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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.

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1  
Are you sure (2) is true? My intuition is that $\pi_2(Y)$ is nonzero, whereas it's zero for $T$. But you should check for yourself. –  Donu Arapura Jul 11 '10 at 14:31
4  
The second space can be constructed up to a homotopy by connecting three points in the sphere with the center with a segment. Doing it cellularly shows that the Euler characteristic os $2$, which is not zero. –  Mariano Suárez-Alvarez Jul 11 '10 at 14:36

2 Answers 2

up vote 4 down vote accepted

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.

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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 $S^2\vee S^1\vee S^1$.

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Heh. Oh, well :-) –  Gregory Arone Jul 11 '10 at 15:02

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