I find a really interesting approach on Excision Axiom with subdivision operator and prism operator defined by a formula (without using induction). The material I followed is public and it has only 5 pages: Peter Kronheimer - Barycentric subdivision https://legacy-www.math.harvard.edu/archive/272a_fall_04/handouts/Barycentric.pdf
The material is very clear. He defines the subdivision operator $S:C_n(X)\to C_n(X)$ via:
$$S\sigma=\sum_{p\in S_{n+1}}(-1)^p\sigma[b_p^0,b_p^1,\dots, b_p^n]$$
where the sum is taken over all permutations $p$ of the set $\{0,1,2,...,n\}$. He also introduced a morphism $K:C_n(X)\to C_{n+1}(X)$ by
$$K\sigma=\sum_{j=0}^n\sum_{p\in P(j)} (-1)^j\cdot (-1)^{p}\sigma[b_p^0,b_p^1,...,b_p^j,e_{p(j)},...,e_{p(n)}]$$,
where $P(j)$ is the set of all permutations $p$ of the set $\{0,1,2,...,n\}$ with $p(j)<p(j+1)<...<p(n)$.
With these definition we have the following essential theorem
$$Id_{C_n(X)}-S=\partial K+K\partial $$
When the author starts to compute $$\partial K\sigma=\sum_{j=0}^n\sum_{p\in P(j)}\sum_{i=0}^{n+1} (-1)^{i+j}(-1)^p \partial^i(\sigma[b_p^0,b_p^1,...,b_p^j,e_{p(j)}, ...,e_{p(n)}])=A+B+C+D+E+F$$
he splits the sum into $6$ smaller sums. I understand how he computes $A,B,D,E,F$. But $C$ remains a mistery for me. This is the only one he doesn't explain here.
$$C=\sum_{j=1}^n\sum_{p\in P(j)} (-1)^{j}(-1)^p \partial^0(\sigma[b_p^0,b_p^1,...,b_p^j,e_{p(j)}, ...,e_{p(n)}])$$
He also claims that $C=-K\partial\sigma$. I tried to prove it but I cannot see how can I transform a sum taken over permutations of $\{0,1,...,n\}$ (first one) and one that is taken over permutations of $\{0,1,2,...,n-1\}$ (in $-K\partial\sigma$).
My questions are:
1) How to prove that $C=-K\partial\sigma$? or if this is too complicated
2) Do you know a reference (book or article) in which the subdivision is defined without induction (with the above formulas) where I may find a complete proof of the excision? In all the books I have (like Hatcher, Munkres and others I saw that proof with induction and it is too complicated for me - I am not an algebrist).