Let $M$ be a connected compact manifold without boundary, $\pi:\widetilde{M}\to M$ be the universal covering map. A fundamental domain of $(\pi,\widetilde{M}, M)$ is a compact subset $D\subset \widetilde{M}$ such that   

1. the union of $\gamma D$ over all $\gamma\in \pi\_1(M)$ covers $\widetilde{M}$,   

2. the collection $\gamma D^o$ are mutually disjoint,   

3. $\pi(D)=M$ and the restriction $\pi|\_{D^o}:D^o\to M$ is diffeomorphic onto its image.


My question is:   

Does there always exist some simply connected fundamental domain?   

Is every fundamental domain simply connected? 


Motivation.  

I saw the following statement in several papers about dynamical systems: let $B^d(0,1)$ be the unit ball in $\mathbb{R}^d$ and $M$ be a $d$-dimensional compact connected manifold without boundary. Then $M\simeq B^d(0,1)/\sim$ where $\sim $ represents some gluing along $S^{d-1}=\partial B^d(0,1)$.  

I think the statement might be related to above question. 


Thanks!



fundamental domain of universal covering
The first part of 3 follows from 1. The second part of 3 follows from 2.


There is always a contractible (in particular simply connected) fundamental domain:
Triangulate $M$. Take the union of all the codimension zero open simplices, together with just enough codimension one open simplices to make it connected. This will be $\pi(D^{o})$. It is an open subset of $M$, dense and contractible (homotopy type of a maximal tree in the dual cell structure). Choose a lifting to $\tilde M$ and let $D$ be its closure.


For an example of a non simply connected fundamental domain in dimension $3$, take $M=S^2\times S^1$, $\tilde M=S^2\times \mathbb R$. One fundamental domain is $S^2\times I$, but you can make another by first cutting $S^2\times I$ into two non simply connected pieces meeting along a closed surface and then letting $D$ be the union of the left piece and a translate of the right piece. 


Edit: This wrong, as pointed out in the comments. My error ws in imagining that the inclusion of the lifted open thing into its closure was necessarily a homotopy equivalence. I wonder if this approach can be fixed.


