Here is the problem:
Suppose $R$ is a field. (a) Show that $h^{n}(?) = Hom_{R}(H_{n}(?; R), R)$ is a cohomology theory defined on (at least) the category of finite CW complexes.
I got a hint to prove the Eilenberg-Steenrod axioms but I am unable to do this, could anyone help me in proving so, please?
To verify that $\text{Hom}_{R-\text{Mod}}(H_n(-;R),R)$ is a cohomology theory we use the fact that $\text{Hom}_{R-\text{Mod}}(-,R):R-\text{Mod} \rightarrow R-\text{Mod}$ preserves short exact sequences of finitely generated $R-$modules if $R$ is a field and actually long exact sequences of finitely generated $R-$modules.
Verifying that if $f \simeq g$ then $h^i(f) = h^i(g)$ is easy since we know that $H_i(f) = H_i(g)$ which means that the homotopy invariance axiom is satisfied.
Verifying exactness of a long exact sequence $...\rightarrow h^i(X) \rightarrow h^i(A) \rightarrow h^i(X,A) \rightarrow ...$ is simple because we know that $... \rightarrow H_i(X,A;R) \rightarrow H_i(A;R) \rightarrow H_i(X;R) \rightarrow ...$ is exact and that all $H_i(X,A;R)$ are finitely generated $R-$modules if $(X,A)$ is a $CW-$pair with $X$ finite.
This means that after applying the functor $\text{Hom}_{R-\text{Mod}}(-,R)$, the sequence is still exact which means that the exactness axiom is satisfied.
You can try to prove that the functor $\text{Hom}_{R-\text{Mod}}(-,R)$ takes coproducts to products which means that the additivity condition is satisfied since $\bigsqcup H_i(X_{\alpha};R) \rightarrow H_i(X;R)$ is an isomorphism if $\bigsqcup X_{\alpha} = X$.
If you have any questions just comment so I can explain my thoughts more clearly! Hope that helped.