Trying to read the section on Poincare duality from Griffiths and Harris is a nightmare. I want to know if there is a place where Poincare duality and intersection theory are done cleanly and rigorously in the order that GH do them (usually, one proves Poincare duality for singular cohomology and then defines the intersection pairing by cup product and proves that (using the Thom isomorphism) that indeed the intersection pairing counts the number of points taken with sign (and convert everything to forms using De Rham's theorem). GH on the other hand define intersection pairing and then proceed further. However, one has to wave their hands at relativistic speeds to make some things work here).
[Math] Griffiths and Harris reference
at.algebraic-topologycomplex-geometry
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The Eilenberg-Zilber theorem says that for singular homology there is a natural chain homotopy equivalence:
$$S_*(X)\otimes S_*(Y) \cong S_*(X\times Y)$$
The map in the reverse direction is the Alexander-Whitney map. Therefore we obtain a map
$$S_*(X)\rightarrow S_*(X\times X) \rightarrow S_*(X)\otimes S_*(X)$$
which makes $S_*(X)$ into a coalgebra.
My source (Selick's Introduction to Homotopy Theory) then states that this gives $H_*(X)$ the structure of a coalgebra. However, I think that the Kunneth formula goes the wrong way. The Kunneth formula says that there is a short exact sequence of abelian groups:
$$0\rightarrow H_*(C)\otimes H_*(D) \rightarrow H_*(C \otimes D) \rightarrow \operatorname{Tor}(H_*(C), H_*(D)) \rightarrow 0$$
(the astute will complain about a lack of coefficients. Add them in if that bothers you)
This is split, but not naturally, and when it is split it may not be split as modules over the coefficient ring. To make $H_*(X)$ into a coalgebra we need that splitting map. That requires $H_*(X)$ to be flat (in which case, I believe, it's an isomorphism).
That's quite a strong condition. In particular, it implies that cohomology is dual to homology.
Of course, if one works over a field then everything's fine, but then integral homology is so much more interesting than homology over a field.
In the situation for cohomology, only some of the directions are reversed, which means that the natural map is still from the tensor product of the cohomology groups to the cohomology of the product. Since the diagonal map now gets flipped, this is enough to define the ring structure on $H^*(X)$.
There are deeper reasons, though. Cohomology is a representable functor, and its representing object is a ring object (okay, graded ring object) in the homotopy category. That's the real reason why $H^*(X)$ is a ring (the Kunneth formula has nothing to do with defining this ring structure, by the way). It also means that cohomology operations (aka natural transformations) are, by the Yoneda lemma, much more accessible than the corresponding homology operations (I don't know of any detailed study of such).
Rings and algebras, being varieties of algebras (in the sense of universal or general algebra) are generally much easier to study than coalgebras. Whether this is more because we have a greater history and more experience, or whether they are inherently simpler is something I shall leave for another to answer. Certainly, I feel that I have a better idea of what a ring looks like than a coalgebra. One thing that makes life easier is that often spectral sequences are spectral sequences of rings, which makes them simpler to deal with - the more structure, the less room there is for things to get out of hand.
Added Later: One interesting thing about the coalgebra structure - when it exists - is that it is genuinely a coalgebra. There's no funny completions of the tensor product required. The comultiplication of a homology element is always a finite sum.
Two particularly good papers that are worth reading are the ones by Boardman, and Boardman, Johnson, and Wilson in the Handbook of Algebraic Topology. Although the focus is on operations of cohomology theories, the build-up is quite detailed and there's a lot about general properties of homology and cohomology theories there.
Added Even Later: One place where the coalgebra structure has been extremely successfully exploited is in the theory of cohomology cooperations. For a reasonable cohomology theory, the cooperations (which are homology groups of the representing spaces) are Hopf rings, which are algebra objects in the category of coalgebras.
Bott and Tu do this completely, in the de Rham theoretic setting of course.
Here's an alternate proof I have used when I teach this material, which I find slightly more clean and direct than using Thom classes in de Rham theory (which require choice of tubular neighborhood theorem, etc) and works over the integers.
Definition: Given a collection $S = \{W_i\}$ of submanifolds of a manifold $X$, define the smooth chain complex transverse to $S$, denoted ${C^S}_*(X)$, by using the subgroups of the singular chain groups in which the basis chains $\Delta^n \to X$ are smooth and transverse to all of the $W_i$.
Lemma: The inclusion ${C^S}_*(X) \to C_*(X)$ is a quasi-isomorophism, for any such collection $S$.
Now if $W \in S$ then "count of intersection with $W$" gives a perfectly well-defined element $\tau_W$ of ${\rm Hom}(C^S_*(X), A)$ and thus by this quasi-isomorphism a well-defined cocycle if the $W$ is proper and has no boundary. It is immediate that this cocycle evaluates on cycles which are represented by closed submanifolds through intersection count.
There are two approaches to show that cup product agrees with intersection on cohomology. Briefly, one is to take $W, V$ over $M$ and consider the special case of $W \times M$ and $M \times V$ over $M \times M$. There some work with the K"unneth theorem leads to direct analysis in this case. But this case is "universal" - cup products in $M$ are pulled back from ``external'' cup products over $M \times M$. A second proof given in https://arxiv.org/abs/2106.05986 uses a variant of the theory, where one fixes a triangulation or cubulation, and assumes $W, V$ transverse to those. There we explicitly see that these products do not agree at the cochain level (they can't since intersection is commutative, but non-commutativity of cup product is reflected in Steenrod operations), but Friedman, Medina and I show a vector field flow leads to a cobounding of the difference.
Best Answer
Poincare duality is very clearly treated, with real coefficients, via de Rham cohomology, in Spivak's Differential geometry vol. 1, the idea of using open covers with contractible intersections, anticipating sheaf theory,, apparently being due to Weil; and also in Bott - Tu's Differential forms in algebraic topology. Both are recommended.
It is also treated over arbitrary coefficient domains in the appendix to Milnor and Stasheff's Characteristic classes. Anything by Milnor is recommended.
If you just visualize a polyhedron, and its first barycentric subdivision, i.e. placing a new vertex in the center of every face, and forming a new face from the union of all new subtriangles adjacent to a given vertex, you may see the duality arising from a triangulation of a manifold. Thus the theorem is a "obvious" generalization of the duality of the Platonic solids.
The simplest argument I ever heard was in a conversation between John Morgan and Simon Donaldson, at Bob Friedman's house. John said he had a simple proof of Poincare duality using Morse functions, and Simon replied, while turning his hands over, "of course you just turn the Morse function upside down".
If you read the description of the homotopy type of a CW complex in terms of the critical points of Morse functions, say in Milnor's notes on Morse theory, you will learn that a single d cell is added each time we pass a critical point of index d. Since turning a function upside down changes a critical point of index d into one of index n-d, where n is the dimension, we are "done", (modulo the non trivial question of tracing the boundary operators, as noted by a comment below).
I have not read Milnor's notes on the h cobordism theorem, so I do not know if this is the same proof given there, but it is a book on applications of Morse theory. I would suggest the moral is that a young person could do worse than to learn Morse theory.