The Steenrod algebra $\mathcal{A}^\*$ is the algebra of cohomology operations for the spectrum $\mathrm{H}\mathbb{F}\_2$, i.e. $\mathcal{A}^\*\simeq (\mathrm{H}\mathbb{F}\_2)^\*(\mathrm{H}\mathbb{F}\_2)$.
The dual Steenrod algebra $\mathcal{A}^\vee \_\*$ can be interpreted as the coalgebra of homology cooperations, i.e. $\mathcal{A}^\vee \_\ast\simeq (\mathrm{H}\mathbb{F}\_2) \_\*(\mathrm{H}\mathbb{F}\_2)$ and it coacts on mod 2 homology groups of cell complexes. This is a consequence of the perfect pairing between mod 2 homology and cohomology. Unfortunately, this does happens because $\mathbb{F}\_2$ is a field, which means that we may in trouble if change our coefficients to something else. The situation may get even worse with generalized cohomology theories.


Now let us say we are given a generalized cohomology theory $E$. The ring $E^\ast E$ acts on cohomology groups of any space $X$: we have map $E^\*E\otimes\_{E^\ast} E^\ast(X)\rightarrow E^\ast(X)$, which is natural in $X$. The coaction may be a little tricky: it is not true that $E\_\ast E$ coacts on $E\_\*(X)$. However, upon requiring that $E\_\ast E$ is a flat $E\_\*$-module, we get a coaction as well. So let us say that we are in this situation. In addition to this we have a pairing $E^\*E\otimes\_{E\_\*}E\_\ast E\rightarrow E\_\ast$.


My questions is: do these pairings work well together or there are some tricky things that has to account for? In particular, $E\_\ast E$ has distinct left and right $E\_\*$-module structures, and I fear this may come into play. 



Homology-Cohomology Pairing
Propositions 17.10 and 17.11 of [Switzer's book](http://books.google.co.uk/books/about/Algebraic_Topology_homotopy_and_Homology.html?id=f1xnOzmX2FYC) seem to be examples of things working nicely.


For those who don't have the book to hand: let $E$ be a commutative ring spectrum such that $E\_\ast(E)$ is flat as a right module over $E\_\ast=E\_\ast(\*)$, and let $X$ be any spectrum.


Let $\psi\_X\colon E\_\ast(X)\to E\_\ast(E)\otimes\_{E\_\ast} E\_\ast(X)$ be the coaction map, and let $c\colon E\_\ast(E)\to E\_\ast(E)$ be the conjugation map (which switches the left and right counits). 


Then for any elements $a\in E^\ast(E)$, $y \in E^\ast(X)$ and $u\in E\_\ast (X)$ with $\psi\_X(u) = \sum\_i e\_i\otimes u\_i$, we have
$$ 
a\cdot u = \sum\_i \langle a,ce\_i\rangle u\_i
$$
and
$$
\langle ay,u\rangle = \sum\_i (-1)^{|y||e\_i|}\langle a,e\_i\langle y,u\_i\rangle \rangle,
$$
where $\langle \cdot , \cdot \rangle\colon E^\ast(X)\otimes\_{E\_\ast} E\_\ast(X)\to E\_\ast$ is the Kronecker pairing.


