When does a map between abstract simplicial complexes induce a homotopy equivalence on their geometric realisations? Let $\left(X_i,S_i\right)$ for $i=1,2$ be abstract simplicial complexes where $X_i$ are the vertex sets and $S_i\in\mathcal P\left(X_i\right)$ are the sets of simplices. Let $F\colon X_1\to X_2$ be a map of abstract simplicial complexes (for any simplex $D\in S_1$ we have that the image $F\left(D\right)\in S_2$). Then is there a way to characterise (using only the simplicial complexes and $F$) when the induced map $\left\lvert F\right\rvert\colon\left\lvert\left(X_1,S_1\right)\right\rvert\to\left\lvert\left(X_2,S_2\right)\right\rvert$ is a homotopy equivalence?
If not, can it be done in the case where the $X_i$ and $F$ are induced from graphs $G_i$ and a graph morphisms $f$ by taking their clique complexes? In this case, I would like to detect if $\left\lvert F\right\rvert$ is a homotopy equivalence from just the graphs and $f$.
EDIT TO CLARIFY WHAT IS ALLOWED: Any process which can be done combinatorically to the simplicial complexes is allowed (such as barycentric subdivision).
 A: Given an abstract simplicial complex $S$, the barycentric subdivision $S^\S$ as an abstract simplicial complex is the set of finite chains of elements of $S$.
Thus, unlike $S$, the elements of $S^\S$ are canonically ordered, so $S^\S$ canonically corresponds to a simplicial set.
More precisely, barycentric subdivision is a functor from the category of abstract simplicial complexes to the category of ordered abstract simplicial complexes, and there is a fully faithful functor from the category of ordered abstract simplicial complexes to category of simplicial sets, and all these functors are compatible with geometric realisation.
Thus, any and all facts about the geometry of simplicial sets can be transferred to abstract simplicial complexes.
(Of course, it is not just a a matter of replacing "simplicial set" with "abstract simplicial complex".
For example, there is a combinatorial characterisation of the morphisms of simplicial sets whose geometric realisation are homotopy equivalences.
(It is, admittedly, very complicated in general, and involves constructing auxiliary simplicial sets unless the morphism is a monomorphism.)
Or we could use Whitehead's theorem to reduce the problem to finding a combinatorial definition of the homotopy groups of a simplicial set (and the homomorphisms induced by a morphism of simplicial sets), which can in turn be solved using the simplicial approximation theorem.
Or perhaps something else still – the homotopy theory of simplicial sets is very well developed.
Restricting to clique complexes makes no difference if you consider barycentric subdivision to be a "free" operation.
This is because the barycentric subdivision functor is isomorphic to the composite of the clique complex functor and a certain functor from the category of abstract simplicial complexes to the category of graphs.
In other words, up to barycentric subdivision, every abstract simplicial complex is a clique complex.

In more detail:
Definition.
An ordered abstract simplicial complex is an abstract simplicial complex $S$ together with a linear ordering of each element of $S$, such that if $D' \subseteq D \in S$ (and $D' \in S$) then $D'$ has the ordering inherited from $D$.
(This does not imply that the vertex set of $S$ is linearly ordered, or even partially ordered!)
A morphism of ordered abstract simplicial complexes is a morphism of abstract simplicial complexes that is compatible with the linear orderings on each element.
Proposition.
Barycentric subdivision is a functor from the category of abstract simplicial complexes to the category of ordered abstract simplicial complexes.
It is compatible with geometric realisation in the sense that there is a natural isomorphism $\left| S \right| \cong \left| S^\S \right|$.
Remark.
The forgetful functor from the category of ordered abstract simplicial complexes to the category of abstract simplicial complexes is surjective on objects: given any abstract simplicial complex, there is a linear ordering of its vertices (by the axiom of choice), hence we may equip it with the structure of an ordered simplicial complex.
Unfortunately, morphisms of abstract simplicial complexes will not be compatible with these arbitrarily assigned orderings, so there is no hope of getting a left inverse for the forgetful functor.
The barycentric subdivision is the best we can do.
Definition.
Given an ordered abstract simplicial complex $S$, the nerve of $S$ is the simplicial set $\mathrm{N} (S)$ defined as follows:

*

*An $n$-simplex of $\mathrm{N} (S)$ is an $(n + 1)$-tuple $(x_0, \ldots, x_n)$ such that $\{ x_0, \ldots, x_n \} \in S$ and $x_0 \le \cdots \le x_n$ (but not necessarily distinct) in the linear ordering of $\{ x_0, \ldots, x_n \}$.

*The $i$-th face operator deletes the $i$-th entry, i.e. the $i$-th face of  $(x_0, \ldots, x_n)$ is $(x_0, \ldots, x_{i-1}, x_{i+1}, \ldots x_n)$.

*The $i$-th degeneracy operator repeats the $i$-th entry, i.e. the $i$-th degeneration of $(x_0, \ldots, x_n)$ is $(x_0, \ldots, x_i, x_i, x \ldots, x_n)$.

Proposition.
The nerve construction is a functor from the category of ordered abstract simplicial complexes to the category of simplicial sets.
It is fully faithful, and it is compatible with geometric realisation in the sense that there is a natural isomorphism $\left| S \right| \cong \left| \mathrm{N} (S) \right|$.
Remark.
A simplicial set is concrete if every $n$-simplex is uniquely determined by the $(n + 1)$-tuple of its vertices.
The nerve of any ordered abstract simplicial complex is concrete, but concrete simplicial sets are slightly more general.
For instance, in a concrete simplicial set, it is possible that there are two distinct edges with the same endpoints, differing only in orientation; pictorially, $\bullet \leftrightarrows \bullet$.
Non-concrete simplicial sets are even more general – for example, it is possible to have an edge that is a loop; pictorially, $\circlearrowleft$.
Proposition.
Given an abstract simplicial complex $S$, let $\mathrm{G} (S)$ be the following graph:

*

*A vertex of $\mathrm{G} (S)$ is an element of $S$.


*Two vertices of $\mathrm{G} (S)$ are connected by an edge if, considered as elements of $S$, one is a subset of the other.
Then the clique complex of $\mathrm{G} (S)$ is the barycentric subdivision of $S$.
