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.
Yes, the normalization axiom suffices. The method to calculate higher Stiefel-Whitney classes from $w_1$ is called the splitting principle. The splitting principle says that for any space $X$ and vector bundle $E \rightarrow X$, there is a space $Z$ and map $Z \rightarrow X$ such that the map $H^*(X) \rightarrow H^*(Z)$ is injective, and the pullback of $E$ splits as a sum of line bundles.
With this fact, we can use the Cartan formula plus definition of $w_1$ to calculate the SW classes of the splitting of $E$ (which is a vector bundle over $Z$), and then we know this uniquely determines the SW classes of $E$ by naturality plus the injectivity of $H^*(X) \rightarrow H^*(Z)$.
The construction of the map $Z \rightarrow X$ is straightforward. Inductively, we define $X'$ to be the projectivization of $E \rightarrow X$, i.e. the points of $X'$ are the 1-dimensional subspaces of the fibers of $E \rightarrow X$. Then the pullback along the projection $X' \rightarrow X$ splits over a point $\ell_x \in X'$ as $\ell_x + F_x/\ell_x$. Now repeat this process with $X'' \rightarrow X'$.
Best Answer
You can use the fact that if you cut the open Möbius strip around center the resulting space is connected.
Any homeomorphism from the Möbius strip to a cylinder will induce an isomorphism on fundamental groups, so if a homeomorphism existed it must send the center of the Möbius strip to a curve homotopic to a circle going around the cylinder, and removing anything homotopic to such a curve from the cylinder disconnects it.