Contractible neighbourhood of $D^2$

Question

I am proving that $D^2$ is not a topological manifold, but I am running into the following. Consider $D^2\subset \mathbb{C}$ and let $1\in D^2$ and $U$ is a contractible open neighbourhood of 1. How can I show that $U\setminus\{1\}$ is still contractible?

Answer 1

It seems that you want to show that manifolds with boundary are no manifolds. To do this you have to show that a boundary point does not have any open neigborhood which is homeomorphic to an open subset of the model space $\mathbb{R}^n$.

A boundary point $x$ of a manifold $M$ is characterized by the existence of a boundary chart at $x$ which is a homeomorphism $h : U \to V$ from an open neighborhood $U$ of $x$ in $M$ to an open subset of the Euclidean halfspace $\mathbb{H}^n = \mathbb{R}^{n-1} \times [0,\infty)$ such that $h(x) \in \mathbb{R}^{n-1} \times \{ 0\}$.

For an interior point $x$ we have a chart at $x$ which is a homeomorphism $h : U \to V$ from an open neighborhood $U$ of $x$ in $M$ to an open subset of $\mathbb{R}^n$.

We conclude that each interior point $x$ has a contractible open neighborhood $U$ such that $U \setminus \{ x \}$ is not contractible.

Conversely, if for any contractible open neighborhood $U$ of $x$ also $U \setminus \{ x \}$ is contractible, then $x$ cannot be an interior point.

So let us consider a boundary point $x$ of $M$. We shall more generally prove that for any open neighborhood $W$ of $x$ the inclusion $i : W \setminus \{x\} \to W$ is a homotopy equivalence.

Choose a boundary chart $h : U \to V$ at $x$. W.lo.g. we may assume $U \subset W$. Let $y = h(x)$. Choose $r > 0$ such that $B_r(y) = \{ z \in \mathbb{H}^n \mid \lVert z - y \rVert \le r \} \subset V$. The set $B_r(y)$ is the compact upper half of a ball in $\mathbb{R}^n$. Let $S_r(y) = \{ z \in \mathbb{H}^n \mid \lVert z - y \rVert = r \}$ be its boundary. $S_r(y)$ is a strong deformation retract of $B_r(y)$. To be explicit, using the notation $z = (z',s) \in \mathbb{R}^{n-1} \times [0,\infty)$, define a homotopy $H : B_r(y) \times I \to B_r(y), H(z,t) = (1-t)z + t(z',\sqrt{r^2-\lVert z'\rVert^2})$. This is well-defined because $B_r(y)$ is convex and $(z',\sqrt{r^2-\lVert z'\rVert^2}) \in S_r(y)$. We have $H(z,0) = z$, $H(z,1) = (z',\sqrt{r^2-\lVert z'\rVert^2}) \in S_r(y)$ and $H(z, t) = z$ for $z \in S_r(y)$. It is obviuos that $H$ restricts to a homotopy on $B_r(y) \setminus \{ y \}$ so that $S_r(y)$ is also a strong deformation retract of $B_r(y) \setminus \{ y \}$.

Hence $S = h^{-1}(S_r(y))$ is a strong deformation retract of both $B = h^{-1}(B_r(y))$ and $B' = h^{-1}(B_r(y)B_r(y) \setminus \{ y \}) = B \setminus \{ x \}$. Extending the deformation retractions by the identity on $W \setminus B$, we obtain strong deformation retractions $r : W \to W \setminus int(B)$ and $r : W \setminus \{ x \} \to W \setminus int(B)$. This shows that $i$ is a homotopy equivalence.