Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question

1 Answer 1

up vote 0 down vote accepted

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.