# Projection map sends closed sets to closed sets

I am reading Algebraic Geometry, a first course by Joe Harris. In the section on projections, he talks about an application of elimination theory to prove that image of a projection map $$\pi: Y \times \mathbb{P}^1 \rightarrow Y$$ is a closed set. I do not have strong background in Algebraic Geometry, so I do not understand elimination theory and resultants. Can someone explain what is the author trying to say in the highlighted part and why is it true? Basically, I am interested in understanding its proof.

Also, does the projection map send an open set $$Z \subseteq Y \times \mathbb{P}^1$$ to an open set $$\pi(Z)$$? If yes, why?

Thank you very much in advance!

Suppose $$k$$ is a field and $$f,g\in k[x,y]$$ are homogeneous polynomials. Claim: the homogeneous resultant of $$f,g$$ is zero iff $$V(f,g)\subset\Bbb P^1_k$$ is nonempty.
Proof: The resultant is the determinant of the map $$(A,B)\mapsto Af+Bg$$ from $$k[x,y]_{\deg g}\times k[x,y]_{\deg f} \to k[x,y]_{\deg f+\deg g}$$. But the image of this map is exactly $$(f,g)\cap k[x,y]_{\deg f+\deg g}$$, so $$(A,B)\mapsto Af+Bg$$ not being surjective is equivalent to $$(f,g)$$ being contained in homogeneous ideal who's radical does not contain the irrelevant ideal of $$k[x,y]$$. Such ideals define nonempty subsets of $$\Bbb P^1_k$$, so by the nullstellensatz we see that $$V(f,g)=V(f)\cap V(g)$$ is nonempty as a subset of $$\Bbb P^1_k$$. $$\blacksquare$$
Now we note that the homogeneous resultant plays nicely with ring homomorphisms, since it's the determinant of a particular matrix: if $$\varphi:R\to S$$ is a ring homomorphism and $$f,g\in R[x,y]$$ are homogeneous, then $$\varphi(Res(f,g))=Res(\varphi(f),\varphi(g))$$. Taking $$R=k[Y]$$, $$S$$ to be the residue field of a point $$y$$, and $$\varphi$$ the evaluation map, we see that $$Res(f,g)$$ vanishes at $$y$$ iff $$f,g$$ have a common zero in the fiber over $$y$$. The highlighted result follows.
As for the question about projections of open sets, this has been addressed on MSE before here. The upshot is that for any scheme over a field, the map $$X\to\operatorname{Spec} k$$ is open, so any projection $$X\times Y\to Y$$ is open because such a projection is the base change of $$X\to\operatorname{Spec} k$$ along $$Y\to\operatorname{Spec} k$$. See Stacks 01TZ for a full proof. A slightly more elementary approach can be found in the comments.