Dominant rational maps necessitate open subsets have dense image

Question

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1. What the problem is asking

Let

(X) and (Y) be irreducible algebraic varieties over an algebraically closed field,
(\varphi) a rational map from (X) to (Y).

A rational map is an equivalence class of morphisms [ (U,\varphi_U),\qquad U\subset X\ \text{non‑empty open},\ \varphi_U:U\to Y, ] where two representatives ((U,\varphi_U)) and ((V,\varphi_V)) are equivalent when they agree on the overlap (U\cap V).

Hartshorne says that (\varphi) is dominant iff for one (hence for every) such pair the image is dense in (Y).

The task is to prove the “hence for every’’ part:

If there exists a representative ((U,\varphi_U)) with (\overline{\varphi_U(U)}=Y), then for any other representative ((V,\varphi_V)) we also have (\overline{\varphi_V(V)}=Y).

Equivalently we must show that the image of the restriction to the common open set [ \varphi_{U\cap V}= \varphi_U|{U\cap V}= \varphi_V|{U\cap V} ] is already dense in (Y).

2. Proof, step by step

2.1 Basic facts we will use

Irreducibility of open subsets – If (X) is irreducible, every non‑empty open subset of (X) is itself irreducible and dense in (X).
Continuity of morphisms – A morphism of varieties is a continuous map for the Zariski topology; the preimage of an open set is open.
Density criterion – A subset (A\subset Y) is dense iff every non‑empty open subset of (Y) meets (A).

2.2 The intersection (U\cap V) is a non‑empty dense open in both (U) and (V)

Because (X) is irreducible, the intersection of two non‑empty opens is again a non‑empty open; moreover it is dense in each of them.
Hence (U\cap V) is an irreducible variety and the restriction maps [ \varphi_{U\cap V}:= \varphi_U|{U\cap V}= \varphi_V|{U\cap V} ] are well defined morphisms (U\cap V\to Y).

2.3 The image of (\varphi_{U\cap V}) is dense in (Y)

Assume, for contradiction, that (\overline{\varphi_{U\cap V}(U\cap V)}\neq Y).
Then there exists a non‑empty open subset (W\subset Y) disjoint from (\varphi_{U\cap V}(U\cap V)).

Because (\varphi_U) is a morphism, the pre‑image [ \varphi_U^{-1}(W)\subset U ] is open (and non‑empty, since (\varphi_U(U)) is dense in (Y); the dense image guarantees that (W) meets (\varphi_U(U))).

Now (U\cap V) is dense in (U), so the open set (\varphi_U^{-1}(W)) must intersect (U\cap V): [ \varphi_U^{-1}(W)\cap (U\cap V)\neq\varnothing . ]

Pick a point (x) in this intersection. Then [ x\in U\cap V,\qquad \varphi_U(x)\in W . ] But on (U\cap V) the two representatives coincide, i.e. (\varphi_U(x)=\varphi_V(x)=\varphi_{U\cap V}(x)).
Thus (\varphi_{U\cap V}(x)\in W), contradicting the definition of (W) (which was chosen to avoid the image of (\varphi_{U\cap V})).

Hence our assumption was false and we must have
[ \overline{\varphi_{U\cap V}(U\cap V)} = Y . ]

2.4 Consequence for any other representative

Because (U\cap V\subset V) we have [ \varphi_V(V) \supset \varphi_{U\cap V}(U\cap V) . ] Taking closures, [ \overline{\varphi_V(V)} \supset \overline{\varphi_{U\cap V}(U\cap V)} = Y . ] Thus (\overline{\varphi_V(V)} = Y); i.e. the image of any representative ((V,\varphi_V)) is dense in (Y).

Therefore a rational map is dominant iff one (equivalently, every) representative has dense image.

3. Final answer

If a rational map (\varphi : X \dashrightarrow Y) has one representative ((U,\varphi_U)) whose image is dense in (Y), then for every other representative ((V,\varphi_V)) we also have (\overline{\varphi_V(V)}=Y). The key point is that the common open set (U\cap V) is dense in both domains, and the restriction of the morphism to this set already has dense image; continuity of morphisms then forces the whole image of any other representative to be dense as well. Hence the definition “dominant if for some (and hence every) representative the image is dense’’ is correct.

4. Common mistakes

Mistake	Why it is wrong	How to avoid it
Thinking “open ⇒ dense’’ in any topological space.	In the Zariski topology an open set need not be dense (e.g. a proper open subset of a reducible variety). The argument works because the ambient space (X) is irreducible; then every non‑empty open is dense.	Always remember to use irreducibility when you claim an open set is dense.
Assuming the image of a restriction is automatically dense because the larger image is dense.	A subset of a dense set can be non‑dense (think of a line missing a point). One must prove density for the restriction, using openness of the pre‑image of a non‑empty open in (Y).	Use the density‑criterion argument above: any open (W\subset Y) meeting (\varphi_U(U)) must already meet the image of the restriction, because the pre‑image of (W) is open and meets the dense open (U\cap V).
Confusing equality of maps with equality of their images.	The two representatives agree point‑wise on the overlap, but their domains are different, so you cannot directly replace (\varphi_U(U)) by (\varphi_V(V)).	Work with the common restriction (\varphi_{U\cap V}); its image is contained in both images, and its density forces the density of each full image.
Neglecting continuity of morphisms.	Without using that morphisms pull back open sets to open sets, the argument that the pre‑image of a non‑empty open meets the dense open subset fails.	Explicitly invoke the fact that a morphism of varieties is continuous for the Zariski topology.

Keeping these points in mind will prevent the typical pitfalls when handling dominant rational maps.

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