From Rotman's Algebraic Topology:
Let $K$ be a finite simplicial complex, and let $s$ be a simplex of highest dimension. Define $K_1 = K – \{s\}$ and $K_2 = \{{s \text{ and all it's proper faces}}\}$. Define $V = s – \{x\}$, where $x$ is an interior point of $s$. Then $H_q(|K_2|, V) \cong H_q(|K|, |K_1|)$.
The proof is as follows.
Let $X_1 = |K_1| \cup V$. Note that $X_1 \cap |K_2| = (|K_1| \cup V) \cap |K_2| = (|K_1| \cap |K_2|) \cup (V \cap |K_2|) = V$ because $|K_1| \cap |K_2| = |K_1 \cap K_2| \subset V \subset |K_2|.$ Furthermore, $|K_1| \subset X_1^{\circ}$ and, since $|K_2| – |K_1|$ is an open subset of $|K_2|$, it follows that $|K_2| – |K_1| \subset |K_2|^{\circ}$. Therefore $X_1^{\circ} \cup |K_2|^{\circ} = |K|$ and singular excision holds and inclusion induces isomorphisms for all $q$ and we have $H_q(|K_2|, V) \cong H_q(|K|, |K_1|)$.
In the last line of the proof, how does excision imply $H_q(|K_2|, V) \cong H_q(|K|, |K_1|)$
We're given $X_1^{\circ} \cup |K_2|^{\circ} = |K|$, but excision is defined as:
Let $X_1$ and $X_2$ be subspaces of $X$ with $X = X_1^{\circ} \cup X_2^{\circ}$. Then the inclusion $j : (X_1, X_1 \cap X_2) \rightarrow (X_1 \cup X_2, X_2) = (X, X_2)$ induces isomorphisms: $j_* : H_n(X_1, X_1 \cap X_2) \cong H_n(X, X_2)$ for all $n$.
But since $|K| = X_1^{\circ} \cup |K_2|^{\circ}$ we should have $H_q(|K_2|, V) \cong H_q(|K|, |X_1|)$. But I don't see how the result follows from this.
Best Answer
Assuming $K - \{s\}$ does not remove the sub-simplices of $s$, then $X_1$ deformation retracts to $|K_1|$ (since $V$ deformation retracts to the boundary of $s$, and this boundary is contained in $|K_1|$). So, $H_q(|K|, X_1) \cong H_q(|K|, |K_1|)$.
Note that, if $X$ is a space and $A \subset B \subset X$, and $B$ deformation retracts to $A$ then $H_q(X, A) \cong H_q(X, B)$. This follows by considering the homology LES of the triple $(A, B, X)$. Every third term $H_q(B, A)$ vanishes. So exactness gives you the result.