General Topology - 2-fold Covering map of Mobius strip (with unit circle)

Question

Given $I=[0,1]$, $S^{1}=\{(x,y)\mid x^2+y^2=1\}$, and the mobius strip M defined by the relation $(0,t)\sim(1,1-t)$ on $I\times I$, How can I prove that there exists a covering map $p:S^{1}\times I\to M$, such that $\forall t\in M: |p^{-1}(t)|=2$ ?

I know that given a covering map p to a connected space B, if there exists $b_0\in B$ such that $|p^{-1}(b_0)|=k$, then p is a k-fold covering of B. (and know how to proove it). So if I were to find (or proove the existence of) a covering map $p:S^{1}\times I\to M$ such that there exist $t\in M: |p^{-1}(t)|=2$, I would be set.

Is this even the right way to approach the question?

Any solutions \ guidance will be of great help.

Answer 1

There is an homeomorphism $S^1\rightarrow [0,2]/(0=2)$ (use for instance polar coordinates on the plane and scale).

Then $$f:(x,y) \in [0,2]\times I \mapsto (x \text{ }(mod\text{ }1),y) \in I \times I$$ is well defined and induces a continuous function when you quotient on the right.

Since $f(0,t)=f(2,t)$ you get a continuous function $S^1\times I \rightarrow M$ which is the desired covering.

You can also convince yourself geometrically that you can cover a Moebius strip by a cylinder if you get around it twice.

Answer 2

Picture proof, the squares correspond to the Möbius strip, the arrows show how to identify, the whole rectangle corresponds to circle times interval: