Then W is A-definable and contains b. By the first paragraph (applied to W in $M^d$ , using that $\dim (b/A)=d$ ), W contains a neighbourhood U of b. In particular, X and Y agree on the preimage of U in $M^n$ , which is a neighbourhood of a.

Proof. Let X be a definable set realizing g. Write $X=Y_b$ for some $\emptyset $ -definable family $\{Y_t:t\in T\}$ of sets. Let $\sim $ be the equivalence relation on $M\times T$ given by $(x,s)\sim (y,t)$ if $x=y$ and $Y_s$ and $Y_t$ agree in a neighbourhood of x. Then $\sim $ is $\emptyset $ -definable, so $T/\sim $ is a sort in $\mathcal M^{eq}$ . Now let $c\in T/\sim $ be the equivalence class of $(a,b)$ . Then one checks easily that c is a code of g.

Proof. Let Y be an A-definable set containing a. By Lemma 2.14(3), we compute:

$$ \begin{align*}\mathrm{germ}_a(Y)\geq\mathrm{germ}(a/A)=\mathrm{germ}_a(X),\end{align*} $$

which is equivalent to the statement of the lemma.

Proof. Let X be a minimal basic witness, corresponding to the projection $\pi :M^n\rightarrow M^d$ sending a to the basis $b=\pi (a)$ . Lemma 2.9 gives that X realizes $\mathrm {germ}(a/A)$ . Now $\pi $ is finite-to-one on X, so by Fact 2.4, $\dim (X)\leq d$ ; while since $a\in X$ , we also have $\dim (X)\geq d$ . So $\dim (X)=d$ , and thus X is a d-approximation.

Proof. The set $U\times M^n$ is a neighbourhood of $(a,b)$ . By the independent neighbourhood property, there is a neighbourhood $W\subseteq U\times M^n$ of $(a,b)$ defined over a parameter t with $\dim (ab/At)=\dim (ab/A)$ . Let $\pi :M^m\times M^n\rightarrow M^m$ be the projection. Then $\pi (W)\subseteq U$ is a t-definable neighbourhood of a.

Assumption 3.3. From now until the end of Section 6, assume $\mathcal M$ has the independent neighbourhood property.

Proof. We show (1); then (2) follows by definition.

To show (1), first note that Y trivially realizes $\mathrm {germ}(a/A)$ , because X does. The point is that we can also show $\dim (Y)=\dim (a/A)$ . For this, by assumption, there is a neighbourhood U of a such that X and Y agree on U. By the independent neighbourhood property, we may assume U is definable over $At$ where $\dim (a/At)=\dim (a/A)$ . Now $Y\cap U=X\cap U$ is $At$ -definable and contains a, so $\dim (Y\cap U)\geq \dim (a/A)$ , and thus $\dim (Y)\geq \dim (a/A)$ as well. Equality then follows since $Y\subseteq X$ and $\dim (X)=\dim (a/A)$ .

Proof.

1. (1) Let Z be an A-definable d-approximation of $\mathrm {tp}(a/A)$ at a (for example, any minimal basic witness of a over A). Then X and Z agree on some neighbourhood of a, say U. By the independent neighbourhood property, we may assume U is $At$ -definable, where $\dim (a/At)=\dim (a/A)$ . Now take $Y=Z\cap U$ . Then Y is $At$ -definable and realizes $\mathrm {germ}(a/A)$ ; while we have $Y=Z\cap U=X\cap U\subseteq X$ , so Y is a d-approximation by Lemma 3.4.

2. (2) Let X be any d-approximation of $\mathrm {tp}(a/A)$ . Then $X\cap U$ is also a d-approximation of $\mathrm {tp}(a/A)$ , by Lemma 3.4. Now apply (1) to $X\cap U$ .

Proof. As in the proof of Corollary 2.22, we automatically have $\dim (X)=d$ . To prove the lemma, we show that $\mathrm {germ}_a(X)$ is cofinal in $\mathrm {tp}(a/A)$ . So let $Y\subseteq X$ be A-definable with $a\in Y$ . We will show that X and Y agree in a neighbourhood of a.

Let $\pi :M^n\rightarrow M^d$ demonstrate that X is a basic witness; so $a\in X$ , $\pi $ is finite-to-one on X and $b=\pi (a)$ is a basis of a over A.

Since $\pi $ is finite-to-one on X, clearly there is a neighbourhood U of a such that $X\cap U$ and $Y\cap U$ agree on the fibre above b (one chooses U small enough that $X\cap U$ has only the single point a lying above b). By the independent neighbourhood property, we may assume U is $At$ -definable where $\dim (a/At)=\dim (a/A)$ . So

$$ \begin{align*} & \dim(a/A)=\dim(a/At)=\dim(ab/At) \le \\ & \dim(b/At)+\dim(a/Abt)=\dim(b/At)\le \\ & \dim(a/At)=\dim(a/A), \end{align*} $$

and it follows that b is still a basis for a over $At$ . Thus, $\dim (b/At)=d$ , and so any $At$ -definable property of b holds on a neighbourhood of b. In particular, this implies that $X\cap U$ and $Y\cap U$ agree on all fibres above points in some neighbourhood of b; and as in Lemma 2.9, this implies the desired statement.

