On the homology groups of a simplicial complex, $K$ and $K-{\sigma}$

Question

If K is a simplicial complex and $\sigma$ a simplex in $K$ that is not a face of any other simplex in $K$.

Then, let $L$ be $K-{\sigma}$.

This is a simplicial complex.

If $\sigma$ has dimension $n$, then, $d_n(\sigma)$ is an $n-1$ cycle and consists of simplices in L, and so, represents a class in $H_{n-1}(L)$. This defines a homomorphism $\phi: \mathbb{Z} \to H_{n-1}(L)$ by sending $1$ to $[d_n(\sigma)]$

I need to find a homomorphism $\psi : H_n(K) \to \mathbb{Z}$ such that the sequence:

$0 \to H_n(L) \to H_n(K) \to \mathbb{Z} \to H_{n-1}(L) \to H_{n-1}(K) \to 0$ is exact with the obvious inclusions.

I understand that, any class, $[\tau]$ in $H_n(K)$, needs to go to $0$ for $\tau \not = \sigma$.

But, I'm not sure where to send $\sigma$. I understand that it's image must be an integer m, say, such that, $m [d_n(\sigma)]$ lies inside the image of $d_n: C_n(L) \to C_{n-1}(L)$

Which means that there needs to be an n-simplex in L , say $\tau$, with boundary, $d_n(\tau)$ such that $[d_n(\tau)] = m [d_n(\sigma)]$. But how is this possible unless $m = 0$?

EDIT: That is, that $\psi$ is trivial. Could I simply argue that this must be the case because $\phi$ is injective?

Have I completely misunderstood what is going on?

Forgive me for the length of the post, and thank you!

Answer 1

Define a map $f:C_n(K)\to\mathbb{Z}$ by sending $\sigma$ to $1$ and every other $n$-simplex to $0$. Since $\sigma$ is not a face of any $(n+1)$-simplex, $f$ vanishes on any boundary, and thus induces a map $C_n(K)/B_n(K)\to\mathbb{Z}$. Restricting this to the subgroup $H_n(K)\subseteq C_n(K)/B_n(K)$, we get a map $\psi:H_n(K)\to \mathbb{Z}$.

You might think this definition can't work, because $\phi\circ\psi$ is supposed to be $0$, but $\phi(\psi([\sigma]))=\phi(1)=[d_n(\sigma)]$. However, this is not a problem, because $[\sigma]$ is not an element of $H_n(K)$ at all: $\sigma$ is just a simplex, not a cycle! You only have to check that $\phi\circ\psi=0$ when evaluated on cycles! You can see that if $c$ is a cycle, then $\phi\circ\psi([c])=m[d_n(\sigma)]$, where $m$ is the coefficient of $\sigma$ in $c$. But in that case, $c=m\sigma+c'$ where $c'\in C_n(L)$, and then $m[d_n(\sigma)]=-[d_n(c')]=0\in H_{n-1}(L)$ since $c$ is a cycle.

(Of course, there are still other things you have to check to prove that this definition of $\psi$ gives an exact sequence, but this much shows that specific the concern you raised is not a problem.)

Answer 2

I'd like to add a more general approach to this problem. Given a subcomplex $L$ of a simplicial complex $K$, the homomorphism $C_n(L)\to C_n(K)$ which sends any simplex in $L$ (as a generator of the simplicial chain group) to the same simplex in $K$ gives a chain map which fits into a commutative diagram below

the relative chain group $C_n(K,L)$ is defined as the quotient $C_n(K)/C_n(L)$, so the columns are exact. Depending on your Algebraic Topology skill, you may know that a short exact sequence of chain complexes induces a long exact sequence of homology groups. So the above induces a long exact sequence of simplicial homology groups $$ \dots \longrightarrow H^\Delta_n(L) \longrightarrow H^\Delta_n(K) \longrightarrow H^\Delta_{n}(K,L) \longrightarrow H^\Delta_{n-1}(L) \longrightarrow \dots $$

If $L$ differs from $K$ only by one simplex and this simplex $\sigma$ has dimension $n$, then $C_k(K,L)$ is trivial for all $k\ne n$ and is $\langle\sigma\rangle\approx\mathbb Z$ for $k=n$, which means that some part of our LES looks like $$ 0 \longrightarrow H^\Delta_n(L) \longrightarrow H^\Delta_n(K) \longrightarrow \Bbb Z \longrightarrow H^\Delta_{n-1}(L) \longrightarrow H^\Delta_{n-1}(K) \longrightarrow 0 $$ The map $\Bbb Z \longrightarrow H^\Delta_{n-1}(L)$ is indeed the map you used in your post. The map $H^\Delta_n(K) \longrightarrow \Bbb Z$ takes a class $[c]$ to the number of instances of $\sigma$ in the chain $c$.