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!

**Edit.** I think I was able to prove it using the nerve complex that Andy Putman suggested below. Here is what I have. The part about retracting simplices down I feel is correct, but perhaps I have not expressed it rigorously enough. Any feedback on this proof would be much appreciated.

Applying Borsuk's nerve theorem to Andy Putman construction and observations,$\Delta_n$ is homotopy equivalent to the simplicial complex $X_n$ whose vertex set is $[n]$ and in which a subset $J\subseteq [n]$ is a face if and only if $J$ is coprime-free. We also know that $X_n$ has $C(n)+2$ connected components, $C(n)+1$ of which are singletons. The nontrivial connected component, call it $X_n'$, consists of all primes in the range $[2,n/2]$ as well as all composite numbers in $[n]$. The claim is that $X_n'$ is contractible.

We will construct the deformation retraction to a single point in steps. First note that the set $V_2$ all even vertices, viewed as a subcomplex of $X_n'$, is a simplex, so we may retract this down to a point without altering the homotopy of $X_n'$. Call this point $\bar 2$. Now in this new complex, consider the set $V_3$ of all vertices whose smallest prime divisor is $3$ ($V_3$ includes all remaining multiples of $3$, since all multiples of $6$ were retracted to the point $\bar 2$). The subcomplex on $V_3\cup \{\bar 2\}$ is a simplex, since in the original complex $X_n'$, the set of vertices divisible by $3$ was a simplex, and all we have done is retract some subsimplex consisting of vertices divisible by $6$ to a point, call it $\bar 3$. The same will be possible when we define the set $V_5$ of all integers whose smallest prime divisor is $5$, and then consider $V_5\cup \{\bar 3\}$. Continuing this reasoning, we may perform a retraction for every prime $p\in [2,p/2]$ until all we are left with is a single point.

4more comments