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Consider the singular homology functors $\mathrm H_n$.

The excision axiom says the commutative square of topological spaces associated to an excisive cover $U\cup V=X$ is taken to an isomorphism on relative homology.

Using the usual binary open cover of the double mapping cylinder, any homotopy pushout is weakly equivalent to the commutative square of a binary open (in particular excisive) cover. Hence, excision combined with the invariance of singular homology under weak homotopy equivalence implies that singular homology takes any homotopy pushout to an isomorphism on relative homology.


Suppose now we only know that singular homology is invariant under weak homotopy equivalence and takes homotopy pushouts to isomorphisms on relative homology. The excision property may be recovered from the following.

Fact 1. The commutative square associated to an excisive cover is a homotopy pushout.


Finally, suppose we know the following properties of singular homology:

  • Long exact sequence;
  • Invariance under weak homotopy equivalence.

Question 1. How can Fact 1 be used in conjunction with the above properties to deduce the excision property? (Without Mayer-Vietoris.)

Question 2. For an inclusion $\overline U\subset \mathring A\subset X$, is the homotopy cofiber of the inclusion $U\subset A$ weakly homotopy equivalent to its cofiber? Is this inclusion by any chance even a cofibration?

If so, a considerable bit of algebra, which furnishes some sort of weak equivalences of chain complexes $$\mathrm{cofiber}\mathrm{Ch}\simeq \mathrm{hocofiber}\mathrm{Ch}\simeq \mathrm{Ch}(\mathrm{hocofiber})\tag{1}$$ can be extended to give a some sort of weak equivalence of chain complexes $$\frac{\mathrm{Ch}(X)}{\mathrm{Ch}(U)}\simeq \mathrm{Ch}(X/U),\frac{\mathrm{Ch}(V)}{\mathrm{Ch}(U\cap V)}\simeq \mathrm{Ch}(V/(U\cap V))$$ which gives isomorphisms on homology. Finally, since the commutative square associated to an excisive cover is a strict pushout, there's a homeomorphism $X/U\cong V/(U\cap V)$ (from the pushout lemma) which finally gives excision.

I do not want to delve into the inner workings of (1).

Question 3. Is there some direct way to prove the extremal complexes in (1) are e.g chain homotopy equivalent (if they are)?

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  • $\begingroup$ Note that the two properties that you assume in Q1 imply immediately that singular cohomology turns homotopy pushouts to homotopy pullbacks (since every cofiber sequence is weakly equivalent to one of the form $X\to M_f \to C_f$). $\endgroup$ – Denis Nardin May 18 '18 at 20:55

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