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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.

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    $\begingroup$ You are talking about a product that involves just homology and not cohomology, so this is not the cap product. Instead it is usually called the intersection product. A classical reference for the intersection product is the textbook by Seifert and Threlfall. A more recent textbook source is Bredon's "Topology and Geometry". Bredon attributes the intersection product to Lefschetz. The intersection product is a perfectly well-behaved product, with the right hypotheses, so it's not clear why you use the words "evil", "fails", and "ill-defined" in reference to it. $\endgroup$
    – Allen Hatcher
    Commented Dec 17, 2014 at 20:42
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    $\begingroup$ Also, $T^*$ is a cellulation (a decomposition into cells), not a triangulation as stated in the question. $\endgroup$
    – Allen Hatcher
    Commented Dec 17, 2014 at 20:44
  • $\begingroup$ Many thanks for the references! As for the choice of words, the first time cup product was presented to me, it was branded as exteremely unrigorous and dangerous to see it (dually) as an intersection. This and the fact that I always saw cup being used and never intersection fueled my unjustified belief in it being ill-defined. I guess that transversality hypothesis make intersection product more cumbersome than cup... $\endgroup$
    – ARG
    Commented Dec 17, 2014 at 21:03

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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!)

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    $\begingroup$ The cap product $\cap: H^k\times H_m\to H_{m-k}$is defined for any (reasonably) behaved space. Only its intersection theoretic interpretation requires the space be oriented manifold. $\endgroup$
    – Liviu Nicolaescu
    Commented Dec 17, 2014 at 16:19
  • $\begingroup$ So if intersection is only counted mod 2, then there would be no problem whatsoever (except compactness)? $\endgroup$
    – ARG
    Commented Dec 17, 2014 at 20:43
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    $\begingroup$ I think you may be missing Liviu's point: the cap product is an algebraically defined operation; on the chain level it is the adjoint to the cup product. Its notation perhaps suggests intersections, but it is not defined in terms of intersecting anything. To interpret cap and cup products in terms of intersections requires being in a space with some kind of transversality properties, typically a manifold. To get integer-valued intersection numbers, as opposed to mod 2 intersections, requires some orientations. $\endgroup$
    – Danny Ruberman
    Commented Dec 18, 2014 at 13:32

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