Proof. (1) $\Rightarrow $ (2): Let X be a basic witness of a over A. If $\dim (a/B)=\dim (a/A)$ , then by definition, X is also a basic witness of a over B (This is a subtle use of independent neighbourhoods: if we only knew that minimal basic witnesses were d-approximations, the same argument would fail). By Lemma 3.6, X is a d-approximation of both $\mathrm {tp}(a/A)$ and $\mathrm {tp}(a/B)$ at a.

(2) $\Rightarrow $ (3): By definition of d-approximations, (2) gives that $\mathrm {tp}(a/A)$ and $\mathrm {tp}(a/B)$ have the same dimension and the same germ at a, and thus they have the same d-approximations at a.

(3) $\Rightarrow $ (4): Clear, since a d-approximation contains the data of the relevant germ.

(4) $\Rightarrow $ (1): Assume $\mathrm {germ}(a/B)=\mathrm {germ}(a/A)$ . Then by Lemma 3.4, we have

$$ \begin{align*}\dim(a/B)=\dim(\mathrm{germ}(a/B))=\dim(\mathrm{germ}(a/A))=\dim(a/A).\\[-35pt]\end{align*} $$

Proof. (1) $\Rightarrow $ (2): Assume (1). Let c be a basis of a over A, and let d be a basis of b over A. Then $ab\in \mathrm {acl}(Acd)$ . If we can show that the concatenated tuple $cd$ is independent over A, it will follow that $cd$ is a basis of $ab$ over A, which implies (2). To show this, assume $cd$ is not independent over A. Then there is some coordinate from $cd$ that is algebraic over A with all other coordinates of $cd$ . Call this coordinate $e\in M$ . If e occurs in c, then c is not independent over $Ad$ , which implies that $\dim (a/Ab)<\dim (a/A)$ . Similarly, if e occurs in d, then $\dim (b/Aa)<\dim (b/A)$ . In either case, we contradict (1).

(2) $\Rightarrow $ (3): Let X and Y be basic witnesses for a and b over A, respectively. Assuming (2), we have

$$ \begin{align*}\dim(X\times Y)=\dim(X)+\dim(Y)=\dim(a/A)+\dim(b/A)=\dim(ab/A),\end{align*} $$

which implies easily that $X\times Y$ is a basic witness for $ab$ over A (As in Lemma 3.7, this argument fails with minimal basic witnesses, so we are really using independent neighbourhoods). This implies (3) by Lemma 3.6.

(3) $\Rightarrow $ (4): Assuming (3), we conclude that $\dim (ab/A)=\dim (a/A)+\dim (b/A)$ and $\mathrm {germ}(ab/A)=\mathrm {germ}(a/A)\times \mathrm {germ}(b/A)$ . This implies (4) by definition.

(4) $\Rightarrow $ (5): By definition of d-approximations.

(5) $\Rightarrow $ (1): Assume $\mathrm {germ}(ab/A)=\mathrm {germ}(a/A)\times \mathrm {germ}(b/A)$ . Then by Lemmas 3.4 and 2.14(4), we have

$$ \begin{align*}\dim(ab/A)=\dim(\mathrm{germ}(ab/A))=\dim(\mathrm{germ}(a/A)\times\mathrm{germ}(b/A))\end{align*} $$

$$ \begin{align*}=\dim(\mathrm{germ}(a/A))+\dim(\mathrm{germ}(b/A))=\dim(a/A)+\dim(b/A).\end{align*} $$

That is, we have shown that (5) implies (2). But (2) easily implies (1) by sub-additivity.

Proof. Combining additivity over A with sub-additivity over B, we have:

$$ \begin{align*}\dim(ab/B)\leq\dim(b/B)+\dim(a/Bb)\leq\dim(b/A)+\dim(a/Ab)=\dim(ab/A).\end{align*} $$

Since $\dim (ab/B)=\dim (ab/A)$ , all of the above inequalities must be equalities. This implies (1), (2) and (3).

Proof. By Lemma 2.21, we may assume X is definable over $At$ where $\dim (ab/At)=\dim (ab/A)$ . By Lemma 3.13, we have $\dim (b/At)=\dim (b/A)$ . It follows by Lemma 3.7 that $\mathrm {germ}(b/At)=\mathrm {germ}(b/A)$ .

On the other hand, note that $\pi (X)$ is an $At$ -definable set containing b. Applying Lemma 2.15(3), we obtain

$$ \begin{align*}\mathrm{germ}(b/A)=\mathrm{germ}(b/At)\leq\mathrm{germ}_b(\pi(X)).\end{align*} $$

Now let Y be any d-approximation of $\mathrm {tp}(b/A)$ at b. Then $\mathrm {germ}_b(Y)\leq \mathrm {germ}_b(\pi (X))$ . It follows that $Y\cap \pi (X)$ contains a neighbourhood of b in Y, so by Lemma 3.4, it is a d-approximation of $\mathrm {tp}(b/A)$ at b. Since $Y\cap \pi (X)$ is contained in $\pi (X)$ , the lemma is now proved.

Proof. Fix A-definable d-approximations X and Y of $\mathrm {tp}(ab/A)$ at $(a,b)$ and $\mathrm {tp}(b/A)$ at b, respectively. Shrinking X if necessary, we may assume $\pi (X)\subseteq Y$ (here we use that $X\cap \pi ^{-1}(Y)$ is also A-definable, so is also a d-approximation). Recall that:

1. (1) Every neighbourhood of $(a,b)$ in X is also a d-approximation of $\mathrm {tp}(ab/A)$ at $(a,b)$ , by Lemma 3.4; and

2. (2) Every d-approximation of $\mathrm {tp}(b/A)$ at b contains a neighbourhood of b in Y, by definition.

In particular, combining (1) and (2) with Lemma 3.14 gives that for every neighbourhood N of $(a,b)$ in X, $\pi (N)$ contains a neighbourhood of b in Y. This is an A-definable property of $(a,b)$ (since it is enough to verify it for the definable neighbourhood basis), so by Lemma 2.21, it holds on a neighbourhood of $(a,b)$ in X; that is, there is a neighbourhood $X'$ of $(a,b)$ in X such that $\pi :X'\rightarrow Y$ is open. Let $Y'=\pi (X')$ , so that $Y'$ is open in Y and $\pi :X'\rightarrow Y'$ is open and surjective. Then by Lemma 3.4, $X'$ and $Y'$ are d-approximations of $\mathrm {tp}(ab/A)$ at $(a,b)$ and $\mathrm {tp}(b/A)$ at b, respectively, and this proves the desired statement.

Proof. Since $a\in \mathrm {acl}(Ab)$ , we have $\dim (ab/A)=\dim (b/A)$ and $\dim (a/Ab)=0$ ; thus $(a,b)$ is additive over A. By Lemma 3.15, we can choose d-approximations X and Y so that $\pi :X\rightarrow Y$ is surjective and open. As a projection, $\pi $ is automatically continuous. The only missing ingredient is ensuring that $\pi $ is injective on X.

In addition to the above, since $a\in \mathrm {acl}(Ab)$ , there is an A-definable set Z containing $(a,b)$ so that $\pi $ is finite-to-one on Z. By Lemma 2.15(3),

$$ \begin{align*}\mathrm{germ}_{(a,b)}(Z)\geq\mathrm{germ}(ab/A)=\mathrm{germ}_{(a,b)}(X).\end{align*} $$

Thus, $X\cap Z$ contains a neighbourhood of $(a,b)$ in X, so is also a d-approximation of $\mathrm {tp}(ab/A)$ at $(a,b) $ by Lemma 3.4.

Next, since $\pi $ is finite-to-one on Z, there is a neighbourhood U of $(a,b)$ so that

$$ \begin{align*}\pi^{-1}(b)\cap Z\cap U=\{(a,b)\}.\end{align*} $$

By Lemma 3.4(1), $X\cap Z\cap U$ is still a d-approximation of $\mathrm {tp}(ab/A)$ at $(a,b)$ . Then by Lemma 3.5(1), there is moreover a d-approximation $X'\subseteq (X\cap Z\cap U)$ which is definable over $At$ for some tuple t with $\dim (ab/At)=\dim (ab/A)$ . Let $Y'=\pi (X')$ . Then $Y'$ is open in Y, so is still a d-approximation of $\mathrm {tp}(b/A)$ at b by Lemma 3.4, and $\pi :X'\rightarrow Y'$ is still open and continuous.

Since $a\in \mathrm {acl}(Ab)$ , it follows as in the proof of Lemma 3.6 that also $\dim (b/At)=\dim (b/A)$ , so that $Y'$ is also a d-approximation of $\mathrm {tp}(b/At)$ at b. Now let $Y"\subseteq Y'$ be the set of $y\in Y'$ so that $\pi ^{-1}(y)\cap X'$ has only one element. Then $Y"$ is $At$ -definable and contains b, and thus $Y"$ contains a neighbourhood of b in $Y'$ . So by Lemma 3.4(1), $Y"$ is still a d-approximation of $\mathrm {tp}(b/A)$ at b. Finally, let $X"=X'\cap \pi ^{-1}(Y")$ . Then $X"$ contains a neighbourhood of $(a,b)$ in $X'$ , so is still a d-approximation of $\mathrm {tp}(ab/A)$ at $(a,b)$ ; and now the map $\pi :X"\rightarrow Y"$ is simultaneously open, continuous, surjective and injective, and is thus a homeomorphism.

Proof. Let $\pi :X\rightarrow M^k$ be a projection whose image has non-empty interior, as provided by the assumption that X is a very weak k-cell. Let $e=\pi (a)$ , and let $b=f(a)$ . The assumptions imply that $\dim (e/A)=\dim (a/A)=\dim (ab/A)=k$ , and thus e is a basis for both a and $ab$ over A. Letting $\Gamma \subseteq X\times Y$ be the graph of f, it now follows that $\Gamma $ and X are basic witnesses of $ab$ and a over A, respectively, and thus are d-approximations of $\mathrm {tp}(ab/A)$ and $\mathrm {tp}(a/A)$ at $(a,b)$ and a, respectively.

Now by Lemma 3.16, there are d-approximations $X'$ of $\mathrm {tp}(a/A)$ and $\Gamma '$ of $\mathrm {tp}(ab/A)$ such that the projection $(a,b)\mapsto a$ induces a homeomorphism $\Gamma '\rightarrow X'$ . Then $\mathrm {germ}_a(X')=\mathrm {germ}_a(X)$ , and $\mathrm {germ}_{(a,b)}(\Gamma ')=\mathrm {germ}_{(a,b)}(\Gamma )$ . So, letting $X"=X\cap X'$ and $\Gamma "=\Gamma \cap \Gamma '$ , it follows that $X"$ is a neighbourhood of a in X and $\Gamma "$ is a neighbourhood of $(a,b)$ in $\Gamma $ . Note that if $x\in X"$ , then $(x,f(x))\in \Gamma "$ (and vice versa) – thus $\Gamma "$ projects homeomorphically to $X"$ . Finally, on $X"$ , f factors as $X"\rightarrow \Gamma "\rightarrow Y$ , a composition of two continuous maps.

