Dimension of most fibers of $X^m \to Y^n$ of irreducible $k$ varieties are pure dimension $m-n$. I am reading Ravi Vakil's notes on algebraic geometry and in Theorem 11.4.1 he claims: "Suppose $\pi: X \to Y$ is a (necessarily finite type) morphism of irreducible $k$-varieties with $dim X = m,$ and $ dim Y = n$. Then there exists a nonempty open subset $U \subset Y$ such that for all $q \in U,$ the fiber over $q$ has pure dimension $m-n$." He quickly reduces to the affine case $\pi: Spec(A) \to Spec(B)$ with $\pi$ dominant and claims that it suffices to show there is a distinguished open subset $U \subset Spec(B)$ for which the restriction factors through $\pi^{-1}(U) \to \mathbb{A}^{m-n}_U \to U$ where $\mathbb{A}^{m-n}_U := \mathbb{A}_k^{m-n} \times U$ and $\psi: \pi^{-1}(U) \to \mathbb{A}^{m-n}_U$ is a finite surjection.
In particular, he argues that using the fact that codimension is the difference of dimension for varieties and the fact that for any morphism with $\pi(p) = q, codim_Xp \leq codim_Yq + codim_{\pi^{-1}(q)}p$, we can argue that any component of the fiber over a point of $U$ has dimension at least $m-n$. While I agree we can show this for some component, what precisely is stopping a situation like $\mathbb{A}^2_k \coprod \mathbb{A}^1_k \to \mathbb{A}^2_k$ from happening?
I looked in other books but none seemed to specifically address the pure dimension part of this statement.
 A: I managed to figure it out! It's corollary 2.
Lemma: If $\pi :X\to Y$ is a morphism of schemes and $\pi(x) = y$ for some $x \in X$ then the map $O_{X,x}/x \to O_{\pi^{-1}(y), x}/x$ induced by the map of schemes $\pi^{-1}(y) \to X$ is an isomorphism.
Proof of Lemma: Pick an affine open $Spec(B) \subset Y$ containing Y and similarly choose an affine open $Spec(A) \subset \pi^{-1}(Spec(B))$ containing $x$. By abuse of notation, identify $x, y$ as prime ideals of $A, B$ respectively. Then base changing the map $y \to Spec(B)$ is the composite $Spec(B_y/y) \to Spec(B_y) \to Spec(B)$ and base changing this map by $Spec(A) \to Spec(B)$, the domain $\pi^{-1}(y)$ is $Spec(A \otimes_BB_y/y) \cong Spec(A/(\pi^{\sharp}(y)) \otimes_BB_y) \cong Spec((A/(\pi^{\sharp}(y))_{\pi^{\sharp}(B\setminus y)})$.
Then the map of local rings is the map $A_x/x \to ((A/\pi^{\sharp}(y))_{\pi^{\sharp}(B\setminus y)})_{x}/x = ((A/\pi^{\sharp}(y))_x/x$. Note though that this map is a surjection of fields, and thus an isomorphism.
Corollary 1: If $\pi : X \to Y$ is a morphism of integral $k-$schemes for some field $k$ and $\pi(x) = y$ then $dim_X\overline{\{x\}} = dim_Y\overline{\{y\}} + dim_{\pi^{-1}(y)}\overline{\{x\}}$.
Proof of Corollary 1: By the lemma, the map $O_{X,x}/x \to O_{\pi^{-1}(y), x}/x$ is an isomorphism. Also note that finite type $k-$schemes are preserved by base change so if $l := O_{Y,y}/y$, then $\pi^{-1}(y)$ is a finite type $l$-scheme. Thus because $l \to O_{\pi^{-1}(y), x}/x \cong O_{X, x}$ we have
$$dim_X\overline{\{x\}} = tr.deg(O_{X,x}/x, k) = tr.deg(O_{X,x}/x, l) + tr.deg(l, k) = dim_{\pi^{-1}(y)}\overline{\{x\}} + dim_Y\overline{\{y\}}.$$
Corollary 2:  If $\pi : X \to Y$ is a morphism of integral $k-$schemes for some field $k$ then for each $y \in Y$ each irreducible component of $\pi^{-1}(y)$ has dimension $\geq dim(X) - dim(Y)$.
Proof of Corollary 2: Let $x$ be a generic point of an irreducible component of $\pi^{-1}(y)$. Then by Vakil 11.4.A. (linked above) we have $codim_X\overline{\{x\}} \leq codim_Y\overline{\{y\}} + codim_{\pi^{-1}(y)}\overline{\{x\}} = codim_Y\overline{\{y\}}$. Since for irreducible $k$ varieties codimension is the difference of dimension, we obtain
$$dim(X) - dim(Y) \leq dim_X\overline{\{x\}} - dim_Y\overline{\{y\}} = dim_{\pi^{-1}(y)}\overline{\{y\}},$$
as desired!
