I have what is--I think--a really elementary topology question, but I'm an ignorant dummy who's never taken a topology class, and I'm more interested at the moment in knowing the answer (for experimental physics reasons) than in figuring it out. It's super simple:
You have some base space $X$, for which we can take $X = S^2$. We also have a continuous map $M: X \rightarrow S^2 \subset \mathbb{R}^3$ and some projection operator $P: \mathbb{R}^3 \rightarrow \mathbb{R}$ (say, we're measuring the $z$-component of some field of unit vectors in the ambient euclidean space of our sphere). Ok, now what I want to know is this: does $P \circ M$ determine the homotopy class of $M$? In particular, if $M$ is not homotopic to a constant map, is $P \circ M$ a certificate of this fact?
Clearly for any homotopy class, you can find a representative for which the range of $P \circ M$ is $[-1, 1]$... I think we have to do something that is somehow "nonlocal", like showing that nontrivial $M$ divide $X$ into regions of positive $P \circ M$ and negative $P \circ M$ separated by a loop with value zero, or something like this.
Anyway, I'm really not sure how to approach this, although just know I'll be smacking my forehead in a minute.
Thank you!
Take $X=S^2$, let $M_1 = \mathrm{id}_{S^2}$ and let $M_2(x,y,z) = (x,\sqrt{1-x^2},0)$. Let $P : \mathbb{R}^3 \to \mathbb{R}$ be the projection map onto the first coordinate. Then $P \circ M_1 = P \circ M_2$, but $M_1 \not\sim M_2$ since $M_2$ is homotopic to a constant map but $M_1$ is not.