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I have a question regarding the answer of the following question: how to define the composition of two dominant rational maps?. I'm sorry to open a new question for this, but I can't comment an answer yet.

Maybe for readabilty I comment/cite whats going on: Let $\phi: X \rightarrow Y, \psi:Y \rightarrow Z$ be dominant rational maps given by the pairs $\langle U,\phi_U \rangle, \langle V, \psi_V \rangle$. The question was how to define $\psi \circ \phi$.

In his answer Matt E says "Let $W$ be the intersection of $U$ (a domain for $\phi$) with $\phi^{-1}(V)$, where $V$ is a domain for $\psi$. This is the intersection of two non-empty open subsets of $X$ (we use dominance of $\phi$ to deduce that $\phi^{-1}(V)$ is non-empty)...

My questions are now: (1) How can one deduce that $\phi^{-1}(V)$ is not empty? I just dont see it.

(2) Do we use irreducabilty of Y to conclude that $\phi^{-1}(V)\cap U$ is not empty as well? ("Any who nonempty, open subsets intersect").

Thanks for your kind help.

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The first question can be explained by a fact from point-set topology.

Fact: As topological spaces, $A\subset B$ is dense iff $A$ intersects every nonempty open subset $C\subset B$ nontrivially. The proof goes as follows (spoiler in case you'd like to try yourself first):

if $C\subset B$ is a nonempty open set that $A$ doesn't intersect, then $C^c$ is a proper closed subset and contains $A$, which means $\overline{A}\subset C^c \neq B$

To use this to solve your issues, note that $V$ is a dense open subset and $\varphi(U)$ is also dense (this is what dominance means). So applying our fact with $A=\varphi(U)$ and $C=V$ we see that their intersection is nonempty. So $\varphi^{-1}(V)$ is nonempty.

For your second question, you've hit the nail on the head. Suppose two nonempty open subsets of $Y$ do not intersect- then their complements are proper nonempty closed sets whose union is $Y$, which means $Y$ is not irreducible, contradiction.

KReiser
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