Proof. We show (1). Then (2) follows by applying (1) twice (the second time with $\tau $ and $\tau '$ reversed).

Now to prove (1), we will show that for cofinitely many $a\in M$ , every definable $\tau $ -neighbourhood of a contains a definable $\tau '$ -neighbourhood of a. Since this is a definable property of a, by compactness, it will suffice to prove it for any $a\in M-\mathrm {acl}(\emptyset )$ . So fix such an a.

Since $a\in M-\mathrm {acl}(\emptyset )$ , we have $\dim (a)=1$ . Now let U be a definable $\tau $ -neighbourhood of a. It is harmless to shrink U, so by the independent neighbourhood property, we may assume U is definable over a tuple t with $\dim (a/t)=1$ . By t-minimality, the $\tau '$ -boundary of U is finite (Remark 2.3), and is thus contained in $\mathrm {acl}(t)$ . So a does not belong to the $\tau '$ -boundary of U, and thus a belongs to the $\tau '$ -interior of U. This shows that $U'$ contains a definable $\tau '$ -neighbourhood of a, as desired.

Proof. (1) $\Rightarrow $ (2): Assume $b\in \mu (a/A)$ . Let X be any $\mathcal M$ -definable d-approximation of $\mathrm {tp}(a/A)$ at a. By definition, $b\in X$ . So to prove (2), it suffices to show that $b\in \mu (a)$ . For this, let U be any $\mathcal M$ -definable neighbourhood of a. By Lemma 3.4, $X\cap U$ is a d-approximation of $\mathrm {tp}(a/A)$ at a, and is also $\mathcal M$ -definable. Thus, $b\in X\cap U\subseteq U$ , as desired.

(2) $\Rightarrow $ (3): Assume (2). We need to show that $b\models \mathrm {tp}(a/A)$ . For this, let X be A-definable with $a\in X$ . By (2), there is an $\mathcal M$ -definable d-approximation Y of $\mathrm {tp}(a/A)$ at a with $b\in Y$ . Then

$$ \begin{align*}\mathrm{germ}_a(X)\geq\mathrm{germ}(a/A)=\mathrm{germ}_a(Y),\end{align*} $$

so there is a neighbourhood U of a with $Y\cap U\subseteq X\cap U$ . As this is an elementary property, and X and Y are $\mathcal M$ -definable, we can choose U to also be $\mathcal M$ -definable. By (2), $b\in \mu (a)$ , and thus $b\in U$ . So $b\in Y\cap U\subseteq X\cap U\subseteq X$ , as desired.

(3) $\Rightarrow $ (1): Assume (3). Fix an A-definable d-approximation Y of $\mathrm {tp}(a/A)$ at a. To show (1), let X be any $\mathcal M$ -definable d-approximation of $\mathrm {tp}(a/A)$ at a. Then $\mathrm {germ}_a(X)=\mathrm {germ}_a(Y)$ , so there is a neighbourhood U of a so that $X\cap U=Y\cap U$ . Since X and Y are $\mathcal M$ -definable, we can choose U to also be $\mathcal M$ -definable. By (3), $b\in Y\cap U$ , and thus $b\in X\cap U\subseteq X$ , as desired.

Proof. It is clear that (1) implies (2), because any d-approximation of $\mathrm {tp}(a/A)$ at a realizes $\mathrm {germ}(a/A)$ . We show only (1). So let X be an $\mathcal M$ -definable d-approximation of $\mathrm {tp}(a/A)$ at a. Then Lemma 4.5(2) gives that $\mu (a/A)$ is the intersection of a small number of open subsets of X – namely those of the form $X\cap U$ where U is an $\mathcal M$ -definable open neighbourhood of a (here ‘small’ is in the sense of $\mathcal N$ , not $\mathcal M$ ). (1) now follows by applying Fact 2.1 in the structure $\mathcal N$ .

Proof.

1. (1) This follows by Lemma 4.5(3), because every realization of $\mathrm {tp}(a/B)$ also realizes $\mathrm {tp}(a/A)$ .

2. (2) First, suppose a is independent from B over A. Then by Lemma 3.7(3), $\mathrm {tp}(a/A)$ and $\mathrm {tp}(a/B)$ have the same d-approximations at a, which by definition gives that $\mu (a/A)=\mu (a/B)$ . Conversely, if $\mu (a/A)=\mu (a/B)$ , then

   $$ \begin{align*}\dim(a/A)=\dim(\mu(a/A))=\dim(\mu(a/B))=\dim(a/B),\end{align*} $$
   so a is independent from B over A.

Proof.

1. (1) By Lemma 4.5(3). Namely, suppose $(x,y)\in \mu (ab/A)$ . Then $(x,y)\in \mu (a,b)$ , so $x\in \mu (a)$ and $y\in \mu (b)$ . Moreover, we have $(x,y)\models \mathrm {tp}(ab/A)$ , and thus $x\models \mathrm {tp}(a/A)$ and $y\models \mathrm {tp}(b/A)$ . Thus, by Lemma 4.5(3), $x\in \mu (a/A)$ and $y\in \mu (b/A)$ .

