Let $X$ be a topological manifold of dimension $n$, equipped with a compatible CAT(0) metric.
Are *sufficiently small* metric spheres in $X$ homeomorphic to metric spheres in Euclidean space $\mathbb{E}^n$?


[In "Ideal boundary of CAT(0) spaces" (1998) by Myung-Jin Jeon, this was unclear to the author; see the bottom of Page 104. That paper examined geodesic completeness for CAT(0) manifolds.]


**EDIT**: It is well-known that sufficiently small metric balls in any CAT(0) space are contractible. Using the manifold property, I believe that my question reduces to a very basic one:


If $U$ is a contractible open subset of $\mathbb{R}^n$, then is $U$ homeomorphic to $\mathbb{R}^n$?



Metric spheres in CAT(0) manifolds
The answer is no. By the [double suspension theorem](http://en.wikipedia.org/wiki/Double_suspension_theorem) of Cannon and Edwards, if $X^n$ is a [homology $n$-sphere](http://en.wikipedia.org/wiki/Homology_sphere), then the double suspension $S^2X$ is a sphere. In particular, the cone $CSX$ on $SX$ will be a simply connected manifold homeomorphic to $R^{n+2}$, since it is obtained from the double suspension by deleting a point. If there is a simplicial complex structure on $SX$ which is flag (e.g. by [barycentric subdivision](http://en.wikipedia.org/wiki/Barycentric_subdivision)), then the cone $CSX$ will have a CAT(0) metric. The point is that one may make each of the simplices of the triangulation of $SX$ into a right-angled $n+1$ spherical simplex, then take the metric cone on this to get a quadrant of $R^{n+2}$, and glue these quadrants together in the manner prescribed by the simplicial complex. The metric ball about the cone point of $CSX$ will not be homeomorphic to a ball, since its boundary will be $SX$ which is not a manifold. Check out the papers of [Davis-Januskiewicz](http://www.ams.org/mathscinet-getitem?mr=1131435) for constructions of CAT(0) complexes. 


