# How to reduce the assertion for quasi-projective varieties to affine irreducible case?

We have this theorem on page 62 of Shafarevich's Basic Algebraic Geometry I:

Theorem: If $$f : X \to Y$$ is a regular map (of quasi-projective varieties) and $$f(X)$$ is dense in $$Y$$ then $$f(X)$$ contains an (non-empty) open set of $$Y$$.

The proof to this theorem starts with The assertion of the theorem reduces at once to the case that both $$X$$ and $$Y$$ are irreducible and affine''.

I'd like to know how to reach this reduction?

I have some feelings for the reduction to the irreducible case: If $$X,Y$$ are not irreducible, they can be decomposed into irreducible components as $$X = \bigcup_{i=1}^n X_i$$ and $$Y = \bigcup_{j=1}^m Y_j$$ with $$X_i$$ irreducible and closed in $$X$$ and $$Y_j$$ irreducible and closed in $$Y$$. Then it follows that $$f(X_i)$$ must fall completely in some $$Y_j$$, otherwise, it would contradicts to that $$X_i$$ is irreducible. Now, suppose we may find an open subset $$U_{ij}$$ of each $$Y_j$$ contained in $$f(X_i)$$, but how can we construct an open subset of $$Y$$ from all the scratches?

As for the reduction to the affine case, I have completely no idea.

Any help will be appreciated. Thank you in advance.

You're on the right track with your ideas.

First, we finish the reduction to the irreducible case. Suppose we know the assertion for $$X,Y$$ irreducible. We first make the reduction to $$X$$ not necessarily irreducible while maintaining $$Y$$ irreducible. If $$f(X)\subset Y$$ is dense, then at least one of $$f(X_i)$$ must be dense in $$Y$$ - if not, then $$f(X)$$ isn't dense (as a union of a finite number of non-dense sets is again non-dense). So there's at least one irreducible component $$X_i$$ so that $$f(X_i)\subset Y$$ is dense and thus $$f(X)\supset f(X_i)$$ must contain a nonempty open by our assumption.

If $$Y$$ isn't necessarily irreducible, then we can apply our logic from the previous paragraph to each irreducible component in turn - for any irreducible component $$Y_j$$, there's an irreducible component $$X_i$$ so that $$f(X_i)$$ is dense in $$Y_j$$, so $$f(X_i)$$ contains a nonempty open subset of $$Y_j$$. We may assume that this open subset misses all the other irreducible components of $$Y_j$$, since $$Y_j\cap Y_k$$ is closed and nowhere dense in both $$Y_j$$ and $$Y_k$$ for $$j\neq k$$. It's then straightforwards to check that the union of all of these open sets over all $$Y_j$$ is again open and contained in the image of $$X$$.

For the affine case, we can apply the reduction to the irreducible case first, so that we have $$f:X\to Y$$ with $$f(X)\subset Y$$ dense and $$X,Y$$ irreducible. There are three key ingredients here, all from point-set topology:

Lemma 1: in an irreducible topological space, any nonempty open subset is dense.

Lemma 2: If $$A\subset B$$ is dense and we have a continuous map of topological spaces $$f:B\to C$$, then $$f(A)\subset f(B)$$ is also dense.

Lemma 3: If $$A$$ is dense in $$B$$ and $$B$$ is dense in $$C$$, then $$A$$ is dense in $$C$$.

Exercise: prove each of these yourself. Now that you've done that, do you see how to piece these together? More details in the spoiler text below:

Pick an open affine $$V\subset Y$$ and then pick an open affine $$U\subset f^{-1}(V)\subset X$$. As $$U$$ is dense in $$X$$ by lemma 1, $$f(U)$$ is dense in $$f(X)$$ by lemma 2 so it must be dense in $$Y$$ by lemma 3. But we know that $$f(U)\subset V$$, so $$f(U)\subset V$$ must be dense. By our assumption that we know the affine case, $$f(U)$$ must contain a nonempty open of $$V$$. But this is also a nonempty open of $$Y$$, and we're done.