2
$\begingroup$

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 :

enter image description here

enter image description here

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?

$\endgroup$
4
  • 2
    $\begingroup$ Be careful! Reduced cohomology theory is not the same as cohomology theory, and homology theory is not the same as reduced homology theory. $\endgroup$ Aug 5, 2020 at 0:53
  • $\begingroup$ so how will the axioms differ between reduced and unreduced in both theories?@user10354138 $\endgroup$
    – user778657
    Aug 5, 2020 at 2:46
  • 1
    $\begingroup$ For one, the reduced homology $\widetilde{H}_0(X)$ of a one-point space $X$ vanishes by definition, but the regular homology $H_0(X)\cong G$ is your coefficient group (typically $G = \mathbb{Z}$). In dimension greater than 0, however, reduced homology and regular homology should evaluate to the same thing. The same statement holds for cohomology/reduced cohomology. $\endgroup$
    – jben2021
    Aug 9, 2020 at 17:23
  • $\begingroup$ @jben2021 and how this will help me in proving the dimension axiom for reduced homology and hence $h^n$? $\endgroup$
    – user778657
    Aug 12, 2020 at 16:06

2 Answers 2

2
$\begingroup$

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".

$\endgroup$
6
  • $\begingroup$ I am sorry. I can not see how your answer answers my questions about the excision axiom of cohomology. Also, I do not see how will I use the exactness axiom you mentioned in 2. $\endgroup$
    – user778657
    Aug 5, 2020 at 2:55
  • $\begingroup$ Are you trying in the above answer to prove the suspension axiom for homology? $\endgroup$
    – user778657
    Aug 5, 2020 at 2:56
  • $\begingroup$ Why you used the cone of Y(CY)? $\endgroup$
    – user778657
    Aug 6, 2020 at 2:32
  • 1
    $\begingroup$ In response to your question about the exactness axiom, when you write out the long exact sequence of the pair $(CY,Y)$ you should get $\dots\to\widetilde{H}_{n+1}(CY)\to\widetilde{H}_{n+1}(CY,Y)\to\widetilde{H}_n(Y)\to\widetilde{H}_n(CY)\to\dots$ Note that $CY$ is contractible, so these sections of the LES turn into isomorphisms $\widetilde{H}_{n+1}(CY,Y)\cong\widetilde{H}_n(Y)$. $\endgroup$
    – jben2021
    Aug 9, 2020 at 17:34
  • 1
    $\begingroup$ We have two reasons to use the cone: (1) the cone is contractible, so we can use it to simplify long exact sequences, and (2) the suspension $\Sigma Y$ can also be realized by gluing two copies of $C(Y)$ together along their nontrivial ends, meaning the ends which have not been collapsed to points. Using cones/cylinders is a very general technique in algebraic topology, and after a while it becomes second nature to try to decompose spaces in this way. $\endgroup$
    – jben2021
    Aug 9, 2020 at 17:40
1
$\begingroup$

The suspension "property" of homology groups can be proved as an application of Mayer-Vietoris sequence of homology groups. So, it is not actually an "axiom".

Theorem $4.2.21$: Let $X$ be a topological space. Then there exists a canonical isomorphism $\widetilde{H_n}(X)\rightarrow \widetilde{H}_{n+1}(\Sigma X)$ for each $n\geq 0$, where $\Sigma X$ denotes the suspension of $X$.

The above theorem is from the book Basic Algebriac Topology by Anant R. Shastri.

$\endgroup$
1
  • 1
    $\begingroup$ Thank you very much! $\endgroup$
    – user778657
    Sep 3, 2020 at 21:47

You must log in to answer this question.