Assume $(X,x_0)$ is a pointed topological space such that $X \backslash\{x_0\}$ is a manifold. Does there always exist a neighbourhood $U$ of $x_0$ which is homeomorphic to the cone $cN := (0,1]\times N/(1,u)\sim(1,v)$ over a manifold $N$?
I am not interested in issues that come from the failure of Hausdorffness as e.g. with the line with two origins. So I'm happy to assume that $X$ itself is Hausdorff or even compact.
I can think of two kinds of examples for $(X,x_0)$ which don't come from artifically attaching a cone to a manifold:
- The wedge sum of two manifolds. In that case there is always a neigbourhood around the wedge point, which is homeomorphic to $c(S^n\coprod S^n)$.
- Spaces $(X,x_0)=(Y/A,A/A)$ for $Y$ a compact manifold and $A\subset Y$ a closed subset. For example the pinched torus has this form, but yet again there is neighbourhood homeomorphic to $c(S^1 \coprod S^1).$ I can imagine that bad things happen when $A$ is really wild, but I couldn't produce a counterexample this way. (Taking a Cantor set inside a circle should leave us with a countable bouquet of circles (?) and hence a wedge sum.)
Motivation: In the context of singular Riemannian manifolds (e.g. in J.Cheegers paper "on the spectral geometry of spaces with cone-like singularities") one studies singularities as above and uses the the term cone-like for a Riemannian metric that has a particular form under the identification $U\backslash\{x_0\} \cong cN \backslash\{\text{cone-tip}\}$. In all references that I have seen in this context, it is first assumed that $U\cong cN$ exists, before the metric is specified and I am wondering whether this is an actual restriction on the local topology of $X$ around $x_0$ or comes automatically.
How about the Hawaiian earring? It is the space consisting of all the circles in the plane with radius $\frac1n$ and center at $(\frac1n, 0)$ for $n\in \Bbb N$ (meaning they all touch at the origin).
Removing the origin makes the space into a disjoint union of open line segments, which is a manifold. But any neighbourhood of the origin will contain infinitely many of the circles, and is therefore not a cone.