# Unable to understand a part of Harris' proof of Chavalley's Theorem: the image of a constructible set contains a non-empty open set

I am reading Harris' Algebraic Geometry, A First Course and could not understand a part of Theorem 3.16.

The setting is this: $$X\subset\mathbb{P}^m$$ is a quasi-projective variety, $$f:X\to\mathbb{P}^n$$ is a regular map, and $$U\subset X$$ is a constructible set.

The part of the proof needs to establish the claim $$f(U)$$ contains a nonempty open subset $$V\subset\overline{f(U)}$$ of the closure of $$f(U)$$.

I am copying the part I don't understand from Harris:

To begin with, we may replace $$U$$ by an open subset, and so may assume that it is affine; restricting to a smaller affine open, we may assume the target space is also affine space. After replacing $$U$$ by the graph of $$f$$ we may realize the map $$f$$ as the restriction to a closed subset $$U\subset\mathbb{A}^n$$ of a linear projection $$\mathbb{A}^n\to\mathbb{A}^m$$, so that it is enough to prove the claim for a locally closed subset $$U\subset\mathbb{A}^n$$ under the projection

$$\pi:\mathbb{A}^n\to \mathbb{A}^{n-1}$$

$$(z_i,...,z_n)\mapsto(z_1,...,z_{n-1})$$

Finally, we can replace $$\mathbb{A}^{n-1}$$ by the Zariski closure Y = $$\overline{\pi(U)}$$ of the image of $$U$$ and $$\mathbb{A}^{n}$$ by the inverse image $$\pi^{-1}(Y) = Y \times \mathbb{A}^{1}$$. It will thus suffice to establish the claim for a locally closed subset $$U$$ of a product $$Y\times\mathbb{A}^{1}$$ (or, equivalently, a locally closed subset $$U\subset Y\times\mathbb{P}^{1}$$) and the projection map $$\pi: Y \times\mathbb{P}^{1}\to Y$$on the first factor, with the further assumption that $$\pi(U)$$ is dense in $$Y$$.

Basically I have trouble understanding what exactly he is trying to say in "we may realize the map $$f$$ as the restriction to a closed subset $$U\subset\mathbb{A}^n$$ of a linear projection $$\mathbb{A}^n\to\mathbb{A}^m$$", I have no idea what he is trying to do, particularly, is the new closed subset $$U$$ related to the old constructible $$U$$ in some way? What is this linear projection from $$\mathbb{A}^n$$ to $$\mathbb{A}^m$$, explicitly?

I am a beginner in algebraic geometry and could speak the language of scheme yet. Any help is really appreciated.

At that stage, you have a regular map from an affine to another affine $$f:U\subset \mathbb{A}^n\to f(U)\subset \mathbb{A}^m$$. The graph of $$f$$ is defined to be the closed subset $$G:=\{(x,f(x))\mid x\in U\}\subset \mathbb{A}^n\times\mathbb{A}^m=\mathbb{A}^{n+m}.$$ It is also canonically isomorphic to $$U$$ by projecting to the first factor $$(x,f(x))\mapsto x$$. Under this isomorphism, the map $$f$$ becomes the projection to the second factor, i.e. it now maps a closed subset $$G\subset \mathbb{A}^{n+m}$$ under this linear projection $$\mathbb{A}^{n+m}\to\mathbb{A}^m$$.