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• CommentRowNumber1.
• CommentAuthorUrs
• CommentTimeNov 11th 2009

added a paragraph and a reference to cup product, archiving the blog comment here

• CommentRowNumber2.
• CommentAuthorEric
• CommentTimeNov 11th 2009

Whenever I see discussion about cup product, it always starts off with "cocycles", but cup product is also defined on cochains that are not cocycles. Is it possible to say something about that on the page? There is more to life than cohomology (isn't there?)

• CommentRowNumber3.
• CommentAuthorUrs
• CommentTimeNov 11th 2009

Okay, maybe I need to make this clearer, because that's precisely the point here:

the cup product on cochains is, while not graded commutative on the nose as its image in cohomology, "homotopy-commutative". Even on cochains, yes.

• CommentRowNumber4.
• CommentAuthorEric
• CommentTimeNov 11th 2009
• (edited Nov 11th 2009)

But there is one glaring reference missing. Hassler Whitney is one of my all time favorite mathematicians and he has a BEAUTIFUL paper called "On Products in a Complex". It might be debatable whether he is the first to construct cup product, but his stuff is very beautiful. I'm not qualified to do it justice though.

PS: Found it:

On Products in a Complex
by Hassler Whitney
(Received June 10, 1937; Revised November 9, 1937)
http://www.jstor.org/pss/1968795

I can at least add this reference to the nLab without much commentary.
• CommentRowNumber5.
• CommentAuthorUrs
• CommentTimeNov 11th 2009
This comment is invalid XHTML+MathML+SVG; displaying source. <div> <blockquote> I can at least add this reference to the nLab without much commentary. </blockquote> <p>You should! Thanks.</p> </div>
• CommentRowNumber6.
• CommentAuthorUrs
• CommentTimeNov 11th 2009

By the way, I think this page is still lacking lots of material and reference. It's not meant to be close to anything satisfactory.

• CommentRowNumber7.
• CommentAuthorColin Tan
• CommentTimeDec 25th 2013

Following the slogan that cohomology is the hom set in an infinity topos, it seems that the cup product for singular cohomology with coefficients in a ring R should be induced by a morphism $B^n R \times B^m R \to B^{n+m} R$. How should this morphism be given?

• CommentRowNumber8.
• CommentAuthorjim_stasheff
• CommentTimeDec 25th 2013
@3 the cup product on cochains is, while not graded commutative on the nose as its image in cohomology, "homotopy-commutative". Even on cochains, yes

yes, in a very very strong sense cf E_\infty

as for :the cup product for singular cohomology with coefficients in a ring R should be induced by a morphism BnR×BmR→Bn+mR. How should this morphism be given?

depends on which model for B^n you are using
• CommentRowNumber9.
• CommentAuthorUrs
• CommentTimeDec 25th 2013

This is discussed at cup product.

(Though looking back at the entry, I see that it could do with some polishing. As usual.)

• CommentRowNumber10.
• CommentAuthorColin Tan
• CommentTimeDec 26th 2013
What I mean is that the general discussion is the top of the entry is not uniform with the discussion at singular cohomology.
• CommentRowNumber11.
• CommentAuthorUrs
• CommentTimeDec 26th 2013

Yes, the paragraph cup product – In singular cohomology does not explain how it relates to the general definition. Some glue involving the words “Eilenberg-Zilber map” needs to be added here. Maybe somebody feels inspired to do so.

• CommentRowNumber12.
• CommentAuthorColin Tan
• CommentTimeApr 8th 2014
• (edited Apr 8th 2014)

How can I write the morphism from $B^n R \times B^m R$ to $B^{n+m} R$ that induces the cup product given that delooping $B$ is defined such that each deloopable space $X$ is the homotopy colimit $pt \times_{BX} pt$?

• CommentRowNumber13.
• CommentAuthorUrs
• CommentTimeApr 8th 2014

I have now written this out in some detail at cup product – Via the Dold-Kan correspondence.

