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Let's follow Hartshorne's definition, according to which, a rational map $\phi:X \rightarrow Y$ from variety $X$ to variety $Y$ is an equivalence class $\langle U,\phi_U \rangle$, where $U$ is open in $X$ and $\phi_U: U \rightarrow Y$ is a morphism of varieties. If $\phi_U(U)$ is dense in $Y$, $\phi$ is called dominant.

Hartshorne says that we can "clearly compose dominant rational maps". So let $\phi: X \rightarrow Y, \psi:Y \rightarrow Z$ be dominant rational maps given by the pairs $\langle U,\phi_U \rangle, \langle V, \psi_V \rangle$. Question: how exactly do we define $\psi \circ \phi$? To be consistent with the definition, we need to construct a pair $\langle W,\sigma_W \rangle$, where $W$ is open in $X$ and $\sigma_W : W \rightarrow Z$ is a morphism of varieties with dense image and somehow this $\sigma_W$ must arise from the "composition" of $\phi_U$ and $\psi_V$. But how can we compose these two?

Remark: i am aware that there are at least two other questions on this forum related to this topic, however the answers and comments do not clarify the issue i am raising.

Manos
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1 Answers1

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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). We then let $\sigma_W$ be the composite of $\phi_W$ (this is just the restriction of $\phi_U$ to $W$) and $\psi_V$.

If you continue to fine this confusing, I would suggest writing down some rational maps, some dominant and some not, and try composing them in the naive way, just by substituing the formula for one into the other. You will quickly see what makes sense and how things can go wrong in the non-dominant context. (The construction of the preceding paragraph, and Hartshorne's definition, is just formalizing the naive notions.)

Matt E
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