$P^1$ not a regular level surface of a $C^1$ function on $P^2$ 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.
 A: 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.
A: I think your argument will work if you consider what happens when you remove the point $y$ from your neighbourhood.
A: Two minor comments:

*

*Say $f\in C^1(P^2, N)$ is an arbitrary map with $P^1=f^{-1}(y)$ for some regular value $y\in N$. Submersivity a priori implies only that the dimension of $N$ is not more than $2$. It would be good form to add a sentence to justify $\dim N = 1$. Then by the local character of the problem one can assume $N$ is an open interval.

*The Inverse Function Theorem (here and in general) does not imply what you say it does. Say $\dim(N)=2$, so that IFT kicks in along $P^1$. Taking small open sets and collating their diffeomorphic images gives that small open sets containing $P^1$ (which are homeomorphic to Mobius bands without boundary, if small enough) is open in the target. We don't have a diffeomorphism though, since $P^1$ is collapsed to a point (or alternatively orientability is a diffeomorphism invariant).

