Isotopy and attaching handles

Recall that a homeomorphism $f: \mathbb {R} \to \mathbb {R}$ is called periodic of period $1$ if either $f(x+1) = f(x) + 1$ (for $f$ order preserving) or $f(x+1) = f(x) - 1$ (for $f$ order reversing). Given an order-preserving homeomorphism $f': [a, a+1] \to [b, b+1]$, define $f: \mathbb {R} \to \mathbb {R}$ by writing $x \in \mathbb {R}$ uniquely as $x = t + n$ with $n \in \mathbb {Z}$, $t \in [a, a+1)$, and setting $f(x) = f'(t) + n$. Show that the map $f$ defined this way is a periodic homeomorphism.

Given two disjoint arcs $I, I'$ in the circle $S^{1}$, choose intervals $J, J' \subset \mathbb {R}$ so that $p(J) = I$, $p(J') = I'$, where $p: \mathbb {R} \to S^{1}$, $p(t) = (\cos 2\pi t, \sin 2\pi t)$. The unique order-preserving affine linear homeomorphism from $J$ to $J'$ induces (via $p$) a standard'' homeomorphism from $I$ to $I'$. Show that this standard'' homeomorphism does not depend on the choice of $J, J'$.

Use the homeomorphism $h: D^{1} \to I$, $h(s) = \frac{1}{2}(s+1)$, and the fact that any order-preserving homeomorphism $f: I \to I$ is isotopic to the identity while any order-reversing one is isotopic to $r'(t) = -t$ (where $I = [-1,1]$), to prove the analogous statement for $D^{1}$:

Define an embedding of $D^{1}$ into $S^{1}$ to be orientation preserving (resp.reversing) if it is the composition of an order-preserving (reversing) embedding of $D^{1}$ into $\mathbb {R}$ and the map $p: \mathbb {R} \to S^{1}$, $p(t) = (\cos 2\pi t, \sin 2\pi t)$. A homeomorphism from $S^{1}$ to $S^{1}$ is orientation preserving if it comes from a periodic homeomorphism of $\mathbb {R}$ which is order preserving.

When a $1$-handle is attached to a single boundary circle of a handlebody, there are two standard ways of attaching it up to homeomorphism, pictured in Figure 2.17: on the left, the new boundary has two components in place of one; on the right, the new boundary still consists of one boundary circle, and there is a Möbius band embedded in the surface using a piece of the collar and the handle $h^{1}$.

Attaching a 1-handle to one boundary circle.

Find the Möbius band that is referred to above on the right surface in Figure 2.17.

Show that the result of attaching a 1-handle to the same boundary circle is never homeomorphic to what you get when it is attached to two different boundary circles.

Theorem: Let $H$ be a handlebody with a handle decomposition with $0$-handles and $1$-handles. Then: (a) $\partial H$ is homeomorphic to the disjoint union of a finite number of circles; (b) there is a neighborhood $N(\partial H)$ of $\partial H$ which is homeomorphic to $\partial H \times I$, where $\partial H \times \left\{ 0\right\} \subset \partial H \times I$ is sent to $\partial H \subset H$ via the standard identification $(x,0) \mapsto x$.

The proof outline is by induction on the number of $1$-handles: the case of $0$-handles alone is immediate, and for the inductive step one writes $H = H_{1} \cup h^{1}$, assumes (a) and (b) hold for $H_{1}$, and uses the existence of a collar where the $1$-handle is attached to determine the new boundary and its neighborhood, treating separately the case where the two attaching arcs of $h^1$ lie on the same boundary circle of $H_1$ and the case where they lie on two different boundary circles of $H_1$.

Fill in the details of this proof.