Prove that $[0,\infty) \times \mathbb{R}^2$ is not homeomorphic to $\mathbb{R}^3$

Question

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.

Answer 1

Consider a slightly generalized variant: $X=[0,\infty)\times\mathbb{R}^n$ and $Y=\mathbb{R}^m$ for any $n\geq 0$, $m>0$.

Your idea about removing points was quite close actually. The proof goes like this:

1. $Y\backslash\{v\}$ is not contractible for any vector $v\in Y$ (the difference is a homotopy sphere)
2. $X\backslash\{(0,v_1,\ldots, v_n)\}$ is still contractible because it is a star subset of $\mathbb{R}^{n+1}$, with a base point for example at $(1,0,\ldots,0)$

You don't need any hard calculations of homologies/homotopies. The only thing you need to know is that a sphere is not contractible which is a well known result of algebraic topology (although analytical proofs do exist as well).

Answer 2

Remove $$\{0\}\times \mathbb {R}^2$$ from $$[0,\infty) \times \mathbb{R}^2$$ and we are left with a simply connected space.

Removing any plane from $$\mathbb{R}^3$$ and the result is not simply connected.

Thus the two spaces are not homeomorphic.

Answer 3

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.