Homology of a simplicial complex union a simplex

Question

Let $L=K\cup\sigma$, where $L, K$ are abstract simplicial complexes, and $\sigma$ is a simplex whose faces are all in $K$. (This condition is necessary for $L$ to be a simplicial complex.)

Furthermore, we have that $\sigma\notin K$, which also implies that $\sigma$ is a maximal face of $L$. (If $\sigma\subsetneq\tau\in L$, then $\tau\in K$ implies $\sigma\in K$ which is a contradiction.)

Is there any relation between the homologies of $L$ and $K$? Is it possible to have a "formula" relating $H_*(L)$ and $H_*(K$)?

Thanks.

Answer 1

Let me elaborate a bit more on Lord Shark the Unknown's answer. We can get a precise answer because the maps in the sequence $$0\to H_n(K)\to H_n(L)\to\Bbb Z\to H_{n-1}(K)\to H_{n-1}(L)\to0$$ can be identified explicitly. Note that the $\mathbb{Z}$ here is generated by the homology class of a generator of $Y_n$, which is just the image of $\sigma\in C_n(L)$ under the quotient map $C_n(L)\to Y$. So, by definition of the long exact sequence, the map $\Bbb Z\to H_{n-1}(K)$ is given by lifting this generator to $\sigma\in C_n(L)$, taking the boundary, and then lifting that boundary to $C_{n-1}(K)$. In other words, it is exactly the class $[\partial\sigma]\in H_{n-1}(K)$.

We conclude that $H_{n-1}(L)$ is exactly the quotient of $H_{n-1}(K)$ by the subgroup generated by $[\partial\sigma]$. We also conclude that the cokernel of $H_n(K)\to H_n(L)$ is the group of $n\in\mathbb{Z}$ such that $n[\partial\sigma]$ is $0$ in $H_{n-1}(K)$. If $[\partial\sigma]$ is torsion this group is isomorphic to $\mathbb{Z}$, and otherwise it is trivial. So we concldue that $H_n(L)\cong H_n(K)\oplus \mathbb{Z}$ if $[\partial\sigma]$ is a torsion class in $H_{n-1}(K)$, and $H_n(L)\cong H_n(K)$ if $[\partial\sigma]$ is not torsion.

Answer 2

Suppose that $\sigma$ has dimension $n$. Consider the complexes of simplicial chains $C(K)$ and $C(L)$. Then there is an exact sequence of complexes $$0\to C(K)\to C(L)\to Y\to 0\tag{*}$$ where the complex $Y$ consists only of a $\Bbb Z$ in dimension $n$. Then (*) has an associated long exact sequence of homology. But the homology of $Y$ consists of just a $\Bbb Z$ is dimension $n$. We conclude that $H_m(K)\cong H_m(L)$ for $m\notin\{n-1,n\}$ and that there is an exact sequence $$0\to H_n(K)\to H_n(L)\to\Bbb Z\to H_{n-1}(K)\to H_{n-1}(L)\to0.$$ We can conclude several things from this, if we know something about the homology groups of $K$. So adding $\sigma$ kills the homology of the cycle $\partial\sigma$ in $H_{n-1}(K)$. Moreover $H_n(L)$ will be either $H_n(K)$ or the sum of $H_n(K)$ with a copy of $\Bbb Z$ etc.