Show $S^2$ is not homeomorphic to the closed unit disk.

Question

How to show unit closed disk is not homeomorphic to sphere $S^2$?

Answer 1

There are more topological invariants than the fundamental group. For example, the Euler characteristic of the disk is $1$ and the Euler characteristic of the 2-sphere is $2$. Another example, $\pi_2(D^2) = 0$ and $\pi_2(S^2) \cong\mathbb{Z}$.

You can think of a topology sequence as, in part, the building of a repertoire of invariants which you can use to check whether spaces are not homeomorphic (or not homotopy equivalent).

Answer 2

The fundamental group of the closed disk is trivial since it is contractible. But if there were a homeomorphism, remove the center of the unit disk and this changes the fundamental group to $\mathbb{Z}$ while $S^2$ with a point removed is contractible.

Answer 3

Also, you can consider arbitrary four distinct points a,b,c,d (in order) on the boundary of the unit disk and their images on S^2. Lets say f is a homeomorphism from unit disk to S^2. Geometrically, we can find two paths from f(a) to f(c) and from f(b) to f(d) on S^2 so that these paths do not intersect. However, If we take the inverse images of those two paths, we can say that they are paths one of which is from a to c and the other is from b to d. Such paths must have at least one intersection point. Lets call this as x in unit disk. Now, f(x) is in the intersection of those two paths, but the images of those paths were taken to be non-intersecting. Hence, we get that there can be no homemorphism between unit disk and S^2.

Answer 4

I don't know if it is too much to invoke Jordan Separation Theorem to show this. $S¹$ doesn't separate the disk but it's homeomorphic image (if a homeomorphism exists) separates the sphere.