Recently, it became apparent to me that I was not the only one who always first thought in terms of cap product before actually computing a cup product. There is no denying this is evil, but I found it hard to get the actual historical source of an example where this line of thought fails.


More precisely, given a smooth manifold and a triangulation $T$, one first considers the barycentric subdivision $B$ of $T$ and produces the dual triangulation $T^\*$ by glueing simplexes in $B$. The cap product is then defined between a chain of $T$ and a chain of $T^\*$ and results in a chain of $B$. 


Could someone be so kind as to provide a reference where, in the above setup, the cap product (in homology) is ill-defined? 


I know this is probably Poul Heegard, but I could not find a reference...


As a bonus, a reference giving a characterisation of this phenomenon would be very welcome.



Cap product à la Poincaré
You need to add the word "oriented", in order to assign signs consistently to the intersections, and "compact" so that there will be finitely many intersections. To see the issue for unoriented manifolds, think about $\mathbb{RP}^2$. We have $H\_0(\mathbb{RP}^2) \cong \mathbb{Z}$ and $H\_1(\mathbb{RP}^2) \cong \mathbb{Z}/2$, so cap product would define a bilinear map $\mathbb{Z}/2 \times \mathbb{Z}/2 \to \mathbb{Z}$. Such a map must be zero, but two non-contractible curves in $\mathbb{RP}^2$ meet an odd number of times. (For example, two lines in $\mathbb{RP}^2$ meet at a single point.)


After that, nothing is wrong with it. For a compact oriented $n$-fold $X$, Poincare duality gives an isomorphism $H^k(X) \cong H\_{n-k}(X)$ so we get a map
$$H\_{n-a} \times H\_{n-b} \cong H^a \times H^b \overset{\cup}{\longrightarrow} H^{a+b} \cong H\_{n-a-b}.$$
Tracing through the definitions on a triangulation, one obtains the map you described.


As far as I know, it is only important to distinguish cap and cup when one wants to move away from the world of compact oriented manifolds. (Cup makes sense when $X$ isn't a manifold at all!)


