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I'm working through the first chapter of Morris Hirsch's "Differential Topology". On Chapter 1, section 3 exercise 11, I encountered the following question.

"Regarding $S^1$ as the equator of $S^2$, we obtain $P^1$ as a submanifold of $P^2$ (Hirsch uses $P^n$ to denote real projective n-space). Show that $P^1$ is not a regular level surface for any $C^1%$ map on $P^2$. Hint: no neighbourhood of $P^1$ in $P^2$ is separated by $P^1$."

I attempted to solve this by contradiction, supposing there was a $C^1$ function of $P^2$ such that $P^1=f^{-1}(y)$ and that $T_pf$ is surjective for each $p\in P^1$. Then the inverse function theorem would imply that every open neighbourhood of $P^1$ is diffeomorphic to some open neighbourhood of $y$. I have a feeling that I'm meant to apply the hint here, and arrive at a contradiction by showing a topological property is not preserved under homeomorphism (path connectedness hopefully).

Unfortunately, I am having trouble visualising why the hint it true. I don't think I can go further without understanding that.

2 Answers 2

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I think your argument will work if you consider what happens when you remove the point $y$ from your neighbourhood.

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    I think I understand now. Thank you very much.2012-08-30
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An other idea is to think of $P^1$ to be the subset in $P^2$ with homogeneous coordinates $[0:b:c]\in P^2$. In this way, you can see that $N=P^2\backslash P^1$ is exactly $\{[1:\alpha:\beta] \in P^2, \alpha,\beta\in \mathbb R \}$ which is clearly homeomorphic to $\mathbb R^2$ and hence connected.

Now suppose that $P^1$ is the regular level set of a smooth map $f:P^2\rightarrow \mathbb R$ , $P^1=f^{-1}(0)$.

By definition $f(N)$ lies in $\mathbb R\backslash \{0\}$, and since $P^2$ is a compact connected surface, one can assume that $f(P^2)=[0,1]$. Then every point in $f^{-1}(0)$ is a minimum for $f$ and has null differential which is a contradiction.