0

My lecturer, to demonstrate that the projective complex space $\mathbb{P}^n(\mathbb{C})$ is Hausdorff, proves that the restriction of the quotient map to $S^{2n+1}$ is a closed map. Here is how the proof begins.

Let $$ \pi_0 := \pi_{\mid S^{2n+1}} : S^{2n+1} \to \mathbb{P}^n(\mathbb{C})$$ be the restriction of the quotient map $\pi: \mathbb{C}^{n+1}\setminus \{0\} \to \mathbb{P}^n (\mathbb{C})$. We want to prove the closedness of $\pi_0$: $\forall C$ closed set in $S^{2n+1}$ $\pi_0(C)$ is a closed set in $\mathbb{P}^n(\mathbb{C})\iff \pi_0^{-1}\left(\pi_0(C)\right)$ is closed in $S^{2n+1}$.

However, being in a quotient topology with quotient map $\pi$, proving that $\pi_0(C)$ is a closed set in $\mathbb{P}^n(\mathbb{C})$ should be equivalent to prove that $\pi^{-1}\left(\pi_0(C)\right)$ is closed in $\mathbb{C}^{n+1}\setminus \{0\}$. Therefore, my question is: does $\pi^{-1}\left(\pi_0(C)\right)$ being closed in $\mathbb{C}^{n+1}\setminus \{0\}$ implies that $\pi_0^{-1}\left(\pi_0(C)\right)$ is closed in $S^{2n+1}$ and viceversa?

Massimo Bertolotti
  • 73
  • 1
    The trick is probably something like factoring the map $\pi:\mathbb{C}^{n+1} \to \mathbb{CP}^n$ as a map $\mathbb{C}^{n+1} \to S^{2n+1}$ (by dividing by modulus) followed by the map $\pi_0:S^{2n+1} \to \mathbb{CP}^n$. – Ethan Dlugie Jan 24 '21 at 20:00
  • 1
    Am I missing something silly? Isn't $\pi_0^{-1}(Z)$ just $\pi^{-1}(Z) \cap S^{2n+1}$? – Ted Shifrin Jan 24 '21 at 20:15
  • @TedShifrin Ah yes, come to think of it, certainly it can be used to prove $\pi^{-1}\left(\pi_0(C)\right)$ closed in $\mathbb{C}^{n+1}\setminus {0}\implies \pi_0^{-1}\left(\pi_0(C)\right)$ closed in $S^{2n+1}$. I don't think that it can be used for the converse, though: if $\pi_0^{-1}(Z)$ is closed in $S^{2n+1}$ and $\pi_0^{-1}(Z)=\pi^{-1}(Z)\cap S^{2n+1}$ I don't think that $\pi^{-1}(Z)$ is necessarily closed in $\mathbb{C}^{n+1}\setminus {0}$. – Massimo Bertolotti Jan 24 '21 at 20:53
  • 1
    Yes, that follows as well. Consider the radial projection $p\colon \Bbb C^{n+1}-{0}\to S^{2n+1}$. Then $\pi^{-1}(Z) = p^{-1}(\pi_0^{-1}(Z))$. I think this is the content of Ethan's remark earlier. – Ted Shifrin Jan 24 '21 at 21:09
  • I see, that's a nice way to see it. Thank you both for your help! – Massimo Bertolotti Jan 24 '21 at 21:31

1 Answers1

1

This community wiki solution is intended to clear the question from the unanswered queue.

As shown in the comments, the answer is "yes". Also have a look at The restriction of a quotient map $p : X \to Y$ to a retract of $X$ is again a quotient map. This shows that $\pi_0 : S^{2n+1} \to \mathbb{P}^n(\mathbb{C})$ is a quotient map because $\pi: \mathbb{C}^{n+1}\setminus \{0\} \to \mathbb{P}^n (\mathbb{C})$ is one. Therefore the following are equivalent for $A \subset \mathbb{P}^n (\mathbb{C})$:

  1. $A$ is closed in $\mathbb{P}^n (\mathbb{C})$.

  2. $\pi^{-1}(A)$ is closed in $\mathbb{C}^{n+1}\setminus \{0\}$.

  3. $\pi_0^{-1}(A)$ is closed in $S^{2n+1}$.

Paul Frost
  • 76,394
  • 12
  • 43
  • 125