I have absolutely no intuition about non-Hausdorff spaces. I would like to understand the topology of non-Hausdorff spaces (in particular spaces obtained by "bad" group actions).

As a first example, I would like to calculate the fundamental group of $\mathbb R^2 / \mathbb R \cong \mathbb C / \mathbb R$, where a real number $q$ acts on $r \operatorname e^{i \theta}$ by multiplication. This is like projective space "without taking out the zero".

Of course the origin is a deformation retract of the plane. We can choose a deformation retraction (scaling the radius), which is "equivariant" w.r.t. the scaling action of $\mathbb R$. Does this show that the fundamental group of $\mathbb R^2 / \mathbb R$ is trivial?

I would also welcome examples of other non-Hausdorff spaces and their fundamental groups and/or (co)homology groups.

  • $\begingroup$ This seems pretty broad. $\endgroup$ – Pedro Tamaroff Feb 5 '15 at 1:17
  • $\begingroup$ @PedroTamaroff The only broad thing about the question is the last paragraph, which is optional (should I say "bonus question"?). $\endgroup$ – Earthliŋ Feb 5 '15 at 1:19

Yes, it is trivial.

To get some intuition for non-Hausdorff spaces, consider this fact: if you take a simplicial complex and quotient every open simplex to a point, you get a space that has the same homotopy groups as the original space. For instance, you can take the boundary of a triangle (which is homeomorphic to a circle) and quotient out the interior of each edge. This gives you a non-Hausdorff space with 6 points, three of which are open (the three squished open intervals) and three of which are closed. This space has fundamental group equal to $\mathbb{Z}$ and its universal cover is the quotient of the real numbers obtained by squishing each open interval between integer points.

To get a 2-sphere (up to homotopy), take the boundary of a tetrahedron and send each open face to a point and each edge to a point. The resulting space is simply connected but with non-trivial $\pi_2$.

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  • $\begingroup$ It should be mentioned that (as far as I know) this fact is not particularly easy to prove. However, it is fairly easy to prove the map in question is a homology equivalence by an inductive Mayer-Vietoris argument. $\endgroup$ – Eric Wofsey Feb 5 '15 at 1:39
  • $\begingroup$ @EricWofsey What is an "inductive Mayer-Vietoris argument"? $\endgroup$ – Earthliŋ Feb 5 '15 at 6:15

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