How can I prove the dimension axiom here? Here is the problem:



I want to prove the dimension axiom in (a). My professor gave me the dimension axiom for reduced cohomology as follows:
$$\widetilde{H}^k(S^n ; \mathbb{Z}) = \mathbb{Z}, \text{ if } k = n \text{ and } \widetilde{H}^k(S^n ; \mathbb{Z})= 0 \text{ if } k \neq n. $$
And the dimension axiom for homology as follows:
$${H}_n(pt.) = 0, \text{ if } n \geq 1 \text{ and } {H}_n(pt.)=  \mathbb{Z} \text{ if }  n = 0. $$
Could anyone explain to me how I can use those 2 definitions to prove the required, please?
 A: I don't think the "dimension axiom" you're citing from your professor is right. Maybe he was talking about the dimension axiom and gave you that result as a consequence of the axiom, but it certainly isn't equivalent to any form of the dimension axiom I've seen. As pointed out in the comments, this axiom isn't restrictive enough to specify the cohomology of spaces built out of suspensions, etc. The one you give for homology is correct, and the natural analog of that axiom for cohomology is $H^n(\text{pt}) = 0$ for $n>0$ and $H^0(\text{pt})$ is your coefficient group.
In your case, your coefficient group will be the underlying group of $R$. Then since $H_n$ is an ordinary homology theory, it satisfies the dimension axiom, so $H_n(\text{pt}) = 0$ for $n>0$. Then $h^n(\text{pt}) = \text{Hom}(0,R) = 0$ for $n>0$. Now, $H_0(\text{pt})\cong R$, so $h^0(\text{pt})\cong\text{Hom}(R,R)\cong R$, as desired.
note: Some of your confusion might be coming from mixing up the definitions of "reduced" theories vs. "ordinary" theories vs. "generalized" theories. These all have slightly different definitions. For example, reduced theories have $\widetilde{H}_n(\text{pt}) = 0$ for $n\geq0$, not just $n>0$. Ordinary theories satisfy all of the usual Eilenberg-Steenrod axioms, and generalized theories don't satisfy the dimension axiom at all, meaning contractible spaces will have nontrivial homology in positive dimension in these theories.
