Computing the homology of two simplicial subcomplexes of $K$

Question

Let $K$ be a simplicial complex on a vertex set $V=\{u_1,\dots,u_n,v_1,\dots,v_m\}$with facets given by the sets $$\sigma_i=V-\{u_1,u_i\},\tau_j=V-\{v_1,v_j\}$$ for each $i,j$.

I want to prove that the subcomplexes $X,Y$ consisting in all the $\sigma_i,\tau_j$ respectively, have no nonzero reduced homology.

I also want to know if there is a sufficient condition for a complex $X$ consisting on facets of dimension $k$, all of them meeting by pairs in faces of dimension $k-1$ to have no nonzero homology. I mean, intuitively it seems I'm taking a sphere, and removing some facets to remove all nonzero homology without adding anything else.

This problem appeared when I wanted to manually compute the homology of $K$. I have another method to do it; it only has reduced homology at dimension $n+m-4$ and it's $\mathbb{Z}$, result which would also hold if we prove that $X,Y$ have no nonzero homology and $X\cap J$ has reduced homology $\mathbb{Z}$ at dimension $n+m-5$ and $0$ elsewhere.

Answer 1

I can prove it inductively by use of Mayer-Vietoris sequences, as the following result.

Proposition: If $X$ is a simplicial complex on a vertex set $V=\{1,2,\dots,k\}$ with facets $\tau_1,\dots,\tau_j$ and there is $i\in \{1,2,\dots,k\}$ such that $$i\in\bigcap_{r=1}^j \tau_j$$ $\newcommand{\h}{\tilde{H}}$ then $\h_n(X)=0$ for all $n$, where $\h_{\bullet}$ denotes reduced homology.

Proof. For $j=1$, $X$ would be a simplex which is contractible. Let $j>1$ and suppose that every complex with less that $j$ facets satisfying the condition has zero reduced homology.

Let $X'$ be the subcomplex generated by $\tau_1,\dots,\tau_{j-1}$. Then there is a Mayer-Vietoris sequence: $$\cdots \to \h_n(X'\cap \tau_j)\to \h_n(X')\oplus \h_n(\tau_j)\to \h_n(X)\to \h_{n-1}(X'\cap \tau_j)\to \cdots$$ Then $i$ belongs to all the facets of $X'\cap \tau_j$ and also to all the facets of $X'$; and both complexes have less than $j$ facets. Then $\h_n(X')=\h_n(X'\cap \tau_j)=0$ for all $n$. Then for each $n$, we have a sequence $$0\to \h_n(X)\to 0$$ which means that $\h_n(X)=0$ for all $n$.

So, in our case, all the facets of $X$ contain a vertex $v_s$ and all the facets of $Y$ contain a vertex $u_{s'}$ (in fact all of them). The result follows.

This is a proposition I already knew about (Is the one I mention there in the case when all the facets contain $U$), but it seems I haven't done anything in maths in a few months so I wasn't able to see it (I'm so rusty that I didn't notice it was the same until I finished proving it).