Let $X \subset \Bbb R^n$ be a non compact, smooth, connected, orientable, algebraic hypersurface such that $H_i(X,\Bbb Z)=0$ for all $i\not=0$. Is $X$ homeomorphic to a hyperplane?
Thanks in advance!
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First, some terminology: An $m$-dimensional manifold $M$ is called an (integer) homology sphere if its homology groups are isomorphic to that of the sphere $S^m$. A topological space $X$ is called acyclic if it has zero reduced homology groups. It is a nice exercise to show that the complement to a point in a homology sphere is acyclic. There are homology spheres which are not simply-connected, the first one was discovered by Poincare. Therefore, the complement to a point in such a homology sphere is not simply-connected either, hence, is not even homotopy-equivalent to $R^n$.
Next, some 3-dimensional homology spheres embed as smooth submanifolds of $S^4$, they bound domains in $S^4$ known as Mazur manifolds, see this Wikipedia article. Furthermore, every smooth compact hypersurface in $R^n$ is isotopic to a nonsingular real-algebraic subset of $R^n$; this argument is due to Seifert, here
S. Akbulut and H. King, The topology of real algebraic sets with isolated singularities, Annals of Math. 113 (1981) 425-446.
for details and generalizations. Thus, there exist 3-dimensional nonsingular hypersurfaces in $S^4$ or in $R^4$, whatever you prefer, which are homology spheres that are not homeomorphic to $S^3$. Lastly, removing a point from such a hypersurface and using the stereographic projection, we obtain a smooth algebraic hypersurface in $R^4$ which is acyclic but not simply-connected. It is likely one can even obtain algebraic hypersurfaces which are contractible and are not homeomorphic to Euclidean spaces. Such examples exist if you do not insist on having a hypersurface, see my answer here.