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I have seen the following result in a few algebraic topology texts (such as Spanier), but only as an exercise:

For any sequence $m_1, \ldots, m_n$ of nonnegative integers, there is a connected simplicial complex $K$ with $H_p(K)$ free abelian of ran $b_p$ for every $1 \le p \le n$ (for the sake of clarity, the $H_p$ notation denotes the $p$-th homology group).

It looks really cool, and I have tried to work out a proof of it, but I am having trouble gaining momentum. I was wondering if anyone visiting today knows how to deduce this result, and if so, would be willing to prove it or give hints towards proving it. Any constructive responses would be greatly appreciated!

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2 Answers 2

up vote 6 down vote accepted

Spheres have homology concentrated in one dimension. So, consider the space $$X=\bigvee_{i=1}^{m_1} S^1\vee\cdots\vee\bigvee_{i=1}^{m_n} S^n$$ This space will have your desired homology.

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You should read about the construction of the wedge $X\vee Y$ of two topological spaces $X$ and $Y$. If $X$ and $Y$ are sufficiently nice connected spaces, then $X\vee Y$ is connected and the homology of $X\vee Y$ is, in positive degrees, the direct sum of the homologies of $X$ and $Y$.

Spheres are nice in that sense, so one can realize a sequence $b_1$, $\dots$, $b_n$ as the ranks of the homology of a space by taking the wedge of $b_1$ $1$-spheres, $b_2$ $2$-spheres and so on.

If you want simplicial complexes, then you should do this simplicially, of course.

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