algebraic-topology
It is know that $\partial M\approx S^1$, where $\partial M$ is the boundary of the mobius band. But how do we prove that? I am trying to define the homeomorphism explicitly, but it is messy.
The Mobius band $M$ is defined by the quotient $q:I\times I\rightarrow M$ with the equivalence relation $(0,x)\sim(1,1-x)$.
$\partial M=q(I\times\{0,1\})$.
I take it that $S^1$, for you, is the unit interval with an identification of the two ends, so that $0 \sim 1$.
The intersection of $U_1$ and $U_2$ is then messy, because it has two components, namely $0 < x < 1/2$ and $1/2 < x < 1$. You want to define $g_{12}$ on both parts, and I suggests that you define it on the first part by
$$ g_{12} (x)(y) = y $$ while on $1/2 < x < 1$ you define it by $$ g_{12} (x)(y) = 1-y. $$
The result should THEN be a mobius band. What you've got a this point is a cylinder (as you've verified).
I'm going to slightly edit your definition, and say that I want $g_{12}(x)(y) = -y$ for $1/2 < x < 1$ (and similarly for the Mobius band), because this leaves a "centerline" at $y = 0$, while in your definition, the centerline is at $y = 1/2$; if you were defining a finite mobius band, using $[0, 1] \times [0, 1]$, that might make sense (although I might argue for $[0, 1] \times [-1, 1]$ instead!). But since you're using all of $\mathbb R$ as a fiber, it'd be nice if the transition maps were, fiberwise, vector-space isomorphisms, which requires sending $0$ to $0$.
Now my space $X$ consists of equivalence classes that look like one of four things: $$ L_{a, s} = \{(a, s, 1), (a, s, 2) \} $$ (where I've added an integer "identifier" to each patch), and where $s$ is any real number, or $$ R_{a, s} = \{(a, s, 1), (a, -s, 2) \} $$ The first is for $0 < a < 1/2$, and the second for $ 1/2 < a < 1$. Then there are two other kinds of equivalence classes: $$ Z_s = \{(0, s, 1), (1, -s, 2) \} $$ and $$ H_s = \{ (\frac{1}{2}, s, 1), (\frac{1}{2}, -s, 2) \}. $$ (The "Z" is mnemonic for "zero or one" and the "H" is for "halfway".)
In the Mobius band $M$, there are two kinds of equivalence classes: $$ U_{a, s} = \{(a, s)\} $$ (the "U" is for "usual") and $$ P_{s} = \{(0, s), (1, -s)\} $$ where the "P" is for "excePtional". :)
Now all I have to do is tell you how my homeomorphism $F$ will send each class in $X$ to a class in $M$. In the following, each line of the definition holds for all $s \in \mathbb R$. So: \begin{align} F(L_{a,s}) &= U_{a, s} & 0 < a < \frac{1}{2} \\ F(R_{a,s}) &= U_{a, -s} & \frac{1}{2} < a < 1 \\ F(Z_{s}) &= P_s \\ F(H_{s}) &= U_{ \frac{1}{2}, s}. \end{align}
It's not hard to see that $F$ is bijective, and I leave that to you; the only question is continuity.
Continuity except at $a = 0, \frac{1}{2}, 1$ seems pretty clear, so I won't discuss that. Let's check continuity at $a = \frac{1}{2}$. To do so, I'm going to look at a "vertical" curve in $X$ and a "horizontal" one, and check that each maps to a nice continuous curve in $M$, That's not really a proof, but I'm hoping it'll suffice to convince you.
The vertical curve is $$ \gamma: I \to X : t \mapsto H_t. $$
Under the map $F$, this becomes $$ \alpha = F\circ \gamma : I \to M : t \mapsto F(H_t) = U_{ \frac{1}{2}, t} $$ which looks perfectly nice (as it should: $F$ is a vector-space isomorphism on the fiber!).
What about a "horizontal" curve in $X$? Let's look at one at "height" $1/3$: $$ \beta: [\frac{1}{4}, \frac{3}{4}] \mapsto X : t \mapsto \begin{cases} L_{t, \frac{1}{3}} & t < \frac{1}{2} \\ H_\frac{1}{3} & t = \frac{1}{2} \\ R_{t, \frac{1}{3}} & t > \frac{1}{2} \end{cases}. $$
It's pretty clear that $\beta$ is a continuous curve, as is $F\circ \beta$. So that part works out OK as well.
Finally, there's continuity at $0$. That's the tricky one. Once again, I'll use $\gamma$ and $\beta$ to denote the vertical and horizontal curves. In fact, I won't even bother with $\gamma$ -- you can give that a shot yourself. But $\beta$ is more interesting. I'll define it for $-1 < t < 1$, to make life a little easier for myself:
$$ \beta: [-1, 1] \to X: t \mapsto \begin{cases} Z_\frac{1}{3} & t = 0 \\ L_{\frac{t}{2}, \frac{1}{3}} & t > 0\\ R_{1+\frac{t}{2}, -\frac{1}{3}} & t < 0 \end{cases} $$
If you look at $F\circ \beta$, you'll see that it's a curve that runs along at height $1/3$ in the mobius band to the right of $x = 0$, and at height $-1/3$ to the left of $x = 1$, which makes it continuous in the Mobius band. Of course, you also need to check that it's continuous in $X$, but I think that's not TOO difficult.
I hope (a) that this explicit description is of some help to you, and (b) that you never have to do such a thing explicitly again. :)
This is a diagram of the Klein bottle, note that the diagonal lines divide it into 2 Möbius strips sharing a boundary:
So the answer is yes.
Best Answer
So first of all you need to know that $\partial M=(I\times\{0,1\})/\sim$. Then consider
$$f:I\times\{0,1\}\to S^1$$ $$f(x,y)=\exp(\pi i (x+y))$$
This function maps $I\times\{0\}$ onto the first half circle and $I\times\{1\}$ onto the second half circle. It is not a homeomorphism, because it is not injective. However it induces a well defined continuous bijection
$$F:\partial M\to S^1$$ $$F([x,y])=f(x,y)$$
on the quotient and thus a homeomorphism since $\partial M$ is compact.