Submanifolds as Inverse Images

[Kreizhn]

Homework Statement

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

2. The attempt at a solution

So we can quite simply show that $\mathbb{RP}^1$ is a submanifold of $\mathbb{RP}^2$, 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 $f : M\rightarrow N$ be a smooth manp and $P \subseteq N$ be a submanifold of N. If $f \pitchfork P$, then $f^{-1}(P)$ is a submanifold of M, provided that $f^{-1}(P)\neq \emptyset$.
Here $f\pitchfork P$ 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?

 

[mighty2000]

Thank you for your interesting question. I have a few thoughts on how to approach this problem. First, let's define the terms "smooth function" and "regular value" to make sure we're on the same page. A smooth function is a function that has continuous derivatives of all orders. A regular value of a function is a point where the derivative is nonzero.

Now, let's consider the function f : \mathbb{RP}^2 \rightarrow \mathbb{R} given in the problem statement. If this function had a regular value q such that f^{-1}(q) = \mathbb{RP}^1, then this would mean that the derivative of f at every point in \mathbb{RP}^1 is nonzero. However, this is not possible because \mathbb{RP}^1 is a closed curve in \mathbb{RP}^2 and any smooth function on a closed curve must have a critical point (where the derivative is zero) by the extreme value theorem. Therefore, f cannot have a regular value q such that f^{-1}(q) = \mathbb{RP}^1.

Another way to approach this problem is to use the theorem you mentioned in your attempt at a solution. If f^{-1}(q) = \mathbb{RP}^1 for some regular value q of f, then f must be transverse to \mathbb{RP}^1. However, this is not possible because \mathbb{RP}^1 is a submanifold of \mathbb{RP}^2 and any smooth function on a submanifold must be tangent to the submanifold. Therefore, f cannot be transverse to \mathbb{RP}^1 and thus cannot have a regular value q such that f^{-1}(q) = \mathbb{RP}^1.

I hope this helps you in your solution. Good luck!