$\mathbb{R}^{2}$ and $\mathbb{R} \times [0, +\infty]$ are homotopy equivalent, but not homeomorphic

Question

So, let's consider $M=\mathbb{R}^{2}$ and $N= \mathbb{R} \times [0, +\infty]$ - two topological spaces.

Since $\pi_{1}(M)=\pi_{1}(\mathbb{R}) \times \pi_{1} (\mathbb{R}) = \{0 \}$ (since $\mathbb{R}$ is path-connected). This time, fundamental group of every convex subset of $\mathbb{R}$ is also trivial, so it's time to conclude that $M$ is homotopy equivalent to $N$.

But how to prove that they are not homeomorphic to each other?

Any help would be much appreciated.

Answer 1

Let $L \subset \mathbb{R}^2$ a straight line; if there exists an homeomorphism $\phi$ between $\mathbb{R}^2$ and $\mathbb{R} \times [0,+\infty]$ then $\phi_{|_{\mathbb{R}^2-L}}: (\mathbb{R}^2-L) \longrightarrow (\mathbb{R} \times [0,+\infty])- \phi(L)$ is a homeomorphism again. But it is not possible because the source has two connected component and the target is connected.

Answer 2

As Mike Miller suggested, it is not enough to show that the fundamental groups are isomorphic. In fact, even if all the homotopy groups were isomorphic, that would still not be enough.

Instead, try to show that both of these spaces are contractible (this is not hard to show).

To show that they are not homeomorphic, look closely at the line $\{0\}\times [0,+\infty]$ in the latter space.

Answer 3

The one-point compactification of $\mathbb R^2$ is $S^2$, but the one-point compactification of $\mathbb R\times [0,\infty]$ is (homeomorphic to) $S^1\times I/\{pt\}\times I$. The latter space deformation retracts onto $S^1$, so these two spaces have different fundamental groups, therefore they are not homeomorphic.