I am reading Hatcher, section 3.2 and when he defines a cup product he says that we need to "consider cohomology with coefficients in a ring R" and I am just confused why we need the coefficients to be in a ring.

The reason I ask is because in section 2.2 under "Homology with Coefficients" he defines


Homology groups with coefficients in G, where G is a fixed abelian group. Why are we switching to rings and modules when we cross over into cohomology?

After some more reading I am understanding a little bit more:

From the very beginning of Chapter 3 in Hatcher: (Which I should have read to begin with but thought it was ok to skip over)

Homology groups are contravariant functors, while cohomology groups are covariant functors.

Contravariance leads to extra structure in cohomology: that's what the cup product is.

Why doesn't this happen for homology? Actually, we can define the cross product (discussed in section 3.B) but this relies on the map $X \times X \rightarrow X$. And, for a general X, the only way we can define a map like that is by using projection onto one of the factors, and since projections "collapse the other factor to a point, the resulting product is rather trivial"

For cohomology on the other hand, we need to rely on a map like this: $X \rightarrow X\times X$, for which we use the "diagonal map" $\Delta(x) = (x,x)$ which is actually waht the cup product is.

I am still a bit confused on this though, and would appreciate further clarification: How does the diagonal map lead to the cup product explicitly?

Is it because we somehow get to switch the arrows in the cross product map? In that case, it seems strange to define the cup product first and then the cross product later.

  • $\begingroup$ Where else would you have the coefficients? $\endgroup$ Dec 10 '17 at 9:01
  • $\begingroup$ @TobiasKildetoft in a group. I guess I am really asking: where are we using the ring structure of R? $\endgroup$
    – user281395
    Dec 10 '17 at 9:02
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    $\begingroup$ In a product one has to be able to multiply. $\endgroup$ Dec 10 '17 at 9:08
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    $\begingroup$ Cohomology groups in general have an external product, which gives internal (cup) products when the "coefficients" have a product. For singular cohomology this is explained in Spanier's text. I can add details later. $\endgroup$
    – Pedro Tamaroff
    Dec 10 '17 at 10:04

Hurkyl has already answered your question, and this is mainly to attempt to answer your following question from the comments. My answer was too long for a comment so I posted it here.

A (singular) cochain with coefficients in some group $G$ is an element $\alpha\in Hom_{\mathbb{Z}}(S_n(X),G)$, where $S_n(X)$ is the singular complex of a space $X$. Given another cochain $\beta\in Hom_{\mathbb{Z}}(S_m(Y),H)$ on a space $Y$ with coefficients in a group $H$, we get a product defined by the composition

$\alpha\smile' \beta:S_{n+m}(X\times Y)\xrightarrow{EZ} S_n(X)\otimes_{\mathbb{Z}} S_m(Y)\xrightarrow{\alpha\otimes\beta}G\otimes_{\mathbb{Z}} H$

where the first arrow is an Eilenberg-Zilber morphism composed with the projection onto the target group. This is now a cochain $\alpha\smile'\beta\in Hom_\mathbb{Z}(S_{n+m}(X\times Y),G\otimes_\mathbb{Z}H)$. If we have some homomorphism $\varphi:G\otimes_{\mathbb{Z}}H\rightarrow K$ into a third group $K$ we can compose with the previous product to get an element $\varphi\circ(\alpha\smile'\beta)\in Hom_\mathbb{Z}(S_{n+m}(X\times Y),K)$. Similarly if we have some map $f:Z\rightarrow X\times Y$ then we can precompose to get a cochain $f^*(\alpha\smile'\beta)\in Hom_{\mathbb{Z}}(S_{m+n}(Z),G\otimes_\mathbb{Z}H)$. Its easy to see that all these definitions descend to give well defined operations on cohomology and lead to the most general forms of the cup product available on singular cohomology.

The cup product comes about when $X=Y=Z$ and $f=\Delta:X\rightarrow X\times X$ is the diagonal map, in which case $\Delta^*(\alpha\smile'\beta)\in Hom_\mathbb{Z}(S_{m+n}(X),G\otimes_\mathbb{Z} H)$. Moreover, when $G=H=K=R$ is a ring and $\varphi=\cdot:R\otimes_\mathbb{Z}R\rightarrow R$ is the ring multiplication we get the standard cup product $\alpha\smile \beta=\varphi_*\Delta^*(\alpha\smile'\beta)\in Hom_\mathbb{Z}(S_{m+n}(X),R)$.

  • $\begingroup$ I've replaced $\cup$ with $\smile$ in the answer. Hope it's not a problem. Regards, $\endgroup$
    – Pedro Tamaroff
    Dec 11 '17 at 15:34

For simplicity, consider just $H^0$ of a connected space. A general product would look like

$$ H^0(X, A) \otimes H^0(X, B) \cong A \otimes B \cong H^0(X, A \otimes B) $$

IIRC, you can consider this more general cup product. However, it is more typical to work with a fixed coefficient group.

So, initially the cup product would land in the wrong group: $$H^0(X, A) \otimes H^0(X, A) \to H^0(X, A \otimes A) $$

To fix this, you additionally need a map $A \otimes A \to A$, so that you can form the composite

$$H^0(X, A) \otimes H^0(X, A) \to H^0(X, A \otimes A) \to H^0(X, A)$$

So that's where your product map comes into play. And you would like it to be well-behaved algebraically; e.g. to satisfy the ring axioms.

  • $\begingroup$ I am curious about this possible "more general cup product." Where exactly can I find more information about such a thing? $\endgroup$
    – user281395
    Dec 10 '17 at 9:11

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