Here is a lemma in tag01FE.
Let $f: X→Y$ be a morphism of locally Noetherian schemes which is flat and locally of finite type such that all fibers are equidimensional of dimension d. Then for every point x in X with image y in Y we have $\dim_x(X)=\dim_y(Y)+d$.
Proof. After shrinking X and Y to open neighborhoods of x and y, we can assume that $\dim_x(X)=\dim X$ and $\dim_y(Y)=\dim Y$, by definition of the dimension of a scheme at a point. It remains to show that $\dim(X)=\dim(Y)+d$.
I think this process should go as follows: Choose an open neighborhood $V$ of $y$ such that $\dim_y(Y)=\dim V$ and choose an open neighborhood $U$ contained in $f^{-1}(V)$ such that $\dim_x(X)=\dim U$. Then we consider the morphism $U\rightarrow V$.
My question is that why the fibers of $U\rightarrow V$ are still equidimensional of dimension d? When we shrink $X$ and $Y$ into $U$ and $V$, I think the underlying topological space of fibers should become smaller.
Thanks.
The shrinking that goes on cannot change the dimension of the fiber in this case. Why? The fiber over $y\in Y$ is given by $\operatorname{Spec} \kappa(y)\times_Y X$, which means that the fiber is locally of finite type over a field. Now use that for an irreducible scheme locally of finite type over a field, any (nonempty) open subset has the same dimension as the whole scheme - applying this to each irreducible component given the reduced induced structure, we see that all of the fibers are still equidimensional of dimension $d$.