Differentiability of Quotient Maps; Open Books . I would appreciate your comments re the differentiability of 
 a quotient map $q$: 
Say I have a quotient manifold $(S\times I )/q ;I=[0,1]$ , where $S$ is
a surface with non-empty boundary, where the quotient is   given by: 
i)$h: S \rightarrow S$ is a diffeomorphism that restricts to Id on the boundary , and q is defined by,
ii)$q: S \times I \rightarrow (S \times I)/q $,
$q :(x,0) :=(h(x),1)$ and $q$ is otherwise the identity.
How does one answer issues of differentiability of q ;can this be
done without using coordinates? I know since $h$ is a diffeo., that the
quotient space is a manifold. This is one of the steps in the open-book 
decomposition of a 3-manifold http://en.wikipedia.org/wiki/Open_book_decomposition ,
but this quotient is not quite the decomposition of the 3-manifold $M^3$ (the decomposition
of the manifold is homeomorphic, diffeomorphic to $M^3$).  Is there a way of defining
the derivative (i.e., the tangent map) of this quotient without using coordinates?
Any Ideas?
Thanks.
 A: As it seems to me, the hard part is convincing oneself that this quotient is a smooth manifold. The following is a discussion about the derivative, assuming one knows that the quotient is smooth. In the end we introduce an alternative way to think of the quotient, which may help to show smoothness of the quotient.
Obviously, for any point $(p,t)\in S\times[0,1]$ with $t\ne0,1$ there is a neighborhood contained in $S\times (0,1)$, and there is no problem to define the derivative. Now consider a point $(p,0)$. The quotient maps this point to the equivalence class $q(p)=\{(p,0),(h(p),1)\}$, and the tangent space to the quotient manifold at $q(p)$ may be thought of as
$$(T_{(p,0)}S\times[0,1]\cup T_{(h(p),1)}S\times[0,1])/\sim\\=((T_{p}S\oplus T_0[0,1])\cup (T_{h(p)}S\oplus T_1[0,1]))/\sim$$ where for any $v\in T_pS,$ $$v\sim dh_p(v),$$ and $T_0[0,1],T_1[0,1]$ are identified in the obvious manner. In other words, the tangent space to the quotient at $q(p)$ can be thought of in two equivalent ways, one of which is just $T_{(p,0)}S\times[0,1]$, and then the derivative $dq$ can be thought of as the identity.
Intuitively, the quotient map glues $(p,0)$ together with (h(p),1), and so the derivative glues together both tangent spaces.
And now an alternative approach: Consider the manifold $S\times\mathbb{R}$, with the $\mathbb{Z}$ action given by $$(n,(p,t))\mapsto(h^n(p),t+n),\qquad n\in\mathbb{Z},(p,t)\in S\times\mathbb{R}.$$ It is obvious that this action is an action of diffeomorphisms, and it is clearly properly discontinuous, thus we can define the quotient manifold $$(S\times\mathbb{R})/\mathbb{Z}.$$ I think the latter is equivalent to the quotient presented in the question, and now it is clearly smooth. This alternative approach may also help to understand the derivative.
