What does it mean to "Glue the boundary circle of a mobius strip to another circle in a 2:1 covering map"

Question

What does it mean to "Glue the boundary circle of a mobius strip to another circle in a 2:1 covering map."?

This wording is confusing to me. Does this mean the preimage of the the circle will be two points on the boundary of the Mobius band? Or, is it implied that wrapping a mobius strip around a circle, the mobius strip goes around the circle twice because a mobius strip's boundary retracts onto it's central circle in a 2:1 fashion.

See why I'm a bit confused?

Answer 1

Here's a general description, of which your particular problem is a special case.

Given two topological spaces $X,Y$, a subset $A \subset X$ and a function $f : A \to Y$, gluing $A$ to $Y$ using $f$ means forming the quotient space of the disjoint union $X \coprod Y$ using the equivalence relation generated by $a \sim f(a)$, for all $a \in A$.

The equivalence classes of this equivalence relation can always be explicitly listed, but if $f : A \to Y$ is a surjective 2--1 map then the equivalence classes are particularly easy to list: all singleton sets $\{x\}$ for $x \in X-A$; and all tripleton sets $f^{-1}(y) \cup \{y\}$ for $y \in Y$. (In fact this description holds in general except that the sets $f^{-1}(y) \cup \{y\}$ need not always be three point sets).

Answer 2

Let $M$ be the Möbius strip and $S$ be its boundary circle we we may assume to be given together with a homeomorphism $h : S \to S^1$. Now let $\mu_2 : S^1 \to S^1, \mu(2)(z) = z^2$ which is the protype of a 2:1 covering map. Then you form the adjunction space $$X = M \cup_{\mu_2 h} S^1 = (M \sqcup S^1)/\sim$$ where $s \sim \mu_2h(s)$ for all $s \in S$. Then $S^1$ embeds a subspace into $X$ by mapping $z \in S^1$ to $[z] \in X$. In fact each of these points $[z]$ has two preimages in $M$.