2. (2) First, suppose a and b are symmetrically independent over A. By (1), we only need to show that if $x\in \mu (a/A)$ and $y\in \mu (b/A)$ , then $(x,y)\in \mu (ab/A)$ . It is automatic that $(x,y)\in \mu (a,b)$ , since $x\in \mu (a)$ and $y\in \mu (b)$ . So by Lemma 4.5(2), it suffices to show that $(x,y)$ belongs to some $\mathcal M$ -definable d-approximation of $\mathrm {tp}(ab/A)$ at $(a,b)$ . Now by Lemma 3.10, there are $\mathcal M$ -definable d-approximations X of $\mathrm {tp}(a/A)$ at a and Y of $\mathrm {tp}(b/A)$ at b, so that $X\times Y$ is a d-approximation of $\mathrm {tp}(ab/A)$ at $(a,b)$ . Since $x\in \mu (a/A)$ and $y\in \mu (b/A)$ , we have $x\in X$ and $y\in Y$ , and thus $(x,y)\in X\times Y$ , as desired.

   Now assume $\mu (ab/A)=\mu (a/A)\times \mu (b/A)$ . Then

   $$ \begin{align*}\dim(\mu(ab/A))=\dim(\mu(a/A))+\dim(\mu(b/A)),\end{align*} $$
   so $\dim (ab/A)=\dim (a/A)+\dim (b/A)$ , and thus a and b are symmetrically independent over A.

Proof.

1. (1) Let $\pi $ be the projection sending $(a,b)$ to b. By Lemma 3.15, there are $\mathcal M$ -definable d-approximations X of $\mathrm {tp}(ab/A)$ at $(a,b)$ , and Y of $\mathrm {tp}(b/A)$ at b, so that $\pi $ induces an open surjective map $X\rightarrow Y$ . By Lemma 4.6, $\mu (ab/A)$ and $\mu (b/A)$ are open in X and Y, respectively, which implies that the projection $\mu (ab/A)\rightarrow \mu (b/A)$ is also open.

   To show $\mu (ab/A)\rightarrow \mu (b/A)$ is surjective, let $y\in \mu (b/A)$ . By compactness, it suffices to show that for any $\mathcal M$ -definable d-approximation Z of $\mathrm {tp}(ab/A)$ at $(a,b)$ , we have $y\in \pi (Z)$ . But by Lemma 3.14, $\pi (Z)$ contains an $\mathcal M$ -definable d-approximation of $\mathrm {tp}(b/A)$ at b; thus $\pi (Z)$ contains $\mu (b/A)$ , and thus $y\in \pi (Z)$ .

2. (2) Repeat the proof of (1), but starting with d-approximations X and Y so that $X\rightarrow Y$ is moreover injective (as provided by Lemma 3.16).

3. (3) Apply (2) to both $\mu (ab/A)\rightarrow \mu (a/A)$ and $\mu (ab/A)\rightarrow \mu (b/A)$ .

Proof. (1) follows from Lemma 4.5(3): we have $x\in \mu (a/Ab)$ if and only if $x\in \mu (a)$ and $x\models \mathrm {tp}(a/Ab)$ ; and this is equivalent to the assertion that $(x,b)\in \mu (a,b)$ and $(x,b)\models \mathrm {tp}(ab/A)$ .

(2) now follows from (1) by applying Lemma 4.6(2) to both sides of the equation. That is, by Lemma 4.6(2), we can rewrite the desired statement as $\mathrm {germ}_a(\mu (ab/A)_b)=\mathrm {germ}_a(\mu (a/Ab))$ . But (1) gives that $\mu (ab/A)_b=\mu (a/Ab)$ , so of course the two sets have the same germ at a.

Now for (3), assume $B\supset A$ and $\dim (ab/B)=\dim (ab/A)$ . By Lemma 3.7, $\mathrm {germ}(ab/B)=\mathrm {germ}(ab/A)$ . Now applying (2) gives

$$ \begin{align*}\mathrm{germ}(a/Bb)=(\mathrm{germ}(ab/B))_b=(\mathrm{germ}(ab/A))_b=\mathrm{germ}(a/Ab).\end{align*} $$

Then by Lemma 3.7 again, $\dim (a/Bb)=\dim (a/Ab)$ .

Proof. Let X be an A-definable d-approximation of $\mathrm {tp}(ab/A)$ at $(a,b)$ . By Lemma 4.10(2), the fibre $X_b$ realizes $\mathrm {germ}(a/Ab)$ . Now if $(x,y)\in \mu (ab/A)$ , then $\mathrm {tp}(xy/A)=\mathrm {tp}(ab/A)$ . So $X_y$ must realize $\mathrm {germ}(x/Ay)$ . Since $\mu (ab/A)$ is open in X, the same germ is also realized by $\mu (ab/A)_y$ .

Proof. Let $\{X_\alpha :\alpha <\kappa \}$ enumerate all $\mathcal M$ -definable d-approximations of $\mathrm {tp}(a/A)$ at a. We may assume that $\kappa $ is a cardinal and $\kappa \leq |M|$ . We will inductively build $b_{\alpha }$ and $Y_{\alpha }$ , for $\alpha <\kappa $ , such that:

1. (1) $b_\alpha $ is a finite tuple from $\mathcal M$ .

2. (2) $Y_{\alpha }\subseteq X_{\alpha }$ is an $Ab_{\alpha }$ -definable d-approximation of $\mathrm {tp}(a/A)$ at a.

