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Homology is not naturally a coalgebra unless you take field coefficients or unless your object has torsion free homology groups over the integers. The basic issue, as mentioned above, is that even though you have a split exact universal coefficient sequence for the homology of a product, the splitting isn't natural. You don't actually need homology to be dual to cohomology because that would involve some additional finiteness properties.

However, if your space has torsion-free homology with integer coefficients, then H_(X;R) = H_(X) ⊗ R for all R, and so you get a coalgebra structure on the homology of X with coefficients in R simply as the base change of the one over the integers. If R is an algebra over a field then you get a coalgebra structure with no assumptions on X by base-change from said field.

I should probably point out that the universal coefficient theorem Kunneth formula is more complicated than stated in a previous answer. There's a universal coefficient an exact sequence

0 → H_(C;Z) ⊗ H_(D;M) → H_(C ⊗ D;M) → Tor(H_(C;Z), H_*(D;M)) → 0

but notice that one side involves integer coefficients and the other coefficients in a general module. If you want the universal coefficient theorem with the same coefficients on both sides it takes the form of a spectral sequence with E_2-term

Tor^R_{p,q} (H_(X;R), H_(Y;R))

converging to H_*(X x Y;R). (The bigrading on Tor comes because we're taking Tor of graded modules.)

In general if E is a generalized homology-cohomology theory then flatness of E_* X over the ground ring E_* guarantees a coalgebra structure on the E-homology of X. This also may or may not have anything to do with duality, because flatness and projectiveness are not the same.

As mentioned, you still do have a coalgebra structure on the chains of X (or the "E-homology object of X" in the stable homotopy category), and this is really just some kind of failure of the homology groups to mimic what's going on behind the scenes.

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Homology is not naturally a coalgebra unless you take field coefficients or unless your object has torsion free homology groups over the integers. The basic issue, as mentioned above, is that even though you have a split exact universal coefficient sequence for the homology of a product, the splitting isn't natural. You don't actually need homology to be dual to cohomology because that would involve some additional finiteness properties.

However, if your space has torsion-free homology with integer coefficients, then H_(X;R) = H_(X) ⊗ R for all R, and so you get a coalgebra structure on the homology of X with coefficients in R simply as the base change of the one over the integers. If R is an algebra over a field then you get a coalgebra structure with no assumptions on X by base-change from said field.

I should probably point out that the universal coefficient theorem is more complicated than stated in a previous answer. There's a universal coefficient sequence

0 → H_(C;Z) ⊗ H_(D;M) → H_(C ⊗ D;M) → Tor(H_(C;Z), H_*(D;M)) → 0

but notice that one side involves integer coefficients and the other coefficients in a general module. If you want the universal coefficient theorem with the same coefficients on both sides it takes the form of a spectral sequence with E_2-term

Tor^R_{p,q} (H_(X;R), H_(Y;R))

converging to H_*(X x Y;R). (The bigrading on Tor comes because we're taking Tor of graded modules.)

In general if E is a generalized homology-cohomology theory then flatness of E_* X over the ground ring E_* guarantees a coalgebra structure on the E-homology of X. This also may or may not have anything to do with duality, because flatness and projectiveness are not the same.

As mentioned, you still do have a coalgebra structure on the chains of X (or the "E-homology object of X" in the stable homotopy category), and this is really just some kind of failure of the homology groups to mimic what's going on behind the scenes.