(co)homology functor and basepoint-preserving homotopy class

Question

We can use $h^n(X)=\langle X,K(G,n)\rangle $ to define a reduced cohomology theory. I wonder if we can use the basepoint-preserving homotopy classes $\langle -,-\rangle $ to define homology?

And I wonder if the cohomology and homology functors are adjoint?

Answer 1

The answer to both questions is no.

For the first question, if $H_n$ were representable it would preserve products in the homotopy category, which are just the usual products : but it doesn't, so it's not representable.

For the second one : neither $H^n$ nor $H_n$ preserves products (see the Künneth formula), which are also products in the homotopy category, so neither of them is a right adjoint.

Answer 2

It is possible to define homology using homotopy classes of maps but it is rather more complicated. Specifically, we can define the reduced homology $h^n(X;G)$ of a pointed space $X$ as the colimit of the sequence $$\langle S^n,K(G,0)\wedge X\rangle \to \langle S^{n+1},K(G,1)\wedge X\rangle \to \langle S^{n+2},K(G,2)\wedge X\rangle \to \langle S^{n+3},K(G,3)\wedge X\rangle \to \dots$$ Here the map $\langle S^{n+m},K(G,m)\wedge X\rangle\to \langle S^{n+m+1},K(G,m)\wedge X\rangle$ is defined by taking a map $S^{n+m}\to K(G,m)\wedge X$ to get a map $S^{n+m+1}\to \Sigma K(G,m)\wedge X$ and then composing with the map $\Sigma K(G,m)\to K(G,m+1)$ which is adjoint to the homotopy equivalence $K(G,m)\to \Omega K(G,m+1)$.

For a simple example, when $X=S^n$, $K(G,m)\wedge X\cong \Sigma^n K(G,m)$ has $(m+n)$th homotopy group $G$ and so $\langle S^{n+m}, K(G,m)\wedge X\rangle\cong G$ for all $m$ and the maps in the sequence above can be checked to be the canonical isomorphism. So, the colimit is just $G$ and we find $h^n(S^n;G)\cong G$.

To prove that this homology theory coincides with singular homology, you can prove that it satisfies the Eilenberg-Steenrod axioms (the nontrivial part being excision, which comes from the homotopy excision theorem) and compute its value on $S^0$.

It's not possible to have a formula as simple as $h^n(X;G)=\langle Y,X\rangle$ for some space $Y$. For instance, this would imply that $h^n(X\times X';G)\cong h^n(X;G)\times h^n(X';G)$ which is not true.