So to show this is a 2 manifold with boundary you have to show that around each point there is a neighborhood that is either homeomorphic to $D^2$ or $D^2_+= \{(x,y)\in \mathbb{R} | \,\,\,\, y\geq 0, \,\,\,\, |(x,y)|<1 \}$.
Let $X$ be the described set $X / \sim$ the quotient and $\pi$ the quotient homomorphisim.
For $x \in \pi( \text{int} \, ( X )) = \text{int} \, (X)$ we are done, this set is homeomorphic to the disk. On $\text{int} \,(X)$, $\pi$ is a homeomorphism.
For $x \in \pi( (-10, 10) \times \{1\})$ consider $\pi((-10, 10) \times [1,-1))$. Similarly for the other side.
For $x \in \pi( \{10\} \times (-1,1) )$ it is more difficult. Here we have to somehow work with the twist. Let $f: [-10,-9) \cup (9,10] \times (-1,1) / \sim \,\, \to (-1,1)^2 $ be:
$$
f(x,y) = \left\{
\begin{array}{lr}
(x-10,y) & : x \in (9,10] \\
(x+10,-y) & : x \in [-10,-9)
\end{array}
\right.
$$
I claim that this is continuous and bijective. Pulling $f$ back to $X$, i.e. considering $f \circ \pi : X \to (-1,1)^2$, it is continuous (this is the universal property of quotients). And it is bijective as $f \circ \pi$ is 1 to 1 except for the points that are identified where is is 2 to 1. But those points are identified so $f$ is 1 to 1 and onto. $f$ is also an open map, any open set in $X / \sim$ is the union of the images of an open sets from $X$ and $f \circ \pi$ is clearly an open map.
Any space constructed in this fashion deformation retracts to the core circle, so they are all homotopy equivalent to $S^1$.
I think the study of homeomorphism classes of these objects is an interesting exercise, so I'll leave it as one for you with the following hints as a guide:
Homeomorphism doesn't depend on any embeddings; it's an intrinsic property. So "number of twists," which is definitely an extrinsic property (in this case, it comes from the relationship between the space and its immersion in $\mathbb{R}^3$) isn't going to be as helpful as you might want in classifying up to homeomorphism.
Instead, it may be helpful to think of these spaces intrinsically, in particular as quotients of the square $[0,1]\times [0,1]$. Do this by thinking about how you would "glue" the square to produce these multiply-twisted strips.
Notice that the square does not "know" how many times you've twisted it. All it knows is how you glue it. Can you formalize this into a proof about homeomorphisms between differently-twisted strips?
Based on the result of (3), can you classify all such twisted bands into homeomorphism types?
Best Answer
Consider the following strips:
The "non-cut" möbius strip is the composition of gluing $A$ and $B$ along the common side $[0,2]\times\{1\}$ and do the classical identification for the mobius strip.
For visualize the "cut" möbius strip, just do the classical identification for the mobius strip for the set $A\bigsqcup B$.
I think a good approach is consider the following quotient map:
where $\$$ denote the "cut" möbius strip.
Exercise: Prove that $\pi\left(A\bigsqcup B\right)$ is connected.