For $R$ a commutative ring and $X$ a topological space the cup product on the singular cohomology $H^{\ast}(X)$ is graded commutative. I have a question about the proof of this claim.
Some definitions first:
- Given topological spaces $X,Y$ and open subsets $A\subseteq X$, $B\subseteq Y$, pick a chain homotopy equivalence $$EZ\colon S_{\ast}(X\times Y,X\times B \cup A \times Y;R)\rightarrow S_{\ast}(X,A;R)\otimes S_{\ast}(Y,B;R).$$ For $\alpha \in S^p(X,A;R)$ and $\beta \in S^q(Y,B;R)$, define the cohomology cross product as $$\alpha \times \beta:=\mu\circ(\alpha\otimes \beta)\circ EZ_{p+q}.$$
The proof I know goes as follows:
We first show that the cohomology cross product is graded commutative. The claim then follows from $\alpha\cup \beta=\Delta^{\ast}(\alpha \times \beta)$, where $\Delta\colon X\rightarrow X\times X$ is the diagonal map and $\alpha \in H^p(X)$, $\beta \in H^q(X)$. Denote by $T\colon X\times Y \rightarrow Y\times X$ the map that swaps the arguments and by $\tau \colon S_{\ast}(Y)\otimes S_{\ast}(X) \rightarrow S_{\ast}(X)\otimes S_{\ast}(Y)$ the usual braiding in the monoidal category of chain complexes (i.e. $x\otimes y \mapsto (-1)^{\vert x \vert \vert y \vert}y\otimes x$). The chain map $\tau \circ EZ \circ T$ agrees with $EZ$ on $S_0$. Therefore it is chain homotopic to $EZ$ and we could have used $\tau \circ EZ \circ T$ to define the cross product. If we use $\tau \circ EZ \circ T$ to define the cohomological cross product (denote this operation by $\times^{\tau}$), we find $\alpha \times^{\tau} \beta= (-1)^{\vert \alpha \vert \vert \beta \vert}\beta\times \alpha.$ The claim follows.
- Where did we use that $R$ is commutative?
- Why does the fact that the chain map $\tau \circ EZ \circ T$ agrees with $EZ$ on $S_0$ imply that it is chain homotopic to $EZ$? I know that if the domain of $EZ$ is a projective resolution, this is just the fundamental lemma of homological algebra. But in this case the domain of $EZ$ is generally not an exact sequence.
Best Answer
Let me first answer your second point: The Eilenberg-Zilber theorem (using the method of acyclic models) states that a natural transformation $\phi: S_*(-)\otimes S_*(-) \to S_*(-\times -) $ is uniquely determined (in the homotopy category) by what it does in degree 0. As you suspected the proof uses the fundamental lemma of homological algebra but to even get there you need acyclic models.
This remains true when working with coefficients $\phi_G : S_*(-;G)\otimes S_*(-;G) \to S_*(-\times -;G)$ (where $G$ is an abelian group, e.g. when $G = (R,+)$ is the additive group associated to a ring $R$.)
For the existence part of the Theorem you will need a ring structure.
Now your proof works if you check that the constructed map $\tau \circ EZ \circ T$ agrees with $EZ$ in degree 0.
For $X=Y=pt$ we identify $S_0(pt,R)=R$. The map $T$ is going to be the identity and $\tau:R\otimes R \to R\otimes R$ is $r \otimes s \mapsto s\otimes r$. The EZ map $\phi_R$ is going to be the ring multiplication. So in total we have to check $$ r\cdot s \overset{?}{=} s \cdot r.$$ EDIT: (after reading comments) The above just 'checks' the required identity for $X=Y=pt$, I'm not saying this is a proof for anything. It is just to highlight that commutativity of $R$ should be necessary for the 'usual' graded commutativity of the cohomology.
Remark: If your ring is graded commutative and $\tau$ is the braiding in the monoidal category of chain complexes over $R$ the corresponding cross-product is still graded commutative... maybe even more general for 'braided rings' (don't know if the concept exists)