Simply Connected Points in Disk

Question

Why is the set of all points $z \in D^2$ for which $D^2 \setminus \{z\}$ is simply connected just $S^2$?

Answer 1

I'm assuming the ordinary Euclidean topology. First note that the boundary of $D^2$ is (homeomorphic to) $S^1$ (not $S^2$, mind you). If a point belongs to the interior of $D^2 $ (which you can embed in $\mathbb R^2 $ as the unitary disc) you may as well assume it is the origin; build up the map $ H : [0,1] \times D^2 \to D^2 $ sending $$ (t,x) \mapsto (1-t)x + t\frac{x}{\|x\|}$$ and this is a strong deformation retraction of $D^2$ onto $S^1$, thus there is an isomorphism on the fundamental groups $$ i_* : \pi_1(S^1) \to \pi_1(D^2) $$ which is induced by the inclusion of spaces.

This cannot be done if the point lies on the boundary, and indeed in that case it's quite apparent that the fundamental group is unchanged.

Answer 2

Intuitively, a disc is simply connected because any loop can always be shrunk to a point. Now if we take a point away from the interior of the disc, this is no longer the case, since when when we shrink the loop it gets caught in the hole we just created. This is not the case when we remove a point from the boundary of $D^2$, that is, $S^1$.