I was trying to understand simplicial excision which sais that if L is the complex we get from excising $U$ from $K$ and $L_0$ is the complex obtained by excising $U$ from the subcomplex $K_0$ then
$H_p(L,L_0) \cong H_p(K,K_0)$
In the proof, we show that
$\frac{C_p(L)}{C_p(L_0)} \cong \frac{C_p(K)}{C_p(K_0)}$ and then it is written
Since the boundary operator is preserved under this isomorphism, it follows that $H_p(L,L_0) \cong H_p(K,K_0)$
But what does it mean to be preserved under the isomorphism and why does it imply the result?
We have a simplicial inclusion map $i : L \to K$ which induces a chain map $i_\# : C_*(L) \to C_*(K)$. For the component in level $p$ let us write $i_{(p)} : C_p(L) \to C_p(K)$.
Thus for each $p$ we have $\partial_p^K i_{(p)} = i_{(p-1)} \partial_p^L$.
Since $i(L_0) \subset K_0$, we get $i_{(p)}(C_p(L_0)) \subset C_p(K_0)$. This means that we obtain an induced $$ i'_{(p)} : C_p(L,L_0) = C_p(L)/C_p(L_0) \to C_p(K)/C_p(K_0) = C_p(K,K_0)$$ which satisfy $$\partial_p^{(K,K_0)} i'_{(p)} = i'_{(p-1)} \partial_p^{(L,L_0)} .$$ Hence we get a chain map $i'_\# : C_*(L,L_0) \to C_*(K,K_0)$. This means that the boundary operator is preserved for the quotient (aka relative) chain complexes.
You said that is was shown that $i'_{(p)}$ are isomorphims. Hence $i'_\#$ is an isomorphism of chain complexes and the claim follows.