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Let $H$ be a $K$-Hopf Algebra, $A$ a right $H$-comodule algebra and $l:B\rightarrow A^{coH}$ a ring morphism.

Then $_Ahom(A\otimes_BA,A)\cong\ _Bend(A)$

I don't even know what this means.

$_Bend(A)$ is the set of $K$-linear maps from $A$ to itself to which we add $$ +:\ _Bend(A)\times\ _Bend(A)\rightarrow\ _Bend(A) : (\phi,\varphi) \rightarrow (x \rightarrow \phi(x) + \varphi(x)) \\ \cdot:\ _Bend(A) \times B\rightarrow \ _Bend(A): (\phi, b) \rightarrow (x\rightarrow\phi(x)l(b))$$ i.e. $_Bend(A)$ is the (right) $B$-vector space of K-linear maps from $A$ to itself?

For $_Ahom(A\otimes_BA,A)$ I don't really know what it could be. $A$ is a vector space over $K$ so we have to see $A\otimes_BA$ as a vector space over K for linear applications from $A\otimes_BA$ to $A$ to make any sense. But for me $A\otimes_BA$ doesn't make any sense since $A$ is not a vector space over $B$.

1) is my definition of $_Bend(A)$ correct?

2) How can we give any sense to $_Ahom(A\otimes_BA,A)$

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    It would be helpful for potential answerers to add some more context. For example what course you are taking and what textbook(s) you are using. – 2017-01-18

1 Answers 1

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This is not an answer, but I'm writing it here since it's too long for a comment.

$A$ is an right $H$-comodule algebra. So in particular $A$ is a right $H$-comodule. Hence we can look at the coinvariants, i.e. $A^{\text{co}H}=\left\{a\in A\mid \rho(a)=a\otimes 1\right\}$. Here $\rho$ denotes the coaction on $A$ that makes it a right $H$-comodule. As $A$ is also an algebra is has a product, hence $A^{\text{co}H}$ also has a product. Show that $A^{\text{co}H}$ is an algebra and hence also a ring.

Now consider a ring morphism $l:B\rightarrow A^{\text{co}H}$. Via this map we can view $A$ as a $B$-module. Indeed, $b\cdot a:=l(b)a$ for $b\in B$ and $a\in A$, $l(b)a$ is defined since $A$ is an algebra. Then we can interpret $_B\text{End}(A)$ simply as the endomorphisms of the $B$-module $A$. Likewise $A\otimes_B A$ is the tensor product of the $B$-modules $A$ and $_A\text{Hom}(A\otimes_B A,A)$ are the $A$-linear maps from the $A$-module $A\otimes_B A$ to the $A$-module $A$.

Now I'm not sure whether this is the correct interpretation, some more information or context could be useful. But this seems to make sense. If you can show that $_A\text{Hom}(A\otimes_B A,A)\cong\ _B\text{End}(A)$ holds with this interpretation, it's probably the correct on.

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    I'm having a really hard time juggling with all these structures which are equivalent. (I can't remember seing the tensor product of two modules but I suppose it's the same idea as tensor product between two vector spaces) – 2017-01-18