3. (3) $\dim (a/A_{\alpha }b_{\alpha })=\dim (a/A_{\alpha })$ , where $A_{\alpha }=A\cup \{b_{\beta }:\beta <\alpha \}$ .

Indeed, to build $b_{\alpha }$ and $Y_{\alpha }$ , one just needs to apply Lemma 3.5 to $X_\alpha $ with parameter set $A_{\alpha }$ . The key point making this work is that $|A_{\alpha }|<|M|$ for all $\alpha $ , and hence (as $\mathcal M$ is saturated) such a $b_{\alpha }$ can be found inside $\mathcal M$ .

Now suppose we have built $b_{\alpha }$ and $Y_{\alpha }$ for all $\alpha $ . Let $B=A_{\kappa }$ be A together with all $b_{\alpha }$ for $\alpha <\kappa $ – equivalently, the union of all $A_{\alpha }$ for $\alpha <\kappa $ . It follows by induction that $\dim (a/A_{\alpha })=\dim (a/A)$ for all $\alpha \leq \kappa $ (for the limit step, use that any drop in dimension is witnessed by a formula with finitely many parameters). In particular, $\dim (a/B)=\dim (a/A)$ .

Now we show that $\mu (a/A)$ is the set of realizations of $\mathrm {tp}(a/B)$ . First, assume $a'\models \mathrm {tp}(a/B)$ . Then for each $\alpha <\kappa $ , we have $\mathrm {tp}(a'/Ab_{\alpha })=\mathrm {tp}(a/Ab_{\alpha })$ , and thus $a'\in Y_{\alpha }\subseteq X_{\alpha }$ . So $a'$ belongs to all $\mathcal M$ -definable d-approximations of $\mathrm {tp}(a/A)$ at a, and thus $a'\in \mu (a/A)$ .

Conversely, suppose $a'\in \mu (a/A)$ . If $\mathrm {tp}(a'/B)\neq \mathrm {tp}(a/B)$ , then there is $\alpha <\kappa $ so that $\mathrm {tp}(a'/A_{\alpha })\neq \mathrm {tp}(a/A_{\alpha })$ . But $A_{\alpha }$ is a small set in $\mathcal M^{eq}$ , and $\dim (a/A_{\alpha })=\dim (a/A)$ ; so we have $\mu (a/A_{\alpha })=\mu (a/A)$ , by Lemma 4.5 and the definition of $Y_\alpha $ . Thus, $a'\in \mu (a/A_{\alpha })$ , and thus $\mathrm {tp}(a'/A_{\alpha })=\mathrm {tp}(a/A_{\alpha })$ , a contradiction.

Proof.

1. (1) For ease of notation, assume $A=\emptyset $ . Since $a\in \mathrm {dcl}(g)$ , we have $\dim (b/g)=\dim (b/ga)$ . Moreover, since g is a code for the germ of a over b, we get $g\in \mathrm {dcl}(ab)$ . Therefore, $\dim (ab)=\dim (abg)$ . Now combining with additivity, we compute:

   $$ \begin{align*}\dim(a)+\dim(b/a) &=\dim(ab)=\dim(abg) \\ &=\dim(a)+\dim(g/a)+\dim(b/ga) \\ &=\dim(a)+\dim(g/a)+\dim(b/g).\end{align*} $$

   Rearranging now gives

   $$ \begin{align*}\dim(b/a)=\dim(g/a)+\dim(b/g).\end{align*} $$

   In particular, $\dim (g/a)=0$ holds if and only if $\dim (b/a)=\dim (b/g)$ , as desired.

2. (2) The right-to-left direction is trivial by (1); we show left to right. So assume $\mathcal M$ is topologically 1-based. Let a be a real tuple, A be a parameter set and $g=\mathrm {germ}(a/A)$ . Let c be a finite tuple from A so that $\dim (a/c)=\dim (a/A)$ , and thus $\mathrm {germ}(a/c)=g$ . Then let b be a real tuple so that $c\in \mathrm {dcl}(b)$ . Arguing by compactness, we may choose b to be independent from a over c; thus $\mathrm {germ}(a/b)=g$ as well. Then since $\mathcal M$ is topologically 1-based, and by (1), we get that

   $$ \begin{align*}\dim(\mathrm{germ}(a/b)/a)=\dim(g/a)=0,\end{align*} $$
   which proves (2).

Proof. We may assume $A=\emptyset $ . Let $g=\mathrm {germ}(a/b)$ . First, assume $\mathrm {tp}(a/b)$ is topologically 1-based, and let $b'\in \mu (b/a)$ . We show that $g=\mathrm {germ}(a/b')$ . Indeed, by 1-basedness, $\dim (b/a)=\dim (b/ga)$ , and thus $\mu (b/a)=\mu (b/ga)$ by Lemma 4.7(2). So $b'\in \mu (b/ga)$ , and thus $b'\models \mathrm {tp}(b/ga)$ (Lemma 4.5(3)). So $\mathrm {tp}(abg)=\mathrm {tp}(ab'g)$ , and since $g=\mathrm {germ}(a/b)$ , we conclude that also $g=\mathrm {germ}(a/b')$ .

Conversely, suppose $y\mapsto \mathrm {germ}(a/y)$ is constant on $\mu (b/a)$ . Then for any $b'\in \mu (b/a)$ , we have $\mathrm {tp}(abg)=\mathrm {tp}(ab'g)$ : indeed, $\mathrm {tp}(ab)=\mathrm {tp}(ab')$ follows by Lemma 4.5(3), and then we can add g to each side because g is defined over both $ab$ and $ab'$ by the same formula.

