Let $G_0, G_1, G_2,\cdots, G_n$ be a sequence of finitely-generated abelian groups. Does there always exist a simplicial complex $K$ such that
(1). the dimension of $K$ is no larger than $n$;
(2). for each $0\leq i\leq n$, $H_i(K;\mathbb{Z})\cong G_i$?
Are there any references or Theorems? Thanks.
For this to be possible, the group $G_n$ must be free, since $H_n(K)$ will just be the group of simplicial $n$-cycles (there are no boundaries to mod out since $\dim K\leq n$), which is a subgroup of the group of simplicial $n$-chains and hence free. The group $G_0$ must also be free, since $H_0$ of any space is free, and $G_0$ must be nontrivial if any other $G_i$ is nontrivial since $H_0(K)=0$ implies $K$ is empty.
These are the only restrictions: given $G_0,\dots,G_n$ with $G_0$ and $G_n$ free and $G_0\neq 0$, it is possible to find $K$ as you ask. To prove this, it suffices to find simplicial complexes $K_i$ for each $i$ such that each $K_i$ has dimension $\leq n$, $K_i$ is connected for $i>0$, $H_i(K_i)\cong G_i$, and $H_j(K_i)=0$ if $j>0$ and $j\neq i$. Indeed, given such $K_i$ we can just take $K=\bigvee K_i$.
Constructing such $K_i$ for $i=0$ and $i=n$ is easy: just take a discrete space for $i=0$ and a wedge of $n$-spheres for $i=n$. So suppose $0<i<n$. Choose a presentation $$0\to \mathbb{Z}^d \stackrel{F}\to \mathbb{Z}^e\to G_i\to 0.$$ Let $X$ be a wedge of $e$ copies of $S^i$. We can then consider maps $f_1,\dots,f_d:S^i\to X$ whose induced maps on $H_i$ are the $d$ homomorphisms $\mathbb{Z}\to\mathbb{Z}^e$ which combine to give the map $F$. By simplicial approximation, we can choose simplicial complex structures on $X$ and $S^i$ such that these maps are all simplicial. We can then put a simplicial complex structure (of dimension $i+1\leq n$) on the space $K_i$ obtained by attaching $(i+1)$-cells to $X$ along each of these maps $f_1,\dots,f_d$. We can consider $K_i$ as a CW-complex whose only cells are a $0$-cell which is the vertex of $X$, $e$ $i$-cells which are the spheres of $X$, and $d$ $(i+1)$-cells which we attached via the maps $f_1,\dots,f_d$. We can then compute the cellular homology of $K_i$ to be $H_i(K_i)\cong G_i$ and $H_j(K_i)=0$ for all other $j>0$, since the cellular boundary map from the $(i+1)$-chains to the $i$-chains is $F$.