This came up in the question about Eilenberg-MacLane spaces. Given the definition of K(G, n), it's easy to prove that there is a map K(G,n) x K(G,n) --> K(G,n) that endows cohomology with an additive structure.
Question: what's the most geometric way to show the existence of maps
K(G,n) x K(G,m) --> K(G,n+m)that endow cohomology with multiplicative structure?
If you form the smash product $X = K(A,p) \wedge K(B,q)$ of two Eilenberg-MacLane spaces, then the resulting space is $(p+q-1)$-connected, and the first non-trivial homotopy group in dimension $p+q$ is $A \otimes B$.
To see this "geometrically", I would model the EM spaces as CW-complexes, whose first non-basepoint cells are in dimensions p and q respectively. Then in the smash product $X$, the bottom dimensional cells are products of the bottom dimensional cells of the EM spaces; this gives the connectivity result, and by looking at the attaching maps of the $(p+q+1)$-dimensional cells in $X$, you can compute $\pi_{p+q}$. Now you can use obstruction theory to produce a map $X \to K(A\otimes B, p+q)$.
There is a geometric (space level) way of realizing Eric's answer. In fact, this is the subject of section 1 in a paper by Ravenel and Wilson (MathSciNet). They used the iterated simplcial bar construction $B^nG$ as a model for $K(G,n)$ so that, when $G$ is a ring, the ring multiplication
$$G\times G \to G$$
induces
$$G\times BG \to BG$$
by fixing the first factor and then
$$BG\times BG \to B(BG)=K(G,2)$$
by fixing the second factor, and so on. An explicit description is found in page 700 in their paper, where they prove the resulting maps give rise to the cup product pairing (Theorem 1.7).
One perspective I like is that homology is abelianization. Specifically, given a based space X you can construct a new space AG(X), the (reduced) free abelian group on X, whose points are finite formal sums
Σ nx xof points of X with integer coefficients, subject to the relation that the basepoint is sent to 0. (The topology takes some describing, but it is roughly a quotient topology from two copies of the infinite symmetric product.) Instead of integer coefficients I could take coefficients in some abelian group M, but let's stick to this for now.
For X a CW-complex, the homotopy groups of AG(X) are actually the reduced integral homology groups of X (this is the Dold-Thom theorem). You might believe this because you can show that the functor X -> π*(AG(X)) is a homology theory satisfying the Eilenberg-Steenrod axioms, because it converts quotient sequences X -> Y -> Y/X of spaces into exact sequences AG(X) -> AG(Y) -> AG(Y/X) of topological groups, which are (almost) fibration sequences.
Under this perspective, you actually have a construction of K(Z,n); it is AG(Sn), the free abelian group on the n-sphere. The map described in previous answers from K(Z,n) ^ K(Z,m) to K(Z,n+m), giving you cup products, is then the map
AG(Sn) ^ AG(Sm) -> AG(Sn ^ Sm) = AG(Sn+m)
given by
(Σ nx x) ^ (Σ my y) -> Σ (nx my) (x ^ y)
which looks exactly like a tensor product map.
Yes, this iterated bar construction model should be better known, and can be expressed even more geometrically than Ravenel and Wilson do.
If $A$ is an abelian group then $K(A,n)$ is modeled by a space whose points are given by finite collections of points in $[0,1]^n$ which are labelled by $A$. If two or more points coincide, that collection is equivalent to one in which those coinciding points are replaced by one point whose label is given by the sum of the coinciding labels. If a point is labelled by the identity or if a point is on the boundary of the cube, it can be removed. This is a model for the free abelian group on $S^n$, with its basepoint as identity. The free abelian group was first studied by Dold and Thom fifty years ago, but I heard about this way to think about it from John Baez.
In this model, the first product which was referred to is the group structure which is "take the union". The second product $K(R,n) \times K(R,m) \to K(R,n+m)$ is to "take products and multiply labels" (note that $R$ now needs to be a ring). That is, one gets a point in $K(R, n+m)$ whose underlying collection of points in $[0,1]^{n+m}$ is the product of underlying collections, and where the label on some $x \times y$ is the product of the label of $x$ with the label of $y$.
