Past exam of the topology course of the University of Oxford / 2021 / 3 / a

The unit circle is Hausdorff and second countable, since both properties pass to topological subspaces of the Euclidean plane. To exhibit its local coordinate charts, consider the four open semicirclesThey cover $S^1$. On $U_{y,\pm }$, the projection map $(x,y)\mapsto x$ is a homeomorphism onto the open interval $(-1,1)$, with continuous inverse functionOn $U_{x,\pm }$, use $(x,y)\mapsto y$, with inverse function $t\mapsto (\pm \sqrt{1-t^2},t)$. These coordinate charts show that $S^1$ is a one-dimensional topological manifold with empty manifold boundary.

Let $o=(0,0)$ be the common point of the two circles. For sufficiently small $\epsilon >0$, the open neighbourhoodhas the following local structure: $W\setminus \{o\}$ consists of four disjoint open arcs, an upper and a lower branch from each circle. These are its four connected components, each open and closed relative to $W\setminus \{o\}$.

Suppose a coordinate chart of some dimension of a manifold $n$ existed at $o$. The case $n=0$ would make $o$ an isolated point, which it is not. For $n\ge 1$, shrink the chart inside $W$ so that its image is an open ball centred at the image of $o$. Its domain of a function $V$ is an open neighbourhood of $o$, so it meets all four arms of $W$. Intersecting those arms with $V\setminus \{o\}$ gives four disjoint nonempty clopen sets partitioning $V\setminus \{o\}$; hence there are at least four connected components.

But a punctured ball in $\mathbb{R}$ has exactly two connected components, and a punctured ball in ${\mathbb{R}}^n$ for $n\ge 2$ is path-connected. A homeomorphism preserves connected components, giving a contradiction in every dimension of a manifold. Thus $X$ is not a topological manifold. The argument uses arbitrarily small coordinate charts; the fact that the whole $X\setminus \{o\}$ has two pieces alone would not be a local obstruction.