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Let me preface this question by saying that I am not an algebraic topologist.

Motivation. I was looking with a colleague at the homotopy type of a family of posets and we were able to show using discrete Morse theory that the order complex of this poset was homotopy equivalent to a space with exactly one-cell in dimensions $0$ to $n$ and no cells of higher dimension. Moreover, we can show that the fundamental group is cyclic. Extensive computer computation shows that it is most likely true that the homology in each dimension from $0$ to $n$ is $\mathbb Z$, Thus it smells like our complex is homotopy equivalent to a wedge of spheres, one from each dimension from $1$ to $n$. Before killing ourselves to prove that the homology is as the computer suggests, we wanted to figure out if this data implies the homotopy type of a wedge.

Question. Is is true that if one has an $n$-dimensional CW complex $X$ with exactly one cell in each dimension (from $0$ to $n$) such that $\pi_1(X)=\mathbb Z$ and $H^q(X)=\mathbb Z$ for $0\leq q\leq n$, then $X$ has the homotopy type of a wedge $\bigvee_{q=1}^n S^q$?

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2 Answers 2

up vote 22 down vote accepted

No, this is false. For example take $\mathbb{CP}^2\vee S^1\vee S^3$. It admits a cell decomposition and cohomology groups as you describe but clearly has a different homotopy type then a wedge of spheres because the cohomology ring structure is different.

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Thanks!!!!!!!!! –  Benjamin Steinberg Dec 16 '11 at 2:27

You can take $S^1\vee S^2\vee S^3$ and then use the Hopf fibration from $S^3$ to $S^2$ as the attaching map for a 4-cell onto the 2-cell. This has to have the cohomology (and homotopy) as you described.

(Having now seen Vitali's answer, I guess this is the same as his, but I thought I might as well give this answer anyway.)

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Thanks Hugh.... –  Benjamin Steinberg Dec 16 '11 at 2:45

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