= Topological tripod
A topological tripod is three copies of the <closed interval> $[0,1]$ with their zero endpoints identified to a single junction $o$. Equivalently, embed three line segments with distinct directions in the <Euclidean plane> and use the <subspace topology>. It is a <compact>, <connected> <metric space>, and is <Hausdorff>, <second countable>, and <paracompact>. Every interior arm point has a <coordinate chart> to an <open interval>, and each outer endpoint has a half-interval chart as in a <topological manifold with boundary>. The junction has neither kind of chart.
\Image[/line-circle-tripod-punctured-neighborhoods.png]
{title=Removing the center of a small neighborhood leaves two pieces on a line or circle, but three at a tripod junction}
Here is a precise local obstruction. Let $S$ be a small <open> <neighbourhood> of $o$ consisting of initial segments of the three arms, excluding their outer endpoints. Every <connected> <open> <neighbourhood> $N$ of $o$ contained in $S$ contains an initial segment on every arm, because $N$ is open at $o$. Its intersection with each arm has no gaps: if a point between $o$ and a point of $N$ were missing, cutting there would separate the part beyond it from the rest of $N$, contradicting <connectedness>. Thus $N\setminus\{o\}$ has exactly three <connected components>, one <interval> per arm.
Suppose a <coordinate chart> $h$ to the <real line> existed at $o$. Choose a small <open interval> $I$ about $h(o)$ lying inside $h(S\cap\operatorname{domain}(h))$, and restrict the chart to $N=h^{-1}(I)$. This is a <connected> <open> <neighbourhood> of $o$ contained in $S$. Removing $o$ leaves three <connected components> in $N$, while removing $h(o)$ leaves two in $I$. A <homeomorphism> preserves <connected components>, so this is impossible. This restriction step is essential: the count is proved for any sufficiently small connected chart neighborhood, not merely one chosen drawing.
A higher <dimension of a manifold> cannot repair the junction. For $n\geq2$, repeat the argument with an <open ball> about $h(o)$: a punctured <open ball> in $\mathbb R^n$ is <path-connected>, whereas $N\setminus\{o\}$ still has three <connected components>. A zero-dimensional chart would make $o$ an <isolated point>, which it is not. A chart at a boundary point of a <topological manifold with boundary> similarly gives a punctured relative half-ball: one <connected component> when $n=1$, and a <path-connected> space when $n\geq2$. Consequently the <topological tripod> fails the local definition even when intrinsic boundary is allowed.
The <circle> provides a useful warning about scale: deleting a point from the whole <circle> leaves a <connected space>, but deleting it from a sufficiently small connected arc leaves two <connected components>. Local chart tests use the latter. The <circle and interval are not homeomorphic> article uses deletion from the whole spaces for a different, global obstruction.
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