Identifying all points of edge of disc If we identify all points of the edge of a disc, do we get the Moebius strip? Why?
 A: It depends how you identify them. If you identify all of them to one point then the resulting space is a sphere $S^2$.
In detail: Define an equivalence relation $\sim$ on $D^2$ with $x \sim y$ if and only if either $x=y$ or $x,y \in \partial D^2$. Then it is not so difficult to show that $D^2 / \sim $ is homeomorphic to a sphere $S^2$.  
A: You definitely don't get a Möbius strip.  The disc has a boundary, which is a circle.  If you identify all the boundary points, the circle disappears, leaving a surface with no boundary.  You might get a sphere, or a torus, or a Klein bottle, or a projective plane, or something else, depending the details of how you make the boundary points go away.  But if you identify all the boundary points, there is no boundary left.
But the Möbius strip has a boundary, which is also a circle.  So if you started with a disc with one circular boundary, and got rid of that circle by squashing it down to a point, but ended up with a Möbius strip with one circular boundary, where could the new circle have come from?
A: For n>0, the following relation holds:
$B^n/ S^{n-1} \cong S^n $
Where $B^n$ is the n-ball (or n-disk), and $S^n$ is the n-sphere (surface)
The case you are asking is n=2. This will result in a 2-sphere in 3 dimensions. There exists an explicit homeomorphism (from $B^2/ S^{1}$ to $S^2$) which isn't too hard to derive, which will take all points on the boundary of the disc to the north-pole of a sphere (since after identification, this boundary IS one point), and the rest of the disk will be wrapped around the surface of the sphere. 
