Dimension of vector spaces under surjection Definition. Let $V$ be a vector space and $X\subseteq V$ be a definable subset in the language of vector spaces. $X$ is independent if for all $a\in X$, $a\notin span(A\setminus\{a\})$.
Definition. Let $V$ be a vector space and $Y\subseteq V$  be a definable subset (in the language of vector spaces). We say that $B$ is a basis for $Y$ if $B \subseteq Y$ is
independent and $span(A) = span(Y)$. The dimension of Y is the cardinality
of a basis for Y .
My question is this: 
Suppose $V$ is a vector space and $X,Y\subseteq V$ are definable subsets. Let $f:X\rightarrow Y$ be a definable surjection. Also, let $\dim(f^{-1}(y))=d$ for all $y\in Y$. Is the following statement always true? why?
$$\dim(X)= \dim(Y) + d$$
 A: With your definition of dimension, almost any definable map you try will be a counterexample. 
For example, let $V$ be a $1$-dimensional vector space. Then any set $X\subseteq V$ has dimension $1$, except for $\emptyset$ and $\{0\}$ (which both have dimension $0$). 
Let $X = Y = V\setminus \{0\}$, and let $f\colon X\to Y$ be the identity map. Then for all $y\in Y$, $\dim(f^{-1}(y)) = \dim(\{y\}) = 1$. But $\dim(X) = 1$,  while $\dim(Y) +1 = 2$. 

Here is an alternative (and more sensible) definition of dimension for definable sets: Let $X\subseteq V^n$ be a definable set, and let's assume $V$ is $\omega$-saturated (if not, take a suitable elementary extension, and evaluate $X$ there). An element of $X$ is an $n$-tuple $a$, and we define $\dim(a)$ to be the size of a maximal independent sub-tuple of $a$, equivalently the dimension of $\text{span}(a)$ in your sense. Then we define $\dim(X)$ to be $\max_{a\in X} \dim(a)$, the maximal dimension of an element of $X$.
This definition of dimension agrees with the Morley rank relative to the theory of infinite $K$-vector spaces, and it satisfies the additivity condition in your question.

In the comments, you asked some follow-up questions. 

(1) Is my definition of dimension coincide to Morley rank? (I used Definitions 6.1.5 and 6.1.8 and 6.1.10 of Marker's book).

No - as I said in my answer, my definition of dimension coincides with Morley rank.  
There's a confusing thing about strongly minimal theories, which is that there are two reasonable meanings of "dimension", which are very different. Let $T$ be a strongly minimal theory. 


*

*Let $M\models T$, and let $A\subseteq M$ (not necessarily definable!). Define $\dim_1(A)$ to be the cardinality of a basis for $\text{acl}(A)$.  

*Let $X$ be a definable set, defined by $\varphi(x)$. Let $\mathbb{M}\models T$ be a monster model of $T$ (or at least $\omega$-saturated). Define $\dim_2(X)$ to be the maximal value of $\dim_1(a)$ over all tuples $a\in X(\mathbb{M})$. 
The fact that $\dim_2$ coincides with Morley rank is Theorem 6.2.19, together with Lemma 6.2.16(i), in Marker.
In vector spaces, $\dim_1$ coincides with the ordinary notion of linear dimension, for closed sets. But it's unsatisfactory for definable sets, since it isn't invariant under elementary extensions. For example, let $T$ be the theory of $\mathbb{R}$-vector spaces, and consider the formula $\phi(x): x = x$. We have $\dim_1(\varphi(\mathbb{R}^2)) = 2$, but $\dim_1(\varphi(\mathbb{R}^{42})) = 42$, despite the fact that $\mathbb{R}^2\preceq \mathbb{R}^{42}$. So $\dim_1$ is really assigning a dimension to a particular subset of a particular model. 
If you're familiar with field theory or algebraic geometry, it might be morehelpful to think about the theory of algebraically closed fields, where both notions of dimension are familiar mathematical notions. 


*

*If $A\subseteq K\models \text{ACF}_0$, then $\dim_1(A)$ is the transcendence degree of the field $\overline{\mathbb{Q}(A)}$ over $\mathbb{Q}$. 

*If $X$ is a definable set (which is a Boolean combination of zero-sets of polynomials), then $\dim_2(X)$ is the Krull dimension of the Zariski closure of $X$. 

(2) Does acl(X)=span(X) in vector spaces? 

Yes, in any infinite vector space $V$, $\text{acl}(X) = \text{span}(X)$ for all sets $X$. The inclusion $\text{span}(X)\subseteq \text{acl}(X)$ is clear, and for the other, note that if $b\notin \text{span}(X)$, then $b$ has infinitely many images under automorphisms of $V$, so it's not in $\text{acl}(X)$. 
Of course, if $V$ is finite (which can only happen if the base field is finite), then $\text{acl}(X) = V$ for any $X$. 

(3) Does the additivity condition in my question hold for Morley rank in any strongly minimal structure (or we need to have extra assumptions)?

Yes, it does. Basically, this comes down to the fact that in any strongly minimal structure, for any tuples $a$ and $b$, we have $$\dim(ab) = \dim(a/b) + \dim(b).$$
Here $\dim$ is what I wrote as $\dim_1$ above. For a proof, see here.
