Cohomology is a stronger invariant than homology because it can be equipped with a ring structure. To be precise, if one starts with the singular cohomology groups $H^\bullet(-; R)$ with coefficients in a ring $R$, then there is a map $$H^n(X; R) \times H^m(X; R) \stackrel{\smile}{\to} H^{m+n}(X; R)$$ which is graded commutative - that is, $\alpha \smile \beta = (-1)^{mn} \beta \smile \alpha$ where $\alpha \in H^n(X; R)$ and $\beta \in H^m(X; R)$. This makes $\bigoplus_i H^i(X; R)$ into a graded commutative ring, which we denote as $H^*(X; R)$, known as the graded cohomology ring of $X$ with coefficients in $R$.

Now, the cup product map above can be defined in several ways.

  • One standard way to define it is as $$H^n(X; R) \times H^m(X; R) \to H^{n+m}(X \times X; R) \to H^{n+m}(X; R)$$ where the first map in the composition is induced from the Eilenberg-Zilber map $C^n(X) \times C^m(Y; R) \to C^{n+m}(X \times Y; R)$ and the second is induced from the diagonal inclusion $\Delta : X \hookrightarrow X \times X$.

  • Originally, Poincaré defined this for smooth manifolds using transversality. That is, if $M$ is a smooth $n$-manifold, $N, L$ are submanifolds of $M$ of codimension $k$ and $l$ respectively, then $[N], [L]$ represent homology classes in $H_{n-k}(M; \Bbb R)$ and $H_{n-l}(M; \Bbb R)$ respectively. Assume further than $N$ and $L$ are transverse. Under the Poincaré duality, $[N]$ corresponds to $[N]^*$ in $H^k( M; \Bbb R)$ and $[L]$ corresponds to $[L]^*$ in $H^l(M; \Bbb R)$. Poincaré then defined cup product of these two classes as $[N]^* \smile [L]^* := [N \pitchfork L]^*$, the dual of the class represented by $N \pitchfork L$.$(\star)$

  • However, there is a much easier definition one can see in standard algebraic topology books (e.g., Hatcher) which goes as follows. Take a $k$-cochain $\psi$ in $C^k(X; R)$, an $l$-cochain $\varphi$ in $C^l(X; R)$, and then define the $k+l$-cochain $\psi \smile \varphi$ by $$(\psi \smile \varphi)(\sigma) := \psi(\sigma|_{[v_0, \cdots, v_k]}) \varphi(\sigma|_{[v_k, \cdots, v_{k+l}]})$$ where $\sigma : \Delta^{k+l} \to X$ is a $(k+l)$-singular simplex. One can easily see that this cochain-level map $C^k(X; R) \times C^l(X; R) \to C^{k+l}(X; R)$ descends to homology. This is precisely the cup product.

All the definitions above are equivalent. The first definition was discovered by Eilenberg & Zilberg in this paper, the second by Poincaré in Analysis Situs. But who discovered the third?

Remark$(\star)$: It is not true however that every sinuglar cycle of a smooth manifold can be represented by a submanifold, see Thom's paper. Although, this is a high dimensional phenomenon and the first counterexample appears at dimension $7$.


1 Answer 1


The history of the cup product is described on pages 135–136 of Never a Dull Moment: Hassler Whitney, Mathematics Pioneer by Keith Kendig.

A key event was the 1934 International Conference in Topology, held in Moscow. Andrey Kolmogoroff and James Alexander independently presented the basic idea, but it apparently wasn't quite right.

Whitney heard Kolmogoroff and Alexander's definitions of product, and on the one hand the idea seemed tremendously significant, but at the same time something didn't smell quite right. Whitney had made friends with the Czech mathematician Edouard Čech, and the two of them agreed that the multiplication idea seemed powerful and promising. … Nevertheless, there seemed to be a problem: If you multiply two sets like a disk and an interval, the result is a solid cylinder—the dimensions add: $2+1=3$. … However, Kolmogoroff and Alexander's definition of product didn't do that, instead giving a dimension that was larger than the sum by one. Whitney and Čech puzzled over this problem, and within a few months they had ironed out the difficulty. … By that time, Alexander had already written a paper using his original definition and submitted it to the Annals. Upon learning of the Whitney–Čech revision, Alexander immediately saw its advantages. He rewrote and resubmitted his paper, which appeared in 1936.

  • $\begingroup$ Fascinating piece of history! Thanks a lot! $\endgroup$ Commented Jul 13, 2021 at 18:52

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