"Face-contraction" of Simplicial Complex Preserving Homology

Let $$X$$ be a simplicial complex on a vertex set $$V=\{1,\dots,n\}$$. Suppose there is some subset $$U\subseteq V$$ such that every facet of $$X$$ contains $$U$$ or is disjoint with $$U$$.

Then, define $$X'$$ on a new vertex set $$V'=(V-U)\cup \{v_U\}$$ and facets $$\sigma'$$ where $$\sigma'=\sigma$$ if $$\sigma$$ is a facet of $$X$$ disjoint to $$U$$, and $$\sigma'=(\sigma-U)\cup \{v_U\}$$ if $$\sigma$$ is a facet of $$X$$ containing $$U$$. Then $$X$$ and $$X'$$ have the same homology.

I was able to prove this by cases (There are a few details I haven't written at the end so it might have some mistakes, but I don't think so): First, if every facet contains $$U$$, then I used a Mayer-Vietoris sequence to prove there is no homology in this case. Second, if no facet contains $$U$$ (Thus every facet is disjoint to $$U$$), then the simplicial complex remains unchanged by the operation, therefore there is also no change in its homology.

Lastly, in the general case, I used Mayer-Vietoris sequences for both $$X$$ and $$X'$$ taking $$X=A\cup B,X'=A'\cup B$$ where $$B$$ consists in the facets disjoint to $$U$$ (Or not containing $$v_U$$ in $$X'$$) and $$A,A'$$ consist in the facets containing $$U,v_U$$ respectively. After using the fact that the homologies of $$A,A'$$ are zero, and that the intersections $$A\cap B=A'\cap B$$ remain unchanged by the operation it's easy to define a chain isomorphism between the exact sequences by the five lemma (Where four of the morphisms are identity maps and the fifth one is induced by the inclusion $$v_U\mapsto v$$ for any fixed $$v\in U$$).

My question is about the following remarks:

First, the operation done is analogous to the operation of edge contraction. We can interpret it in some ways. It identifies the vertices in $$U$$, and faces not disjoint with $$U$$, causing some of them to reduce their dimension (If they contain more than one vertex of $$U$$). In this case if I'm not mistaken we can also interpret it like removing all the vertices of $$U$$ but one.

Second: It doesn't always preserve homology, but sometimes it does. In this case the map $$v_U\mapsto v$$ (for a fixed $$v\in U$$) even seems to be a good candidate for a homotopy equivalence.

My question is the following: Is there any literature about the "edge contraction" operator applied to simplicial complexes? It seems like a rigid version of Proposition 0.17 in Hatcher's book.

$$\newcommand{\cU}{\mathcal{U}}$$ This actually is related to the construction of the "Nerve" of a simplicial complex.

Let $$\Delta$$ be a simplicial complex with facet set $$F(\Delta)=\{\sigma_1,\dots,\sigma_r\}$$ and vertices $$v_1,\dots,v_n$$. For each facet of $$\Delta$$, consider a small enough open set around it. Now you have an open cover $$\cU=\{U_1,\dots,U_r\}$$. The nerve $$N=N(\cU)$$ is defined to be a simplicial complex with vertex set $$\cU$$ given by $$N=\{\tau\subseteq\cU:\cap\tau\neq\varnothing\}$$, i.e. the subcollections of open sets with nonempty intersection.

You can make a similar discrete construction, by setting a complex $$D=D(\Delta)$$ with vertices $$V=\{\sigma_1,\dots,\sigma_r\}$$ defined by $$D=\{\tau\subseteq V: \cap \tau\neq \varnothing\}$$.

You can choose $$\cU$$ appropiately, so that the open sets are not "too big" so that they only intersect each other if the corresponding facets do: And in this case, it's clear that $$N$$ and $$D$$ are isomorphic by the map $$\sigma_i\mapsto U_i$$. So we'll call $$D$$ the nerve of $$\Delta$$.

Now, let $$D^2$$ be $$D(D(\Delta))$$. Naively, the facets of $$D$$ are given by $$F(D)=\{\tau_1,\dots,\tau_n\}$$ where $$\tau_i=\{\sigma\in F(\Delta):v_i\in\sigma\}$$, so the vertices of $$D^2$$ are $$\tau_1,\dots,\tau_n$$. The facets of $$D^2$$ are thus $$\rho_1,\dots,\rho_r$$ where $$\rho_j=\{\tau\in F(D): \sigma_j\in\tau\}$$.

Now, $$\sigma_j\in \tau_i$$ iff $$v_i\in \sigma_j$$, therefore after renaming $$\tau_i$$ by $$v_i$$ you get $$D^2=\Delta$$.

Is this correct? Actually, it's not. We're missing that $$\tau_i$$ may equal $$\tau_{i'}$$ for $$i\neq i'$$, so $$D^2$$ will be a little smaller than $$\Delta$$. When you remove repeated elements, you'll find that $$D^2$$ is what results of $$\Delta$$ after contracting all the vertex sets with the condition that every facet of $$\Delta$$ contains it or is disjoint with it. And then you have the Nerve Theorem of Leray which proves that $$D$$ is homotopically equivalent to $$\Delta$$ which means that of course $$D^2$$ is also homotopically equivalent to it. I expect the proof of that theorem should help to prove that the intermediate steps of contracting those sets one by one also preserve homotopy.