Question about the fundamental group and homotopy equivalence - MathOverflow

Question about the fundamental group and homotopy equivalence

2010-07-11T14:22:33Z

Let T be a two-dimensional torus and Y be the one point compactification of a two dimensional sphere ($S^2$) minus three points. I have to prove:

1)they have the same fundamental group

2)they are homotopically equivalent

This is what i thought: i can see Y this way, let A,B,C three distinct points on the sphere and P a point not on the sphere, Y is homotopically equivalent to $S^2\cup\overline{AP}\cup\overline{BP}\cup\overline{CP}$. My hope for showing that they have the same fundamnetal group was to use van kampen theorem with open sets one $U$ homeomorphic to an open disk and the other one $V$ homotopically equivalent to a bouquet of 2 circles and $U\cap V$ homotopically equivalent to a circle. Now i would like to show that the generator of $\pi_1(U\cap V)$ is sent by inclusion map to the commutator of the generators of $\pi_1(V)$ but i can't see how.

For the homotopy equivalence i can't see the homotopy between these two spaces.

Please can anyone help?

Thank you in advance.

Answer by Gregory Arone for Question about the fundamental group and homotopy equivalence

2010-07-11T14:36:38Z

I think Y is homotopy equivalent to $S^2\vee S^1\vee S^1$. Proof: $Y$ is homotopy equivalent to the homotopy cofiber or the map from $3$ point to $S^2$. This map is null-homotopic, so $Y$ is equivalent to the wedge sum of $S^2$ with the suspension of three points. Suspension of three points is equivalent to $S^1\vee S^1$.

It is easy enough to construct the homotopy equivalence directly.

In conclusion: both (1) and (2) are false.

Answer by Jeff Strom for Question about the fundamental group and homotopy equivalence

2010-07-11T14:39:17Z

They are clearly not homotopy equivalent, and they have different homotopy groups. The nontorus is the cofiber of any injective map from a discrete $3$-point set to $S^2$, which is homotopy equivalent to $S^2\vee S^1\vee S^1$.