This question was part of 1 of the exercises in differential geometry course and I am struggling with this part:
Question: Let $ s: S^{1} \to$ Mobius strip be a section of the vector bundle ( The vector bundle is defined as: Mobius strip is a vector bundle over over $S^{1} $ with typical fiber $\mathbb{R}$. I have managed to show that this is actually a vector bundle ) that is at least continuous. Show that there exists $m\in S^{1}$ such that $s(m)=0$.
But I am not able to show the existence of such an $m\in S^{1}$ such that $s(m)=0$.
Can you please give some leads?
The specifics heavily depend on the what definition you use for your Mobius strip.
We'll define the circle as $S^1=[0,1]/\sim_{0=1}$ and the Mobius strip as $M=[0,1]\times \mathbb R/\sim$ with $(0,x)\sim (1,-x)$ for each $x\in\mathbb R.$ The map $M\to S^1$ sends $\langle x,y\rangle_{\sim}\to \langle x\rangle_{\sim_{0=1}}.$
Then $m:S^1\to M$ is defined on $(0,1)$ so that $m(t)=\langle t,m'(t)\rangle_{\sim}.$ At $0=1,$ $m(0_{\sim_{0=1}})$ can be written as $\langle 0,a\rangle=\langle 1,-a\rangle,$ for some $a.$
If $a=0,$ then we are done. So assume $a\neq 0.$
Now, continuity of $m$ means we get $\lim_{t\to 0^+} m'(t)=a,\lim_{t\to1^{-}} m'(t)=-a.$
Thus $m'$ is a continuous function $(0,1) \to \mathbb R$ with negative and positive values, and thus must have a zero value.
Basically, if you think of a Mobius strip as a glued rectangle, a section is a path $(t,f(t))$ for $t\in [0,1]$ and $f(t)\in[-1,1]$ such that $f(1)=-f(0).$ Then the intermediate value theorem means some $f(t)=0.$
Intuitively: If you've ever seen the demonstration of cutting a Mobius strip on the center line, resulting in a single-piece loop homeomorphic to a cylinder of twice the circumference, then any loop on the Mobius strip that never hits the middle line remains a loop, but then the loop could not have been a section.
Another definition of $M$ might be $S^1\times \mathbb R/\sim$ where $(x,y)\sim (-x,-y).$
The map $M\to S^1$ sends $\langle x,y\rangle\mapsto x^2.$
Then a section $m:S^1\to M$ is defined as $m(x)=\langle f(x),g(x)\rangle$ where $f(x)$ is a principle square root of $x$ which is continuous everywhere but $-1,$ and with $f(x)\to i$ as $x\to -1$ in the upper half plane and $f(x)\to -i$ as $x\to -1$ in the lower half plane.
Then you get the $g(x)\to g(-1)$ as $x\to -1$ from the upper half-plane, and $g(x)\to -g(-1)$ as $x\to -1$ from the lower half plane.
So either $g(-1)=0$ or the intermediate value theorem applies.