I am reading preliminaries for a paper The worst way to collapse a simplex. It is written there:
A [simplicial] complex $X$ is said to be $R$-acyclic [$R$ is a ring] if $H_i(X;R) = 0$ for all $i ≥ 0$. Regarding the cases $R = \mathbb{Q}$ and $R = \mathbb{Z}$, we have that if X is $\mathbb{Z}$-acyclic then it is necessarily $\mathbb{Q}$-acyclic, however, the converse need not hold.
Could anyone please help me with an example of a simplicial complex, which is $\mathbb{Q}$-acyclic but not $\mathbb{Z}$-acyclic? Or explain where the problem occurs? Thank you.
Note that the stated definition of acyclic is a little funny around $H^0$, you should probably be looking at reduced homology or nothing is acyclic.
First, the motivating example,which is pure algebra, take $$ 0 \to \mathbb{Z} \stackrel{\cdot 2}{\to} \mathbb{Z} \to 0 $$
This resulting complex is acyclic if you tensor with $\mathbb{Q}$ but not if you tensor with $\mathbb{Z}$ (which does nothing).
So to get an example, we need a simplicial complex whose (reduced) chain complex looks like the above. Now, that's not enough literal simplicies to build an interesting space out of, but if you have seen cellular homology, you know that this is the chain complex for (the reduced cellular homology of) $\mathbb{RP}^2$. Since simplicial and cellular homology coincide, any simplicial decomposition of $\mathbb{RP}^2$ will give your example.