Why there is no suspension axiom for homology ? and why there is no excision axiom for cohomology theory

algebraic-topologyfunctorsgroup-cohomologyhomological-algebrahomology-cohomology

Here are the axioms of reduced cohomology theory as given to me in the lecture:

1- $\tilde{H}^n(-;G): J_{*} \rightarrow Ab_{*}$ is a contravariant functor.

2- $\tilde{H}^n(X;G) \cong \tilde{H}^{n+1}(\sum X;G).$

3- Homotopy Axiom. homotopic maps induce the same map in cohomology.

4- Exactness. cofibre sequence induced a LES.

5- Dimension Axiom:

$$\tilde{H}^k(S^n ; \mathbb{Z}) = \mathbb{Z}, \text{ if } k = n \text{ and } \tilde{H}^k(S^n ; \mathbb{Z})= 0 \text{ if } k \neq n. $$

My professor was depending on the book " Modern classical homotopy theory " by Jeffery Strom.

While the homology theory axioms from Rotman book(on pg.231) "Introduction to algebraic topology " are as below :

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My questions are:

1-Why there is no suspension axiom for homology or what is its equivalence? and why there is no excision axiom for cohomology theory or what is its equivalence?

2-If I transformed the dimension axiom given by my professor to homology theory, I do not understand how the statement I obtained is the same as the statement mentioned in Rotman. Here is the statement I obtained for homology dimension axiom:

$$\tilde{H}_k(S^n ; \mathbb{Z}) = \mathbb{Z}, \text{ if } k = n \text{ and } \tilde{H}^k(S^n ; \mathbb{Z})= 0 \text{ if } k \neq n. $$

Could anyone explain to me how they are equivalent, please?

Best Answer

I work with reduced homology here. Keep in mind that it is a straightforward consequence of definitions that for $A$ nonempty $\widetilde H_*(X, A) = H_*(X,A)$.

The fact that $\widetilde H_{k+1}(\Sigma Y; \Bbb Z) = \widetilde H_k(Y; \Bbb Z)$ follows immediately from:

  1. the homotopy axiom, which shows that because the cone $CY$ is contractible, we have $\widetilde H_*(CY) = 0$; this is also used to show that $\widetilde H_*(\Sigma Y, CY) \cong \widetilde H_*(\Sigma Y, *)$.

  2. the exactness axiom, applied to the pair $(CY, Y)$, which shows that $\partial: H_{k+1}(CY, Y) = \widetilde H_{k+1}(CY, Y) \to \widetilde H_k(Y)$ is an isomorphism for all $k$;

  3. the excision axiom applied to $X = \Sigma Y, A = CY$, and $U = \Sigma Y \setminus CY$, which provides an isomorphism $$\widetilde H_*(\Sigma Y) = \widetilde H_*(\Sigma Y, *) \cong \widetilde H_*(\Sigma Y, CY) \cong \widetilde H_*(CY, Y).$$

Putting this together this gives your desired "axiom".