• CommentRowNumber14.
• CommentAuthorDmitri Pavlov
• CommentTimeDec 2nd 2018
• (edited Dec 2nd 2018)

The article cup product contains the following claim:

for the tensor product of abelian groups there are canonical natural projection maps p_{A,B}:A×B→A⊗B.

This is then applied to the case of simplicial abelian groups A=Γ(V) and B=Γ(W).

From the context, p_{A,B} must be a morphism of (simplicial) abelian groups.

It appears to me that such a natural map cannot exist: in the category of abelian groups products are isomorphic to coproducts, so a map A×B→A⊗B is a pair of homomorphisms of abelian groups A→A⊗B and B→A⊗B, and it is easy to show that the only natural maps of this type are the zero maps.

Am I missing something here?

• CommentRowNumber15.
• CommentAuthorTodd_Trimble
• CommentTimeDec 2nd 2018

I think what was obviously meant was canonical bilinear (not linear) maps.

• CommentRowNumber16.
• CommentAuthorDmitri Pavlov
• CommentTimeDec 2nd 2018

Re #15: No, we need to apply the forgetful functor U to this map, so it must be a homomorphism of simplicial abelian groups. Otherwise the displayed formula for ∪_DK does not make sense. Perhaps Urs can clarify this?

• CommentRowNumber17.
• CommentAuthorRichard Williamson
• CommentTimeDec 2nd 2018
• (edited Dec 2nd 2018)

Could do with some more details certainly! Maybe something like the following is intended. The gadgets $\Gamma(V)$ and $\Gamma(W)$ are both defined levelwise by a Hom out of (at each level) the same object, which will preserve products. So one reduces to needing a morphism $V \times W \rightarrow V \otimes W$, of which there is a canonical one.

• CommentRowNumber18.
• CommentAuthorUrs
• CommentTimeDec 3rd 2018
• (edited Dec 3rd 2018)

[ removed ]

• CommentRowNumber19.
• CommentAuthorUrs
• CommentTimeDec 3rd 2018
• (edited Dec 3rd 2018)

Thanks everyone.

In #18 I had first missed Dmitri’s actual point, as maybe Todd and Richard did, too.

Todd and Richard are both right in that $p_{A,B}$ is meant to be the universal bilinear map and meant to be applied degreewise.

But then its domain and codomain were indeed denoted incorrectly, forgetting the forgetful functor $U$ (which is easy to do, we do this by default on the page on tensor product of abelian groups, which should maybe be improved) and this was Dmitri’s point.

So I have fixed it now (I think) as follows:

The declaration of $p$ I changed from

$(wrong) \;\;\;\; p_{A,B} \;\colon\; A \times B \longrightarrow A \otimes B$

to

$p_{A,B} \;\colon\; U(A) \times U(B) \longrightarrow U(A \otimes B)$

and then the definition of the cup product itself I changed form

$(wrong) \;\;\;\; \cup_{DK} \;\colon\; DK(V_\bullet)\times DK(W_\bullet) = U(\Gamma(V_\bullet)) \times U(\Gamma(W_\bullet)) \stackrel{\simeq}{\to} U(\Gamma(V_\bullet) \times \Gamma(W_\bullet)) \stackrel{U(p)}{\to} U(\Gamma(V_\bullet)\otimes \Gamma(W_\bullet)) \stackrel{U(\gamma)}{\longrightarrow} U(\Gamma(V_\bullet \otimes W_\bullet)) \stackrel{U(\Gamma(f))}{\to} U(\Gamma(Z_\bullet)) = DK(Z_\bullet) \,.$

to

$\cup_{DK} \;\colon\; DK(V_\bullet)\times DK(W_\bullet) = U(\Gamma(V_\bullet)) \times U(\Gamma(W_\bullet)) \stackrel{ p_{\Gamma(V_\bullet), \Gamma(W_\bullet)} }{\to} U(\Gamma(V_\bullet)\otimes \Gamma(W_\bullet)) \stackrel{U(\gamma)}{\longrightarrow} U(\Gamma(V_\bullet \otimes W_\bullet)) \stackrel{U(\Gamma(f))}{\to} U(\Gamma(Z_\bullet)) = DK(Z_\bullet) \,.$

(just removing the first $\overset{\simeq}{\to}$ and its codomain and then fixing and expanding the notation for $U(p)$).

