Understanding step four in the excision theorem (Hatcher - algebraic topology). I'm studing the proof of the excision theorem (Proposition 2.21 p119) in Hatcher's book (freely available on https://pi.math.cornell.edu/~hatcher/AT/AT.pdf). This proof gives four steps. In the fourth step, on p123, we consider the following fragment.

I guess here $S^m(\Delta^n)$ refers to the operator $S$ constructed in the second step. I can see that if we consider $S^m(\Delta^n)$ for large enough $m$, then every simplex that occurs in its decomposition has image contained in some $\sigma^{-1}(U_j)$.
Why does it follow that $S^m(\sigma)$ is in $C_n^\mathcal{U}(X)$? Maybe we have the relation $S^m(\sigma) = \sigma_\sharp(S^m(\Delta^n))?$ (I can't prove this). Any help on the matter will be appreciated!
 A: On p. 122 Hatcher defines

(3) Barycentric Subdivision of General Chains. Define $S : C_n(X) \to C_n(X)$ by setting $S(\sigma) = \sigma_\sharp(S(\Delta^n))$ for a singular $n$-simplex $\sigma : \Delta^n \to X$.

Thus $S$ is defined on the generators of $C_n(X)$; this extends uniquely to a group homomorphism.
As you guess correctly, we have $S^m(\sigma) = \sigma_\sharp(S^m(\Delta^n))$, but this is not completely obvious.
The first problem is that Hatcher does not explicitly define $S(\Delta^n)$. He only explains what the barycentric subdivision of $\Delta^n$ is (it is a simplicial complex consisting of "smaller" $n$-simplices contained in $\Delta^n$ and all their faces). In "(2) Barycentric Subdivision of Linear Chains" he introduces the subdivision homomorphism $S : LC_n(Y) \to LC_n(Y)$. The missing ingredient is this: Take $Y = \Delta^n$ and define $$S(\Delta^n)  = S(id_{\Delta^n}) . \tag{1}$$
This is a linear $n$-chain in $\Delta^n$.
The next problem is that Hatcher misses to show that for each $\lambda \in LC_n(Y)$ the subdivision homomorphism $S$ satisfies
$$S(\lambda) = \lambda_\sharp(S(id_{\Delta^n})) = \lambda_\sharp(S(\Delta^n)).\tag{2}$$
Note that on the LHS of $(2)$ we use $S : LC_n(Y) \to LC_n(Y)$ and on the RHS we use $S : LC_n(\Delta^n) \to LC_n(\Delta^n)$. I do not prove this here, you can do it using the inductive formula $S(λ) = b_λ(S∂λ)$. Clearly $(2)$ is the minimal requirement in order that
$$S(\sigma) = \sigma_\sharp(S(\Delta^n)) \tag{3}$$
is well-defined for all $\sigma \in C_n(X)$.
Once we know that $(2)$ is true, we can easily prove that for $\lambda \in LC_n(\Delta^n)$ and $\sigma \in C_n(X)$
$$S (\sigma_\sharp (\lambda)) = \sigma_\sharp (S(\lambda)) . \tag{4}$$
In fact, we have $S (\sigma_\sharp (\lambda)) = S(\sigma \lambda) = (\sigma \lambda)_\sharp(S(\Delta^n)) = \sigma_\sharp(\lambda_\sharp(S(\Delta^n))) = \sigma_\sharp (S(\lambda))$.
But $(4)$ shows that on $LC_n(\Delta^n)$ we have
$$S \sigma_\sharp = \sigma_\sharp S .\tag{5}$$
By induction we now easily get $S^m(\sigma) = \sigma_\sharp(S^m(\Delta^n))$. In the proof note that $S^m(\Delta^n) \in LC_n(\Delta^n)$ for all $m$.
