- #1

Kreizhn

- 743

- 1

## Homework Statement

Show that there exists no smooth function [itex] f : \mathbb{RP}^2 \rightarrow \mathbb{R} \text{ such that } f^{-1}(q) = \mathbb{RP}^1[/itex] for some regular value q of f.

**2. The attempt at a solution**

So we can quite simply show that [itex] \mathbb{RP}^1 [/itex] is a submanifold of [itex] \mathbb{RP}^2 [/itex], which is what the first part of the question actually asked but I have omitted here since I don't think it's important. I've tried looking at a few things, but I have one method that really seems to scream out at me but I'm not sure how to use it.

Theorem: Let [itex] f : M\rightarrow N [/itex] be a smooth manp and [itex] P \subseteq N [/itex] be a submanifold of N. If [itex] f \pitchfork P [/itex], then [itex] f^{-1}(P)[/itex] is a submanifold of M, provided that [itex] f^{-1}(P)\neq \emptyset [/itex].

Here [itex] f\pitchfork P[/itex] means f is transverse to P. I want to use contradiction, but this theorem seems to have the implication in the wrong direction in order for me to fully utilize it. Any ideas?