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Let $B = (B, \nabla, \eta, \Delta, \epsilon )$ be a bialgebra over a commutative ring $k$. Let $M$ and $N$ be two left $B$-modules. Then the tensor product $M \otimes_k N$ becomes a left $B$-module with multiplication rule given by

$$ B \otimes M \otimes N \xrightarrow{\Delta \otimes 1} B \otimes B \otimes M \otimes N \xrightarrow{1 \otimes \tau \otimes 1} B \otimes M \otimes B \otimes N \xrightarrow{\mu \otimes \mu} M \otimes N$$

where the $\mu$'s are the $B$-module multipication on $M$ and $N$. I am trying to show that $H$-$\mathsf{Mod}$ forms a monoidal category with tensor product $- \otimes_k -$. We have a $k$-linear associator $$ \alpha_{MNP} \colon (M \otimes_k N) \otimes_k P \xrightarrow{\sim} M \otimes_k (N \otimes_k P)$$ and I think this is supposed to be the associator in $H$-$\mathsf{Mod}$, so I need to show it is $B$-linear, i.e. that $$b [m \otimes (n \otimes p)] = \alpha_{MNP} (b[(m \otimes n) \otimes p])$$ but I am not really getting anywhere with it. Does anyone have any advice or at least be able to point me to a souce where this is covered? I am currently using Pareigis's Advanced Algebra, but proving monoidal structure is left as an exercise.

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What you want should follow straight from the coassociativity of $B$, which you have not used yet. In Sweedler notation, both terms are equal to $\sum b_{(1)}m \otimes b_{(2)}n \otimes b_{(3)}p$, although depending on your style, you may want a proof that is more categorical. –  Aaron Sep 14 '12 at 15:56
    
@Aaron I can see by coassociativity we have $$ \sum_{(b)}\sum_{(b_{(2)})} b_{(1)} \otimes \left( (b_{(2)})_{(1)} \otimes (b_{(2)})_{(2)} \right) = \sum_{(b)}\sum_{(b_{(1)})} (b_{(1)})_{(1)} \otimes \left( (b_{(1)})_{(2)} \otimes (b_{(2)}) \right) $$ but why is it true if I slot in $m, n$ and $p$? –  Paul Slevin Sep 14 '12 at 17:28
    
Are we just using the fact that $M \otimes_k N \otimes_k P$ is a $B \otimes_k B \otimes_k B$ module, and the action of the two algebra elements above on $m \otimes n \otimes p$ gives what we want? –  Paul Slevin Sep 14 '12 at 19:09
    
Yes, exactly. Essentially, we're using that $\mu \otimes (\mu \otimes \mu) = \alpha \circ (\mu \otimes \mu) \otimes \mu$ –  Aaron Sep 14 '12 at 19:44
    
@Aaron Please consider converting your comment into an answer, so that this question gets removed from the unanswered tab. If you do so, it is helpful to post it to this chat room to make people aware of it (and attract some upvotes). For further reading upon the issue of too many unanswered questions, see here, here or here. –  Julian Kuelshammer Jun 11 '13 at 21:30
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