Now since $\dim (\mu (b/a))=\dim (b/a)$ , by compactness, there is an element $b'\in \mu (b/a)$ with $\dim (b'/abg)=\dim (b/a)=\dim (b'/a)$ . Because $a\in \mathrm {dcl}(g)$ , it follows that $\dim (b'/ag)=\dim (b'/g)=\dim (b'/a)$ as well, and since $\mathrm {tp}(abg)=\mathrm {tp}(ab'g)$ , this implies $\dim (b/g)=\dim (b/a)$ .

Proof. Before starting the proof, we should note that $\mathcal M_C$ is also t-minimal with the same topology. Moreover, $\mathcal M_C$ also has independent neighbourhoods. Indeed, given any a and A, the independent neighbourhood property for $\mathcal M$ gives us arbitrarily small neighbourhoods U of a defined over parameters t with $\dim (a/CAt)=\dim (a/CA)$ ; and this is equivalent to the independent neighbourhood property for $\mathcal M_C$ . Thus, for the rest of the proof, we can use all of the results collected to this point on independent neighbourhoods.

We now proceed with the proof. First, suppose $\mathcal M_C$ is topologically 1-based. To show $\mathcal M$ is topologically 1-based, let a and b be real tuples. We wish to show that $\mathrm {tp}(a/b)$ is topologically 1-based. As the definition of topologically 1-based only depends on $\mathrm {tp}(ab)$ , we may replace $(a,b)$ with any other realization of the same type. In particular, we may assume that $(a,b)$ is independent from C, i.e. $\dim (ab/C)=\dim (ab)$ . By Lemma 4.10, this implies that $\dim (a/Cb)=\dim (a/b)$ and $\dim (b/Ca)=\dim (b/a)$ .

Now let $g=\mathrm {germ}(a/b)$ . Since $\dim (a/Cb)=\dim (a/b)$ , we also have $g=\mathrm {germ}(a/Cb)$ . Then the fact that $\mathcal M_C$ is topologically 1-based gives $\dim (b/Cg)=\dim (b/Ca)$ . But now we compute:

$$ \begin{align*}\dim(b/g)\geq\dim(b/Cg)=\dim(b/Ca)=\dim(b/a),\end{align*} $$

and this implies that $\mathrm {tp}(a/b)$ is topologically 1-based.

Now assume $\mathcal M$ is topologically 1-based; we show that $\mathcal M_C$ is topologically 1-based. To do this, fix real tuples a and b, and set $g=\mathrm {germ}(a/Cb)$ . We want to show that $\mathrm {tp}(a/Cb)$ is topologically 1-based over C – i.e. that $\dim (b/Cg)=\dim (b/Ca)$ .

Let c be a finite tuple from C such that $\dim (a/Cb)=\dim (a/cb)$ , and $\dim (b/Cg)=\dim (b/cg)$ . Then let d be a real tuple such that $c\in \mathrm {dcl}(d)$ . Replacing d with an independent realization of $\mathrm {tp}(d/c)$ , we may assume that $ab$ and d are symmetrically independent over c (for example, this is, essentially, [Reference Johnson11, Corollary 2.16]).

Let $g=\mathrm {germ}(a/Cb)$ . By the choice of c, we also have $g=\mathrm {germ}(a/cb)$ . Now by the choice of d, we have $\dim (ab/d)=\dim (ab/c)$ , and thus Lemma 4.10(2) gives $\mathrm {germ}(a/bd)=\mathrm {germ}(a/bc)=g$ . Since $\mathcal M$ is topologically 1-based, it follows that $\dim (bd/g)=\dim (bd/a)$ . By Lemma 4.10(3), this implies $\dim (b/dg)=\dim (b/da)$ .

Next, recall that $\dim (ab/d)=\dim (ab/c)$ . By Lemma 4.10(3) again, this implies $\dim (b/da)=\dim (b/ca)$ . Then, combining with the previous paragraph, we have:

$$ \begin{align*}\dim(b/Cg) & =\dim(b/cg)\geq\dim(b/dg) \\ & =\dim(b/da)=\dim(b/ca)\geq\dim(b/Ca),\end{align*} $$

which implies that $\mathrm {tp}(a/Cb)$ is topologically 1-based over C.

Proof. By Lemma 5.7, we may assume all basic relations of $\mathcal M^-$ are definable without parameters in $\mathcal M$ . Now we begin with:

The claim also implies the analogous statement for germs: the dimension of a definable germ in $\mathcal M^-$ remains the same when viewed as a definable germ in $\mathcal M$ (on the other hand, the data of which germ is $\mathrm {germ}(a/A)$ for fixed a and A may change).

