Let $X$ and $Y$ be topological spaces and consider cohomology over a ring $R.$ Hatcher (in his standard Algebraic Topology text) defines the cross product of cohomology classes
$$H^k(X) \times H^l(Y) \to H^{k+l}(X\times Y),$$
by $a\times b = p_1^*(a) \smile p_2^*(b),$ with $p_1$ and $p_2$ the projection maps from $X\times Y$ onto $X$ and $Y.$ Here $\smile$ is the cup product of cohomology classes.
My question: while Hatcher gives some idea for how to visualize the $\textit{cup product}$ (in terms of intersections of maps on simplices -see in particular pages 187-189), he does not give much intuition into the $\textit{cross product}$. How should a student who encounters this cross product for the first time thing of it?
I'd be particularly interested in simple examples/ nice pictures/ any general comments to build intuition. Comments on the cup product are also welcome.
Best Answer
Sorry, this isn't a full answer, rather some algebraic hints at how you might relate the cross product to your intuition about the cup product (wrt intersections) and to linear algebra. It should really be a comment but it's too long.
First, a quick review! The cross product, $\times$, is actually part of the cup product:
$$x \smile y := \Delta^*(x \times y)$$
The diagonal embedding $X \xrightarrow{\Delta} X \times X$, is simply a canonical way to embed a space X into an ambient space endowed with the product topology, $\Delta X := \{(x,x) \in X \times X\}$. It is useful when want to look in the neighborhood of a space $X$ (e.g., at germs of functions on $X$), but $X$ sits in no ambient space. The word, "diagonal embedding," comes from the example of embedding of $R^1 \hookrightarrow R^2$ taking $x \mapsto (x,x)$, that is, taking the line $R^1$ and embedding it into $R^2$ as the line $y=x$.
This is the reason we have a cup product in cohomology and not in homology. The map induced by $\Delta$ on homology, $H_*(X) \xrightarrow{\Delta_*} H_*(X \times X)$, being the "wrong direction" for a product (from 1 to 2), whereas the induced map on cohomology, $H^*(X \times X) \xrightarrow{\Delta^*} H^*(X)$, is the right direction (from 2 to 1), and lends itself to the following precomposition.
The cup product (where $p+q=k$), $$H^k(X) \longleftarrow H^p(X) \times H^q(X)$$ is actually $$H^k(X) \leftarrow H^k(X \times X) \leftarrow H^p(X) \times H^q(X)$$
What is this mysterious map $H^p(X) \times H^q(X) \to H^{p+q}(X \times X)$? It's called the "cross product", $\times$, and, as you know, is defined more generally on $H^p(X) \times H^q(Y) \to H^{p+q}(X \times Y)$.
The cross product relates the cohomology groups of two different spaces to the cohomology groups of their product space.
I also think in pictures, so I understand the desire for a cartoon picture, but the algebra is actually quite enlightening here.
The tensor product of two graded abelian groups is $$(A^* \otimes B^*)^n := \bigoplus_{i+j=n} A^i \otimes B^{\text{ }j}$$ which, if you prefer thinking in terms of smash rather than tensor, will remind you of how we get from an $i$-cell in a pointed CW-complex X and a $j$-cell in a pointed CW-complex in Y, to an $i+j$-cell in $X \wedge Y$.
$$(X \wedge Y)_{n} := \times_{i+j=n}(X_i \wedge Y_j)$$
Returning our gaze to the cross product: $$H^p(C^*) \times H^q(D^*) \xrightarrow{\times} H^{p+q}(C^* \otimes D^*)$$
where $\alpha(w_i)$ is 0 when $w_i$ and $\alpha$ are of different degrees, similarly with $\beta(z_i)$.
This might strike you as a bit strange: how does this definition of the cross product match up with yours? Let's revisit Hatcher's definition in terms of the map induced by the projections of the Cartesian product.
$$X \leftarrow X \times Y \rightarrow Y$$ $$H^*(X) \rightarrow H^*(X \times Y) \leftarrow H^*(Y)$$
Taking the pullback:
where the group labeled by the notational atrocity $H^*(X) \times_{H^*(X \times Y)} H^*(Y)$ is defined as the friendly pairs of objects $\{(a,b) \text{ such that } p_1^*(a) = p_2^*(b)\}$ for all $a \in H^*(X), b \in H^*(Y)\}$.
Hatcher's definition of $a \times b$ is then:
$$H^*(X \times Y) \times H^*(X \times Y) \to H^*(X \times Y)$$ $$\text{( }p_1^*(a), p_2^*(b) \text{ )} \mapsto p_1^*(a) \smile p_2^*(b)$$
So, recall that I defined the cup product as $p_1^*(a) \smile p_2^*(b) := \Delta^*(p_1^*(a) \times p_2^*(b))$. Well, wait a minute, how do we define the diagonal embedding on a $X \times Y$, simple: $\Delta(X \times Y) := {(x,y) \text{ such that } x = y} = X \cap Y$, the diagonal embedding is the embedding $X \cap Y \hookrightarrow X \times Y$ (which, since $X \cap X = X$, reduces to the usual diagonal embedding).
Assuming that the definitions of cup product are equivalent, $a \times b = p_1^*(a) \smile p_2^*(b) = \Delta^*(p_1^*(a) \times p_2^*(b))$. From the point of view of the cup product, now have a hint as to what is going on: we're shoving $a$ and $b$ into the same range s.t. we can use our intuition about the cup product (which is only defined for X = Y) to recon with them.
Some other comments:
Bott says something like: a cocycle is a creature which lurks over spaces, pounces on cycles, eats them, and spits out numbers. In other words, it's a lot like a linear functional. In fact, I encourage you to recall that maps $f$ between modules satisfy the same rules that linear maps do, e.g., $f(ax +y) = af(x) + f(y)$. Indeed, this notation, i.e., $(C^k, d^k):= (Hom_R(C_k, R), d_k^*)$, a bit abhorrent at first glance, is actually telling us something. It's just borrowing from familiar linear algebra, for example, for a linear map between R-vector spaces $f: V \to W$, when we look at the dual spaces $V^*:=Hom_{R-Vect}(V, R)$ and $W^*:=Hom_{R-Vect}(V, R)$, we take the transpose $f^*: W^* \to V^*$.
The cross product, I assume, gets its name from being a generalization of the cross product of vectors in $R^3$ we know and love. For any two 1-forms, $\alpha$ and $\beta$, their wedge product is equivalent to the Hodge dual of their cross product:
$$\alpha \wedge \beta = *(\alpha \times \beta)$$
This is only true for 1-forms, but in general, the wedge product of p-form and a q-form is a p+q-form. I suspect that we might picture the cross product on cohomology groups via looking at simplicial groups rather than chain complexes, but I don't understand them well enough to describe how we might take their cohomology without geometrically realizing them or passing back to chain complexes anyway.