In Chapter IV of the book, Algebraic Geometry; An Introduction by Daniel Perrin, he showed the following (the base field is algebraically closed):

Corollary 3.8. Let $\varphi:X\to Y$ be a morphism of algebraic varieties.

(1) Assume that all the fibres of $\varphi$ are of dimension $\leq r$. Then $\dim X \leq r + \dim Y$.

(2) Assume that $\varphi$ is dominant and all the nonempty fibres of $\varphi$ are of dimension $r$. Then $\dim X = r + \dim Y$.

I understand (1) but I do not understand the proof of $(2)$.

Proof of $(2)$ We decompose $Y$ into irreducible components, $Y = Y_1 \cup \cdots \cup Y_n$. Let $Y_i$ be a component of $Y$ of dimension $\dim Y$. The restriction $\varphi:\varphi^{-1}(Y_i) \to Y_i$ is dominant (consider the nonempty open set $Y_i \setminus \bigcup_{j\neq i}Y_j$). We then decompose $\varphi^{-1}(Y_i) = \bigcup_j X_{i,j}$ into irreducible components. After possibly removing components we can assume that all the sets $X_{i,j}$ dominate $Y_i$. If the dimensions of all the $X_{i,j}$ are $< r + \dim Y$, then dimension theorem of generic fibres of irreducible varieties applied to the restriction $\varphi:X_{i,j}\to Y_i$ shows that the general fibre of $\varphi^{-1}(Y_i)\to Y_i$ is of dimension $<r$ (originally $>r$ but should be a typo, comments by me), which is impossible.

Why can we assume all the sets $X_{i,j}$ dominate $Y_i$?

I have understood that there exists $X_{i,k}$ such that $\overline{\varphi(X_{i,k})}=Y_i$ since $Y_i$ is irreducible. Now assume $\dim X_{i,j}<\dim Y + r$ for all $X_{i,j}$. For a general $y\in \overline{\varphi(X_{i,j})}$, we have $$\dim Y + r > \dim X_{i,j} = \dim \overline{\varphi(X_{i,j})} + \dim (\varphi^{-1}(y)\cap X_{i,j})$$ and thus $$\dim (\varphi^{-1}(y)\cap X_{i,j}) < \dim Y_i - \dim \overline{\varphi(X_{i,j})} + r.$$ If $\overline{\varphi(X_{i,j})}=Y_i$, $$\dim (\varphi^{-1}(y)\cap X_{i,k}) < r.$$

However if $\overline{\varphi(X_{i,j})}\subsetneq Y_i$, the above inequality does not seem to necessarily hold. How can we get $\dim\varphi^{-1}(y)<r$?


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