# Alexander-Whitney map gives a coalgebra?

Let $$R$$ be a unital ring with multiplication $$\mu\colon R\otimes R \rightarrow R$$. Consider the category $$\mathcal{Ch}(R-\text{mod})$$ of chain complexes of $$R$$-modules. This category becomes a monoidal category with the tensor product of chain complexes as monoidal product and the chain complex $$R[0]$$ (concentrated in degree $$0$$ on $$R$$) as the monoidal unit. This monoidal category is braided via $$x\otimes y\mapsto(-1)^{\vert x\vert \vert y \vert}y\otimes x$$.

Let $$X$$ be a topological space. Consider $$S_\ast(X):=S_\ast(X,R)$$, its singular chain complex with coefficients in $$R$$. This is an object in $$\mathcal{Ch}(R-\text{mod})$$. There is a special morphism in $$\mathcal{Ch}(R-\text{mod})$$, aka a chain map $$AW(X)\colon S_\ast(X)\rightarrow S_\ast(X)\otimes S_\ast(X).$$ This map is sometimes called Alexander-Whitney diagonal map and defined on a chain $$c\in S_n(X)$$ as $$AW(X)_n(c):=\sum_{p+q=n}F^p(c)\otimes R^q(c)$$, where $$F^p(c)=c\circ \iota$$ and $$R^q(c)=c\circ \tilde{\iota}$$ for the inclusions $$\iota\colon \Delta^p\rightarrow \Delta^n, (t_0,\ldots, t_p)\mapsto (t_0,t_1,\ldots,t_p,0,\ldots,0)$$ and $$\tilde{\iota}\colon \Delta^q\rightarrow \Delta^n, (t_0,\ldots, t_q)\mapsto (0,\ldots,0,t_0,t_1,\ldots,t_q)$$.

Define the chain map $$\epsilon(X)\colon S_\ast(X)\rightarrow R[0]$$ by letting $$\epsilon(X)_{i}=0$$ for all $$i \neq 0$$ and $$\epsilon(X)_{0}$$ be the $$R$$-module map that sends each singular $$0$$-simplex in $$X$$ to $$1_R$$. This makes the triple $$(S_{\ast}(X),AW(X), \epsilon(X))$$ into a coassociative, and counital coalgebra in $$\mathcal{Ch}(R-\text{mod})$$, I think. This coalgebra is not cocommutative.

Is what I have said so far correct? (When) can $$(S_{\ast}(X),AW(X), \epsilon(X))$$ be made into a bialgebra or even a Hopf algebra?

Additionally, the cup product $$\cup \colon S^p(X)\otimes S^q(X)\rightarrow S^{p+q}(X)$$ is defined as $$\alpha\otimes \beta \mapsto \mu \circ (\alpha \otimes \beta) \circ \pi_{p,q} \circ AW(X)_{p+q}$$, where $$\pi_{p,q}\colon \oplus_{k+l=p+q}S_k(X)\otimes S_l(X)\rightarrow S_{p}(X)\otimes S_q(X)$$ is the projection map. This definition somehow looks like a convolution product in a bialgebra. Can this be made precise?

• I'll just toss in a reference to Mariano's comment that graded bialgebras are automatically hopf, since I found this very surprising when i first heard about it: sbseminar.wordpress.com/2011/07/07/… Commented Sep 10, 2023 at 17:16
• Graded connected bialgebras are automatically Hopf. Commented Sep 10, 2023 at 17:56
• Surely, $R$ has to be a field for this to work. It is an extremely important restriction; often one wants to work with coalgebras, but if you need to work with integer coefficients you are out of luck. The issue in your exposition is that what you call the Alexander-Whitney map is actually the composition of the diagonal map with the inverse of the Kunneth isomorphism composed with what is usually called the Alexander-Whitney map. Of course, for Kunneth to be an isomorphism, we need field coefficients. Commented Sep 15, 2023 at 3:21
• Where in my exposition do I need field coefficients? I don't see the problem. The map can be defined as I did. And one checks the axions of a coalgebra (comonoid object in $\mathcal{C}(R-\mathsf{mod})$). Do you maybe mean the Eilenberg-Zilber theorem instead of the Kunneth theorem? I mean, I am talking singular chain complex not homology. Commented Sep 15, 2023 at 5:34