A very specific example: for $G=\mathbb{Z}$ and $n=m=1$, it is the map $S^1 \times S^1 \to CP^\infty$ which maps $S^1 \times S^1$ to the 2-skeleton $S^2$ of $CP^\infty$ by a degree 1 map.
Here's a general but less geometric way of understanding it. $K(G,n)$ can be constructed as the realization of the simplicial abelian group associated to the chain complex with $\mathbb{Z}$ in degree n and 0 elsewhere under the Dold-Kan correspondence. By a form of Eilenberg-Zilber, the levelwise tensor product of two simplicial abelian groups is, up to natural weak equivalence, the same as the ordinary tensor product of chain complexes. But the ordinary tensor product takes $K(G,n)$ and $K(G,m)$ to $K(G \otimes G,m+n)$, and there is a natural map from $K(G,n) \times K(G,m)$ to the levelwise tensor product given levelwise by the natural map from a product to a tensor product. Composing all of this together with a ring multiplication map $G \otimes G \to G$ giving $K(G \otimes G,m+n)\to K(G,m+n)$ should give the cup product. With some unraveling that I don't have time to do right now but I invite someone else to figure out, you should be able to turn this into a reasonably explicit map on the simplicial abelian group level coming from the Alexander-Whitney map.
I like Charles' answer, but it, and the others, treat $H^k(X;A)=[X,K(k,A)]$ as a definition and gives you the map that induces a product.
Here is an answer if you want to use singular cohomology in the usual way as the definition of $H^k(X;A)$, and the cup product defined using a diagonal approximation, and you consider the equality of functors $H^k(-;A)=[-,K(k,A)]$ as a theorem proved by obstruction theory.
Thus the cup product is a well defined, natural in $X$ pairing $[X,K(A,k)]\times [X,K(A,n)]\to [X,K(A,k+n)]$.
Substitute $X=K(A,k)\times K(A,n)$ and look where $(p_1, p_2)$ goes, where $p_i, i=1,2$ is the projection. It goes to the map you asked for, and hence shows "existence".
Consider the evaluation map $S^1 \times [S^1,K(Z,n)] \to K(Z,n)$. Since $S^1$ is a $K(Z,1)$, and $[S^1, K(Z,n)]=\Omega^1(K(Z,n))$ is a $K(Z,n-1)$, up to homotopy we get a map $K(Z,1)\times K(Z,n-1) \to K(Z,n)$. I'm not actually sure if this induces the product on cohomology. If it does, there is a natural generalization:
Consider the space of pointed maps $[K(A,n),K(A,m+n)]$. Then $\pi_k([K(A,n),K(A,m+n)])=0$ for $k>m$, and $=A$ for $k=m$. To see this, note that (all maps are pointed) $$[S^k, [K(A,n),K(A,m+n)]] = [K(A,n), \Omega^k(K(A,m+n))]$$ $$ =[K(A,n),K(A,m+n-k)]= {\check H}^{m+n-k}(K(A,n),A) = 0 $$
if $k>m$, and $=A$ if $k=m$ (by Hurewicz).
Thus, we have a map $i: K(A,m) \to [K(A,n), K(A,m+n)]$ by obstruction theory sending $\pi_m(K(A,m))\to \pi_m([K(A,n),K(A,m+n)])$ isomorphically, which of course is equivalent to a map (by evaluation)
$$ K(A,m) \times K(A,n) \to K(A,m+n).$$
Maybe someone could explain to me if this gives the correct cohomology operation?
The picture for the thom spectrum $TO$, as opposed to the spectrum $\{K(G,n)\}$ , with the maps $\Sigma TO(n)=T(\xi(BO(n)) + \epsilon)\to TO(n+1)$ is much clearer. Namely there are easy to see maps $TO(n)\wedge TO(m) \to T(O(n+m))$ that give the multiplication on unoriented framed bordism.
It is given by the bundle maps $ TO(n)\wedge TO(m) \to B(O(n) \times O(m))$ which is isomorphic to the bundle $T(\xi(O(n)) \times \xi(O(m))) \to B(O(n) \times O(m))$ which maps to the bundle $T(O(n+m)) \to B(O(n+m))$.
This gives the obvious ring structure on $\pi_*(TO)$.
