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I was reading the proof here. I understood almost all the parts after Exercise 4, such as defining $bs_n ^X$ inductively, a chain homotopy $(R^X )$, and that we get a small chain $(bs_n ^X)^k $ by Baire Category (I guess). However, I can't understand that if $(bs_n ^X)^k (\alpha ) $ is a small chain, and $ R_n ^X (\alpha )$ is small, then inclusion of $C'(X) \rightarrow C(X) $ induces isomorphisms in homology. (Exercise 4) I get this intuitively, but it is not explicit.

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This isn't a trivial step in proving excision property: in my personal opinion this is a tricky part of the proof and a little work is required to find a left inverse to the inclusion map $C_\bullet(X) \to C'_\bullet(X)$ which is also a right homotopy inverse (i.e. an inverse up to chain homotopy).

A proof of this fact can be found in Hatcher at page 119 Proposition 2.21.

Actually Hatcher proves a little more general fact: for every family $\mathcal U = \{\mathcal U_i\}_{i}$ of subspace of $X$ such that $\bigcup_i \mathcal{\stackrel{\circ}U_i} = X$, the sub-chain complex of $C_\bullet(X)$ generated by all the singular simplexes which have image contained in one of the $\mathcal U_i$ is such the inclusion induce isomorphism in homology.

Hope this helps.

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