Equivalence of singular and CW cup products in cohomology of a space $X$ Let $X$ be a topological space, and let $C^*(X)$ denote the singular cochains of $X$ (with integral coefficients). The cup product in singular cohomology is defined (in e.g. Hatcher) in the following way:
For $\phi \in C^k(X), \psi \in C^l(X)$, the cup product $\phi \smile \psi \in C^{k+l}(X)$ is defined to be the cochain whose value on a singular simplex $\sigma \in C_{k+l}(X)$ is given by
$$(\phi \smile \psi)(\sigma)= \phi(\sigma|[v_0,...,v_k])\psi(\sigma|[v_k,...,v_{k+l}]).$$
That is, the multiple of $\phi$ evaluated on the restriction of $\sigma$ to the first $k+1$ vertices with $\psi$ on the remaining $l+1$ vertices.
This then induces the cup product in (singular) cohomology $H^*(X)$.
On the other hand, suppose that we have a CW complex $X$, and $H^*(X)$ denotes instead the cellular cohomology. We can now define a "cup product" (e.g. in May) using a diagonal approximation $\tilde{\Delta}$ (and the map it induces in cohomology):
$$H^*(X) \otimes H^*(X) \rightarrow H^*(X \times X) \xrightarrow{\tilde{\Delta}^*} H^*(X) $$
Singular and cellular cohomology are isomorphic, but these multiplications (both "cup" products) are defined differently. Why do they both induce the same multiplication on cohomology?
Either a proof or reference for this would be great.
 A: You can rephrase this as a purely singular homology question by defining a product on singular homology given by the same formula you've given for cellular homology, but with singular homology. Then by naturality, if the cup product on singular homology is the same as this new product, the cup product coincides with the cellular product.
You can do this axiomatically (see, for example, Kirk and Davis's chapter on products), but you can also do it using Eilenberg-MacLane spaces.
A product on cohomology groups gives rise to a map $K(\mathbb{Z}, n) \times K(\mathbb{Z},m) \rightarrow K(\mathbb{Z},n+m)$, by Yoneda lemma. Again by Yoneda, this is classified by a class $H^{n+m}(K(\mathbb{Z}, n) \times K(\mathbb{Z},m)) \cong H^n(K(\mathbb{Z},n)) \otimes H^m(K(\mathbb{Z},m))$ because the kth Eilenberg-MacLane space is k-1 connected. By the universal coefficient theorem this is then a map $\mathbb{Z} \otimes \mathbb{Z} \rightarrow \mathbb{Z}$. This has to be given by $(a,b) \rightarrow k(ab)$.
Transferring back over to cellular cohomology (since these products are defined in the same manner), we can figure out what k is by calculating the product on $S^n \times S^m $ with its 4 cell structure. It is possible to show geometrically the Kunneth formula for cellular cohomology with ring structure given by this product, so we have that the product of the n-cell with the m-cell is the (n+m)-cell, so $k=1$.
So since these two products agree on all products of spheres, we have that the coefficients are the same in the universal case, which implies they agree for all cases.
