Is there an $m$-dimensional simplicial complex $S$ with the following properties:


* *The cone over $S$ is homeomorphic to $\mathbb{E}^{m+1}$.* Here $\mathbb{E}^{m+1}$ denoes the $(m+1)$-dimensional Euclidean space.
* *There is a vertex $v$ in $S$ such that the complement $S\backslash\{v\}$ is not simply connected.*


**Comments.** 


* If such example exist, the cone over $v$ has to from a wild half-line in the Euclidean space. It has to be an embedding of $[0,\infty)$ which complement is not simply connected.
* The first condition is equivalent to the following: *The spherical suspension over $S$ is homeomorphic to $\mathbb{S}^{m+1}$.*



Wild half-line in a Euclidean space
I have got the following answer from Alexander Lytchak:


An example can be constructed the following way.
Start with a nontrivial homology sphere,
pass to its spherical suspension.
Now shrink one of the meridians of suspension to the point, which we denote by $v$.
The obtained space $S$ is the example;
it admits a natural triangulation.


The first condition is easy to check.
To check the second condition note that the spherical suspension $\Sigma(S)$ has a triangulation coming from $S$.


The space $\Sigma(S)$ is a homotopy equivalent to the sphere;
it is homological manifold 
which is also manifold everywhere except maybe the poles of suspension.
Each pole admits a simply connected punctured neighborhood.
Therefore by disjoint disc property the poles are also manifold point.
It remains to apply generalized Poincaré conjecture.


