Is the projection map $\pi: \mathbb A^{n+1}\setminus\{0\} \to \mathbb P^n$ open?

Question

I'm wondering about the topological properties of $\mathbb P^n$ that can be derived using the usual projection mapping $\pi: \mathbb A^{n+1}\setminus\{0\} \to \mathbb P^n$ which sends a point away from the origin to its representant. My intuition tells me that $\pi$ should be neither closed nor open, but I wasn't able to provide a counterexample for the open case.

For the case of $\pi$ being closed, I found a simple counterexample (as I was writing this post!): the line $x_1 = 1$ in $\mathbb A^2$ maps to $\{[x_0:1]:\ x_0 \in k\}$ in $\mathbb P^1$ which is precisely $\mathbb P^1$ minus the point at infinity. Since points are closed in the classical setting and $\mathbb P^1$ is irreducible ($\implies$ connected), the image would have to be clopen, which is impossible.

I'm still wondering about the open case. Maybe it's true: perhaps you could argue that each open set has a covering by sufficiently small open affines and somehow these are mapped to open sets in $\mathbb P^n$... who knows!

Edit: This question on the projective space being a quotient space and this proof (for the differential case) seem to do the trick, but I'm a bit unsure!

Answer 1

No matter what context you're working in, this is true.

• Scheme-theoretic: over any standard affine open $D(x_i)\subset\Bbb P^n$, the map is the spectrum of $k[x_0,\cdots,x_n]_{(x_i)}\to k[x_0,\cdots,x_n]_{x_i}$ and therefore flat of finite presentation. So $\pi$ is flat and locally of finite presentation, hence open (EGA IV Prop 2.4.6 or Stacks 01UA).
• Variety-theoretic: working over an algebraically closed field, use the fact the natural map from a variety $V$ to it's schemeification $V^{sch}$ induces a bijection between the posets of open subsets and that schemeification is a functor to derive the result from the scheme-theoretic standpoint. Or reprove all of the flat + locally finite presentation stuff for the case of varieties (not hard, the same proofs will work, but there's a reason serious people use schemes). Or use going-down based on the local description of the ring maps.
• Differential: $\Bbb CP^n$ is the quotient of $\Bbb C^{n+1}\setminus 0$ by the action of $\Bbb C^\times$, so since quotient maps are open, there you go.

Answer 2

@KReiser's answer solves the problem, but I think this fact is too basic (but not so trivial!) to use all the machinery from schemes. He also gave an answer to another question that my main source to understand the proof. I'll explain what I understood with more detail.

First of all, this is trivial if the base field is finite, since the topology is discrete, so we'll deal with the infinite case. We'll prove that $\pi: \mathbb A^{n+1}\setminus\{0\} \to \mathbb P^n$ is open by showing that the complement of images of opens are closed. Since the distinguished opens are a basis, we might as well take $U = D(f)$ a basic open without loss of generality.

Given $p \in \mathbb P^{n+1}$ with representant $p' = (t_0,\ldots,t_n)$, $p$ is not in $\pi(D(f)) \iff$ every other representant of $p$ in $\mathbb A^{n+1}$ does not lie in $D(f)$. That is, every other one is in $V(f)$, and since every other representant of $p$ is given by $\lambda p'$ for some nonzero $\lambda$, we have that $f(\lambda t_0, \ldots,\lambda t_n) = 0$ for all $\lambda \ne 0$. $V(f)$ must contain the (punctured) line given by the preimage of $p$.

À la Nullstellensatz, this should obviously translate to some closed condition and our only problem is describing it. If we write $f = \sum g_i$ as the sum of its homogeneous components, $p \not \in \pi(D(f)) \iff f(\lambda p') = \sum g_i(p') \lambda^i = 0$. Now using the infinitude hypothesis for the base field, this means that $g_i(p') = 0\ \forall i$. But since the $g_i$ are homogeneous , $g_i(p') = 0 \iff g_i(p) = 0$. Therefore $\pi(D(f)\cap U_0)^C = V(g_0,\ldots, g_d)$, and thus we have proved (and given a nice description of the images) that every open distinguished set is open under $\pi$. Thus $\pi$ is open.

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For future readers trying their own proofs: my initial strategy was trying to homogenize $f$ in small enough neighborhoods to show that the image of basic opens is open.

It's not hard to show that $D(f^{h_i})\cap V_i \subseteq \pi(D(f)\cap U_i)$. Attempting to prove the reverse inclusion doesn't work since not all images of basic opens are basic, so you have to adopt another strategy.