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.

  • 2
    $\begingroup$ That is not a proof that these two spaces are homotopy equivalent. There are plenty of simply-connected spaces that are not homotopy equivalent to $\Bbb R^2$; $S^2$ is a popular one. You need to write down an actual homotopy equivalence. $\endgroup$
    – user98602
    May 24 '15 at 20:54
  • $\begingroup$ Instead of $[0,\infty]$ it might be easier to think about $[0,1]$. $\endgroup$
    – Asaf Karagila
    May 24 '15 at 21:23
  • 1
    $\begingroup$ Another approach, for the sake of completeness. $\endgroup$ Jun 8 '15 at 20:12

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.

  • $\begingroup$ How do you prove that the target is connected? To be honest I don't know what your answer contributes compared to tomasz's answer. $\endgroup$ Jul 20 '15 at 15:03
  • $\begingroup$ Suppose $\phi$ is a homeomorphism and take $L$ a straight line on the plain; so the image of $L$ cannot be closed; then the image do not disconnect the target which is a cylinder with boundary. Otherwise suppose $\phi$ homeomorphism; the preimage of the boundary needs to be a closed curve in the plane that disconnects it; when the cylinder without boundary is still connect. $\endgroup$
    – InsideOut
    Jul 20 '15 at 15:09
  • $\begingroup$ The curve could disconnect $\mathbb{R} \times [0, \infty]$ without touching the boundary, a priori... $\endgroup$ Jul 20 '15 at 15:17
  • $\begingroup$ Mmm ok; honestly I don't remember the reason why I gave that answer; however $\phi$ cannot be a homeomorphism because if so; then the preimage of the boundary of the cylinder needs to be a close curve in the plane. So If I restrict my function in the same way I did above I have a homeomorphism between a space with 2 connected component and a cylinder without boundary which is clearly connected. In this I give to Arteom a way to show that these space are not homeomorphic. $\endgroup$
    – InsideOut
    Jul 20 '15 at 16:04

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.

  • $\begingroup$ I don't really understand where you're going with that last sentence... $\mathbb{R}^2$ also has a subspace homeomorphic to $[0,1]$, I don't understand what that tells you. $\endgroup$ May 25 '15 at 9:29
  • $\begingroup$ @NajibIdrissi: more or less the same thing as looking at a circle will tell you with respect to difference between the plane and the 3-dimensional space. $\endgroup$
    – tomasz
    May 25 '15 at 9:40
  • $\begingroup$ Are you referring to the Jordan curve theorem? It feels a bit overkill. To be honest it's not really "obvious" to me (well, it's intuitive but I don't know how to prove it off the top of my head) that $\mathbb{R}^2$ with a subspace homeomorphic to $[0,1]$ removed is still connected. $\endgroup$ May 25 '15 at 9:46
  • $\begingroup$ @NajibIdrissi: You don't need Jordan curve theorem to see that a standard circle disconnects the plane. All proofs (that I recall) that the three-space is not homeomorphic to a plane boil down to showing that no circle disconnects a three-space (or something closely related). $\endgroup$
    – tomasz
    May 25 '15 at 9:53
  • $\begingroup$ "If $\mathbb{R}^2$ and $\mathbb{R}^3$ were homeomorphic, then they would still be when you remove a single point from both, but $S^1$ and $S^2$ are not homotopy equivalent." is the (summary of the) proof that I know that the plane and the space aren't homeomorphic. How do you prove that a circle cannot disconnect $\mathbb{R}^3$...? (I don't think I've ever read a proof for that, even though it's intuitive... I'm genuinely interested.) $\endgroup$ May 25 '15 at 9:58

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.


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