Prove that $[0,\infty) \times \mathbb{R}^2$ is not homeomorphic to $\mathbb{R}^3$. I want to do this using the tools of algebraic topology.
Some thoughts I have on the matter is that if we remove a point from $\mathbb{R}^3$ then it becomes homeomorphic to $S^3$ minus two points.. Is $S^3$ minus two points contractible? I don't know. I'd appreciate if you guys gave me a few different approaches to this problem! Thanks!
edit: $\mathbb{R}^3$ minus a point is not contractible, as $\mathbb{R}^3 \cong S^2 \times (0,\infty)$ and $S^2$ is not contractible.
Best Answer
Use local homology. The $n$-th local homology at a point $x$ in a topological space $X$ is the relative homology group $H_n(X,X-\{x\})$. Every point in $\Bbb R^3$ has third local homology $\cong\Bbb Z$. But in $[0,\infty)\times\Bbb R^2$ the points in $\{0\}\times\Bbb R^2$ have third local homology zero. So these spaces are not homeomorphic.
All this holds for general manifolds with boundary.