# Find a CW complex with prescribed homology groups

A past qual question asks to construct a CW-complex $X$ with $H_0(X) = \mathbb{Z}$, $H_5(X) = \mathbb{Z} \oplus \mathbb{Z}_6$, and $H_n(X) = 0$ for $n\not= 0, 5$.

One can build a CW-complex $Y$ by taking a filled-in hexagon (divide it up into 2-simplices, if you'd like) and identify all the exterior edges (orient them so that they are all clockwise, say) and the vertices. Also, add another loop at any (and hence at all) of the vertices. Then, $H_0(Y) = \mathbb{Z}$ and $H_1(Y) = \mathbb{Z} \oplus \mathbb{Z}_6$. Is there a process by which we can construct $X$ from $Y$? Or must we make some strange identification of lower-dimensional cells on a 5-cell?

Thanks in advance for you help!

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You're looking for something called a Moore space - a space X with $H_n(X) = G$ for $G$ an arbitrary abelian group, and $\tilde{H_i}(X) = 0$ for $i \neq n$ (i.e. $H_i(X) = 0$ for $i \neq 0, n$, and $H_0(X) = \mathbb{Z}$). There is a general construction of a CW complex that satisfies this requirement: see Example 2.40 in Chapter 2 of Hatcher's Algebraic Topology, available for free online here.

The construction is relatively straightforward in your case: take a cell complex consisting of the wedge of two $S^5$'s, $S^5 \vee S^5$, and attach a $6$-cell to the second factor by a map of degree $6$. Cellular homology shows that this space has the desired homology. In the example you gave for $H_1(X) = \mathbb{Z}_6 \oplus \mathbb{Z}$, this is effectively the same construction, but it's $S^2 \vee S^1$ with a $2$-cell attached to one of them by a map of degree $6$ (since each of the six edges is identified to become the entire boundary)

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First, I think dorebell's solution is the "right" one to the problem. That said, yes, you can build $X$ from $Y$ - the key process is suspensions.

Recall that if $Z$ is a topological space, the suspension $\Sigma Z$ is the topological space obtained by Collapsing $Z\times [0,1]$ down to a point at both $t = 0$ and $t=1$. An easy Mayer-Vietroris argument shows that $H_{n+1}(\Sigma Z) \cong H_n(Z)$ for all $n$.

So, once you have $Y$, just construct $X= \Sigma^4Y.$

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Nice - this shows the relationship more clearly. Just for sake of completeness, I'd like to point out that in the case of finitely generated $G$, we can easily see an $M(G,n)$ as an (iterated) suspension of the type of $M(G,1)$ constructed in the question: the suspension just takes all the circles to $n$-spheres. – Dorebell Jul 18 '14 at 1:23