I am trying to prove a certain statement that seems true based on computational data, and there is a nice argument that proves it, assuming all cycles are the simplest ones (e.g., when the only 1-cycles are triangles and never squares, pentagons, etc.). In real life these are not the only cycles, so this is not a real proof. I'll state the proposition and the "proof" to begin with.

First some definitions. A set $S$ of integers is said to be coprime-free if for all $i,j\in S$ with $i\ne j$, $\gcd(i,j)>1$. A coprime-free subset $S\subseteq [n]$ is said to be maximal if $S\cup \{x\}$ is not coprime free for all $x\in [n]\setminus S$. Let $\Delta_n$ be the simplicial complex whose vertices are the maximal coprime-free subsets of $n$, with a face for every set of vertices with nonempty intersection. In the following, when we speak of homology of $\Delta_n$ and write $H_k(\Delta_n)$, we mean homology over ${\bf Z}$.

Proposition. For all $k\ge 1$, $H_k(\Delta_n)=0$.

"Proof". Let $k\ge 1$ be given and suppose that $x_1,x_2,\ldots,x_{k+2}$ are vertices in $\Delta_n$ such that $\{x_1,\ldots,x_{k+2}\}$ is not a face but for all $1\le i \le k+2$, the set $F_i = \{x_1,\ldots,x_{k+2}\}\setminus\{x_i\}$ does form a face. In other words, $\bigcap_{i=1}^{k+2} x_i = \emptyset$, but for all $i$, the set $\bigcap F_i$ does contain some element, call it $a_i$. Now consider the set $A=\{a_1,\ldots,a_m\}$. If for some pair $i\ne j$ we have $\gcd(a_i,a_j) = 1$ for some $i$ and $j$, then we use the fact that $k+2\ge 3$ to pick $t\in [k+2]\setminus\{i,j\}$ and note that $a_i$ and $a_j$ are both in $x_t$, which would imply that $x_t$ is not coprime-free. So we conclude that $a_i$ and $a_j$ are not coprime for all $i\ne j$. Thus $\{a_1,\ldots,a_{k+2}\}$ is a subset of some maximal coprime-free subset of $[n]$, call it $x$. The above reasoning shows that $\{x,x_1,\ldots,x_{k+2}\}$, viewed as a simplicial subcomplex of $\Delta_n$, is a $(k+2)$-simplex missing its interior as well as exactly one of its $(k+1)$-faces, namely $\{x_1,\ldots,x_{k+2}\}$. The $k$th homology of this simplicial subcomplex is zero, so the $(k+2)$-tuple $(x_1,\ldots,x_{k+2})$ contributes nothing to the $k$th homology of $\Delta_n$. Since $x_1,\ldots,x_{k+2}$ were arbitrary, we are done."▮"

In the general case where a $k$-cycle may be formed by more than $k+2$ vertices (in the way that a square is still a $1$-cycle, being a formal sum of two triangles), I am wondering if there is some way of taking this result that we have and stacking it upon itself in some way. Or if there is some other way of amending the proof above. The reason that I would like to amend the proof above rather than start from scratch is that, under some delusion many months ago, I wrote some other "proofs" just like the one above, forgetting about the general case of $k$-cycles, and it is a naive hope of mine that there is some general schema for taking all of my false homology proofs and promoting them to real ones.

Sorry if the second part of my question is a bit wobbly and ill-defined, but really any help on this would be much appreciated. Thanks in advance!