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So the cup product is not well defined over co-chain groups, but all the books claim it is well defined over co-homology groups. The only thing I am not clear on is invariance under ordering/re-ordering of simplices when we go to the co-homology level. Every book seems to gloss over this, and after doing a few examples, I can't seem to figure out how to get this to work out right. Can someone fill me in?

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How is the cup product "not well defined over cochain groups"? –  Mariano Suárez-Alvarez Mar 14 '11 at 3:23

2 Answers 2

Strictly speaking, the cup product is not commutative, though it is commutative up to sign on the level of cohomology.

There is an abstract way of seeing this: namely, we can use the method of acyclic models. Consider the following two functors from the category of spaces to the category of chain complexes. The first is $X \mapsto C_*(X \times X)$; the second is $X \mapsto C_*(X)\otimes C_*(X)$. Since these are free and acyclic functors on the subset of standard simplices (this means that a) both can be represented as a sum of free abelian groups on sets which are representable functors in $X$, represented by simplices and b) evaluated on a simplex, they lead to acyclic complexes), there is a natural chain equivalence between two $$ C_*(X \times X) \simeq C_*(X) \otimes C_*(X)$$ which itself is unique up to chain homotopy. This is the acyclic model theorem (as in Spanier, for instance).

Now the category of chain complexes over a commutative ring is not just an abelian category; it is a monoidal category. We can tensor two chain complexes and get a new chain complex. Moreover, it is a symmetric monoidal category because there is an isomorphism $A_* \otimes B_* \simeq B_* \otimes A_*$ for chain complexes $A, B$.

Thus if we are given one chain equivalence (fixed throughout the following) $C_*(X \times X) \simeq C_*(X) \otimes C_*(X)$, we get another by composing it with the swap map on the latter. These are both natural in $X$ and so, by the uniqueness (up to chain homotopy) in the acyclic model theorem, we find that the two maps $$C_*(X \times X) \rightrightarrows C_*(X) \otimes C_*(X)$$ are naturally chain homotopic.

But the dual of this means that the two maps $$C^*(X) \otimes C^*(X) \rightrightarrows C^*(X \times X)$$ are naturally chain homotopic. On the level of cohomology, the two maps $H^*(X) \otimes H^*(X) \rightrightarrows H^*(X \times X)$$ are thus equal.

Now the map $C_*(X \times X) \to C_*(X) \otimes C_*(X)$ is precisely the homology cross product, and its dual is the cohomology cross product. So we have seen that if we consider the cross-product map $H^*(X) \otimes H^*(X) \to H^*(X \times X)$, it is invariant under switching the two factors.

You might now object that I said that the cross-product (and thus the cup-product, which is obtained from the cross product by pulling back by the diagonal) is skew-commutative, not commutative. This comes from a feature of how the tensor product of complexes is actually defined: as a result, when you define the swap morphism, you have to introduce a sign (for it to be a chain map).

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Akhil, thank you for the time, but i'm not sure you have answered my question. Very simply, for example a simplicial complex with maximal simplex [a,b,c] how is $(\alpha \cup \beta)([a,b,c]) = (\alpha \cup \beta)([c,a,b])$? Similarly if by $[a,b,c]$ I mean the cohomology class to which $[a,b,c]$ belongs. I'm asking about this in the general sense, but, I am looking for a specific meaningful computation to explain it. –  rhl Mar 13 '11 at 15:39
    
@rhl: Dear rhl, for instance, if you mean that $[a,b,c]$ is the standard 3-simplex, then the cohomology groups are all zero. (In general, it will not be true that $(\alpha \cup \beta)([a, b, c]) = \pm (\alpha \cup \beta)([c, a, b])$; the point is that the difference will differ by a coboundary, which can be constructed explicitly via the acyclic model business; of course, in the case of the standard 3-simplex both will already be coboundaries.) You can, however, get explicit examples in simplicial cohomology by using the torus. –  Akhil Mathew Mar 13 '11 at 15:45

right, ok. I asked about if cup product is well defined, here is why I am confused.

You wrote:


Strictly speaking, the cup product is not commutative, though it is commutative up to sign on the level of cohomology.

There is an abstract way of seeing this...


The cup product being commutative should have nothing to do with it being well defined.

Commutativity of the the product should be talking about if $a \smile b = b \smile a$. This seems irrelevant for a conversation about the cup product being well defined.

So again, what I want to know if why is the cup product well defined. What I am asking is: is it the case that $(\alpha \smile \beta)(a)$ is the same as $(\alpha \smile \beta)(b)$ whenever $a$ is equivalent to $b$. Where equivalent is as as co-homology classes.

Now in your second comment, you have said that that it is not the case that that the cup product is well defined, and it's again unclear if you mean over co-chain groups or co-homology groups. I

n either case, what I want to know is: When is the cup product well defined? Does your definition of well defined differ from the classical one? I do not at this point want to underst

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Dear rhl, I apologize for having misunderstood your question (I interpreted the discussion of "reordering" as an indication that you were confused about the commutativity). The cup product is defined as a natural transformation of functors $H^p(X, M) \otimes_R H^q(X, N) \to H^{p+q}(X, M \otimes_R N)$ whenever $M, N$ are modules over a commutative ring $R$. This follows because there is (as I discuss in my answer) a natural transformation of chain complexes $C_*(X, M) \otimes_R C_*(X, N) \to C_*(X \times X, M \otimes_R N)$ (which is actually a chain equivalence)... –  Akhil Mathew Mar 14 '11 at 1:00
    
...and if we dualize this, we get a map on the cochain complexes in the inverse direction. Since cohomology is functorial on the category of cochain complexes, we get the natural transformation that is the cup product. (Also we are using the fact that whenever $Q, Q'$ are complexes, there is a natural morphism $H^p(Q) \otimes H^q(Q') \to H^{p+q}(Q \otimes Q')$; this is easy to check.) –  Akhil Mathew Mar 14 '11 at 1:02

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