Now towards a proof of the lemma, fix real tuples a and b. We want to show that $\mathrm {tp}_{\mathcal M^-}(a/b)$ is topologically 1-based. We can freely replace $(a,b)$ by any other realization of $\mathrm {tp}_{\mathcal M^-}(ab)$ . In particular, by a compactness argument (using the above claim), one finds a realization such that $\dim _{\mathcal M}(ab)=\dim _{\mathcal M^-}(ab)$ . Now let c be a basis of $(a,b)$ in the sense of $\mathcal M^-$ ; then one still has $(a,b)\in \mathrm {acl}_{\mathcal M}(c)$ , and it follows that c is still a basis in the sense of $\mathcal M$ . In particular, if X is an $(a,b)$ -witness for c in the sense of $\mathcal M^-$ , then X is also an $(a,b)$ -witness for c in the sense of $\mathcal M$ . It follows that $\mathrm {germ}_{\mathcal M^-}(a/b)$ and $\mathrm {germ}_{\mathcal M}(a/b)$ have a common realization, and thus are equal. Call this common germ g. By Lemma 4.10(1), $\mathrm {germ}_{\mathcal M^-}(b/a)$ and $\mathrm {germ}_{\mathcal M}(b/a)$ are both the fibre germ $g_a$ , and thus

$$ \begin{align*}\dim_{\mathcal M^-}(b/a)=\dim(\mathrm{germ}_{\mathcal M^-}(b/a))=\dim(\mathrm{germ}_{\mathcal M}(b/a))=\dim_{\mathcal M}(b/a).\end{align*} $$

Now since $\mathcal M$ is topologically 1-based, we have $\dim _{\mathcal M}(b/g)=\dim _{\mathcal M}(b/a)$ . If $\mathrm {tp}_{\mathcal M^-}(a/b)$ were not topologically 1-based, then we would have

$$ \begin{align*}\dim_{\mathcal M^-}(b/g)<\dim_{\mathcal M^-}(b/a)=\dim_{\mathcal M}(b/a)=\dim_{\mathcal M}(b/g).\end{align*} $$

So there is some g-definable set in $\mathcal M^-$ which contains b and has dimension less than $\dim _{\mathcal M}(b/g)$ . But this assertion is still valid in $\mathcal M$ , so one obtains $\dim _{\mathcal M}(b/g)<\dim _{\mathcal M}(b/g)$ , a contradiction.

Proof. First, note that if $a\in M^n$ and $t\in M^m$ and $\dim (a/t)$ is either 0 or n, then $\dim (a/t)$ is automatically topologically 1-based. Indeed, in this case, $g:=\mathrm {germ}(a/t)$ is either a single point or the full germ in $M^n$ at a. In either case, this implies $g\in \mathrm {dcl}(a)$ , and thus g and a are interdefinable, and thus automatically $\dim (t/g)=\dim (t/a)$ . In particular, this shows that topological 1-basedness is automatic if $n=1$ (i.e. all 1-types are topologically 1-based).

Now let $a\in M^n$ and $t\in M^m$ so that $\mathrm {tp}(a/t)$ is not topologically 1-based over $\emptyset $ , and so that the value of n is minimized among all such pairs. Moreover, write $d:=\dim (a/t)$ . By the above remarks, we have $n\geq 2$ and $0<d<n$ . Write $a=(e_1,\dots ,e_d,f_1,\dots ,f_{n-d})$ where $e:=(e_1,\dots ,e_d)$ is a basis for a over t.

Let us first verify that our choice of n assures that $d=n-1$ . By Lemma 4.9(3), $g=\mathrm {germ}(a/t)$ is realized by the graph of a function $\mu (e)\rightarrow \mu (f_1)\times \dots \times \mu (f_{n-d})$ (i.e. g equals the germ at a of the function’s graph). This function is determined by its coordinate components $\mu (e)\rightarrow \mu (f_i)$ , and each of these components realizes $g_i:=\mathrm {germ}(ef_i/t)$ . In other words, $g\in \mathrm {dcl}(g_1, \dots , g_{n-d})$ . Now by assumption (and Lemma 5.5), we have $\dim (g/a)>0$ , and thus some $\dim (g_i/a)>0$ . Then $\mathrm {tp}(e_1,\dots ,e_d,f_i/t)$ is not topologically 1-based over $\emptyset $ . By the minimality of n, we must have $a=(e_1,\dots ,e_d,f_i)$ , so that $d=n-1$ .

We will show that $n=2$ and thus $d=1$ . So for now, assume towards a contradiction that $n\geq 3$ . Let $t'\in \mu (t/a)$ be independent from t over a. A short computation (using that $d=n-1$ and $n\geq 3$ ) yields that $a\notin \mathrm {acl}(tt')$ . So we can write $a=(b,c)$ where $b\in M$ , $c\in M^{n-1}$ and $\dim (b/tt')=1$ .

Let X be a $\emptyset $ -definable d-approximation of $\mathrm {tp}(cbt)$ at $cbt$ . By Lemma 4.10, the fibre $X_{bt}$ realizes $\mathrm {germ}(c/bt)$ . Since $\mathrm {tp}(cbt)=\mathrm {tp}(cbt')$ , $X_{bt'}$ realizes $\mathrm {germ}(c/bt')$ .

Now by the minimality of n, $\mathrm {tp}(c/bt)$ is topologically 1-based, so by Lemma 5.6, $X_{bt}$ and $X_{bt'}$ agree on some neighbourhood U of c. By the independent neighbourhood property, we may assume U is definable over a tuple u with $\dim (cbtt'/u)=\dim (cbtt')$ . By additivity of $(c,btt')$ over $\emptyset $ (i.e. Lemma 3.13), $\dim (btt'/u)=\dim (btt')$ , and thus $\dim (b/tt'u)=\dim (b/tt')=1$ . Thus, there is a neighbourhood V of b such that the fibres $X_{vt}$ and $X_{vt'}$ agree on U for all $v\in V$ . This means the fibres $X_t$ and $X_t'$ agree on $U\times V$ , so that $\mathrm {germ}_a(X_t)=\mathrm {germ}_a(X_{t'})$ .