Understanding Hempel's proof of uniqueness of cube with handles In Hempel's 3-Manifolds book, Theorem 2.2 says that if $P$ and $Q$ are two cubes, both with $n$ handles, and both are orientable, then they are homeomorphic.
He defines a cube with handles as a 3-manifold that contains a collection $\{D_1,\ldots,D_n\}$ of embedded 2-disks such that, cutting along all of them, you're left with a 3-cell.
The proof proceeds by getting these 3-cells for $P$ and $Q$ (call them $R$ and $S$ respectively).  He notes that since both $P$ and $Q$ are orientable, the two disks on the boundary of $R$ (or $S$) corresponding to $D_i$ are identified in an orientation-reversing way. He then quotes a PL theorem to show there is a homeomorphism $F$ from $R$ to $S$, sending the disks on $\delta R$ corresponding to $D_i$ to the same corresponding disks on $\delta S$. This theorem is below:

Theorem: If $M$ is a PL $m$-manifold, and $C_1$ and $C_2$ are PL $m$-cells in $\rm{Int}\ M$, and $X$ is any closed subset of $M$ such that $C_1\cup C_2$ lies in a component of $M-X$, then there is a PL isotopy $\phi: M\times I\rightarrow M$ with $\phi_0\equiv1$, $\phi_t{|\ X}\equiv1$, and $\phi_1(C_1)=C_2$.

I don't understand how this theorem applies.  Specifically, what should I be using for $M$, $C_1$, $C_2$, and $X$ in the above theorem? Also, there's no mention of orientability, so I'm not sure why Hempel emphasizes it so much.  
I thought I might glue $R$ and $S$ via these disks on their boundaries, but then if I let $X$ be these glued disks in the theorem above, $R$ and $S$ lie in different components.

The proof is finished by saying this homeomorphism $F$ can be extended to go from $P$ to $Q$, taking $D_i\times I$ in $P$ to the corresponding "tube" in $Q$.  To get this extension, he quotes another PL theorem:

Theorem: If $M$ is a PL $m$-cell or $m$-sphere, then any orientation-preserving PL homeomorphism of $M$ onto itself is isotopic to the identity.

Again, I have no idea what $M$ is supposed to be when applying this theorem.  I'm guessing I can successively use each $D_i$ as $M$, and the "tube" is the image of the isotopy, but then this seems to only depend on each $D_i$ being orientable, not all of $P$ or $Q$.
Please help me clear up how to use these two PL theorems.
 A: Here's how to answer your question about how the first theorem applies. 
Fix orientations of $D_1,...,D_n$. 
In the 3-cell $R$, denote some objects as follows. Let $\partial R$ denote the boundary of the 3-cell $R$, which is a PL 2-manifold homeomorphic to the 2-sphere $S^2$. Fix an orientation on $\partial R$ (e.g. fix a homeomorphism $\partial R \to S^2$). In $\partial R$ there is a pairwise disjoint collection of $2n$ PL 2-cells that I'll denoted $D^{+R}_1,D^{-R}_1,D^{+R}_2,D^{-R}_2,...,D^{+R}_n,D^{-R}_n$ such that $D^+_i$ is identified with $D_i$ in an orientation preserving manner and $D^-_i$ is identified with $D_i$ in an orientation reversing manner.
In the 3-cell $S$, denote similar objects as in the previous paragraph, replacing all $R$'s by $S$'s.
In the context of the 1st theorem, one takes $M=S^2$, and one also identifies $M$ with $\partial R$ and $\partial S$ by choosing orientation preserving homeomorphisms $\partial R \rightarrow S^2 \leftarrow \partial S$. It is probably best to choose these homeomorphisms so that union of the images in $S^2$ of the discs $D^{\pm R}_i$ is disjoint from the union of the images in $S^2$ of the discs $D^{\pm S}_i$; these homeomorphisms can always be chosen to have this additional property.
So now we can simply think of $\partial R = S^2 = \partial S$, and each of the 2n discs $D^{\pm R}_i$ is disjoint from each of the 2n discs $D^{\pm S}_i$, and so we have a pairwise disjoint collection of $4n$ PL discs in $S^2$.
The first theorem is now applied inductively, a total of 2n applications. 


*

*In the first application of the theorem, $C_1 = D^{-R}_1$, $C_2 = D^{-S}_1$ and $X$ is the union of all the other discs whose subscript is $i\ge 2$. After this first application of the theorem has been completed, we have arranged that $D^{-R}_1=D^{-S}_1$, and the other $2n-2$ discs have not been moved. 

*In the second application of the theorem, $C_1 = D^{+R}_1$, $C_2 = D^{+S}_1$, and $X$ is the union of the other discs, namely the discs $D^{-R}_1=D^{-S}_1$ as well as all other discs whose index is $\ge 3$. After this second application has been completed, we have arranged that $D^{-R}_1=D^{-S}_1$ and that $D^{+R}_1=D^{+S}_1$, and the other $2n-4$ discs have not been moved.


Now continue in this fashion. After all $2n$ applications of the theorem, we have arranged that the sets of discs $D^{\pm R}_i$ and $D^{\pm S}_i$ are identical, and are identically signed and subscripted.

Next, here's how to answer your question about how the second theorem applies. 
There will be $n$ applications of the theorem, one for each $i=1,...,n$. For the $i^{\text{th}}$ application take $M=D^{-R}_i=D^{-S}_i$. There is an orientation reversing homeomorphiam $f^R_i : D^{-R}_i \to D^{+R}_i$ which describe how points are identified in the first cube with handles. Similarly, there is an orientation reversing homeomorphism $f^S_i : D^{-S}_i \to D^{+S}_i$ which describes how points are identified in the second cube with handles. Because of how we carefully applied the first theorem, the homeomorphisms $f^R_1,f^S_i$ have the same domain $D^{-R}_i=D^{-S}_i$ and the same range $D^{+R}_i=D^{+S}_i$. Therefore, we can form an orientation preserving homeomorphism
$$(f^R_i)^{-1} \circ f_i : D^{-R}_i \xrightarrow{f_i} D^{+R}_i=D^{+S}_i \xrightarrow{(f^R_i)^{-1}} D^{-S}_i = D^{-R}_i
$$
Now apply the second theorem to this homeomorphism, for each $i$.