Hope that’s right now.

• CommentRowNumber20.
• CommentAuthorRichard Williamson
• CommentTimeDec 3rd 2018
• (edited Dec 3rd 2018)

My original thinking was that it would be: $U(\Gamma(V_{\bullet}) \times \Gamma(W_{\bullet})) \cong U(\Gamma(V_{\bullet} \times W_{\bullet})) \rightarrow U(\Gamma(V_{\bullet} \otimes W_{\bullet}))$, the last arrow coming from the canonical morphism $V_{\bullet} \times W_{\bullet} \rightarrow V_{\bullet} \otimes W_{\bullet}$.

[Edited twice] But I think this amounts to the same as you described (i.e. there is a commutative square up to isomorphism involving your approach and mine) :-).

• CommentRowNumber21.
• CommentAuthorDmitri Pavlov
• CommentTimeDec 3rd 2018

Thanks, Urs, this resolves the problem.

• CommentRowNumber22.
• CommentAuthorDmitri Pavlov
• CommentTimeDec 3rd 2018
• (edited Dec 3rd 2018)

Next question: the article claims that

the lax monoidal structure γ_{A,B}:Γ(A)⊗Γ(B)→Γ(A⊗B) being induced dually by the Eilenberg-Zilber map

I do not see how the Eilenberg-Zilber map NP⊗NQ→N(P⊗Q) could possibly induce the map γ. Indeed, the (symmetric) Eilenberg-Zilber lax monoidal structure on N induces a (symmetric) oplax monoidal structure on Γ, i.e., a map Γ(A⊗B)→ΓA⊗ΓB, given by the adjoint of A⊗B→NΓA⊗NΓB→N(ΓA⊗ΓB).

However, the map γ goes in the opposite direction, and is induced by the Alexander-Whitney (nonsymmetric) oplax monoidal structure N(P⊗Q)→NP⊗NQ, namely, as the adjoint of the map N(ΓA⊗ΓΒ)→NΓA⊗NΓB→A⊗B. In particular, the map γ is not symmetric, just like the Alexander-Whitney map, and unlike the Eilenberg-Zilber map and the induced oplax map for Γ.

• CommentRowNumber23.
• CommentAuthorUrs
• CommentTimeDec 4th 2018

Thanks. Will need to remind myself. Am on a phone right now. Does the adjective “symmetric” need to be remived?

• CommentRowNumber24.
• CommentAuthorDmitri Pavlov
• CommentTimeDec 4th 2018

Re #23: No, rather in the definition of γ_{A,B} it should say that it is induced by the Alexander-Whitney map, not the Eilenberg-Zilber map. This is also consistent with other constructions of cup product, many of which use the A-W map, but none (as far as I am aware of) the E-Z map.

• CommentRowNumber25.
• CommentAuthorUrs
• CommentTimeDec 4th 2018
• (edited Dec 4th 2018)

Sounds good. Could you be so kind and fix it in the entry? I am still on my phone. Otherwise I can try to look into it later.

• CommentRowNumber26.
• CommentAuthorDmitri Pavlov
• CommentTimeDec 4th 2018

I fixed it, the construction looks good to me now.

• CommentRowNumber27.
• CommentAuthorUrs
• CommentTimeDec 4th 2018

Thanks a million!

• CommentRowNumber28.
• CommentAuthorDmitri Pavlov
• CommentTimeFeb 22nd 2020

Corrected Unicode problems.