The functors $h^n(X)=\text{Hom}(H_n(X),\Bbb Z)$ do not define a cohomology theory I am trying to show that the functors $h^n(X)=\text{Hom}(H_n(X),\Bbb Z)$ do not define a cohomology theory on CW complexes. If a contravariant functor $h^n(X)$ is a cohomology theory, by definition it must satisfy the followings:
(1) If $f,g:X\to Y$ are homotopic, then $f^*=g^*:h^n(Y)\to h^n(X)$. 
(2) For a CW pair $(X,A)$, there is a long exact sequence of the form $$ \cdots  \to h^n(X,A)\to h^n(X)\to h^n(A)\to h^{n+1}(X,A)\to \cdots $$ 
(3) Excision holds.
For our definition of $h^n(X)$, (1) and (3) clearly hold, so I should show that (2) fails, and I think (2) should indeed fail because  $\text{Hom}(-,\Bbb Z)$ is not exact in general. But I can't find such an example for a pair $(X,A)$. Any hints? 
 A: You have to find an example where $H^n(X)$ is in fact different from $\hom(H_n(X),\mathbb Z)$. The easiest example is, I think, $X = \mathbb RP^2$.
You can then take something like $A=S^1\subset \mathbb RP^2$ (corresponding to the nontrivial element in $\pi_1(\mathbb RP^2)$, which is what makes cohomology differ from the naive formula)
Then you have the long exact sequence in (usual) homology : $0\to H_2(X,A) \to H_1(A) \to H_1(X) \to H_1(X,A)\to H_0(A) \to H_0(X)$. The last map is an isomorphism, so we may change it to $0\to H_2(X,A) \to H_1(A) \to H_1(X) \to H_1(X,A)\to 0$
Moreover, $H_1(A)\to H_1(X)$ is the same map as $\pi_1(A)\to \pi_1(X)$ (both are abelian), so it's the usual projection $\mathbb{Z\to Z}/2$, so in particular it's surjective, so that $H_1(X,A)=0$. Finally, $H_2(X,A)=\mathbb Z$.
It follows that your long exact sequence looks something like $0\to \mathbb Z\to\mathbb Z\to \mathbb Z/2\to 0$. 
So if you take $\hom(-,\mathbb Z)$ of it, you get $0\to 0\to \mathbb Z\overset{2}\to \mathbb Z\to 0$, which is not exact. 
A: A fun answer using more machinery: Any cohomology theory that has the cohomology of a point the same as singular cohomology is isomorphic to singular cohomology. If $\operatorname{Hom}(H_n(-); \mathbb{Z})$ were a cohomology theory, it must then be isomorphic to singular cohomology. However, we know spaces such as $\mathbb{R}P^2$ where they disagree which yields a contradiction.
