# Dual of a finite dimensional algebra is a coalgebra (ex. from Sweedler)

Let $(A, M, u)$ be a finite dimensional algebra where $M: A\otimes A \rightarrow A$ denotes multiplication and $u: k \rightarrow A$ denotes unit.

I want to prove that $(A^*, \Delta, \varepsilon)$ is a colagebra where $\Delta: A^*\rightarrow A^* \otimes A^*$ is a composition: $$A^* \overset{M^*}{\rightarrow}(A\otimes A)^* \overset{\rho^{-1}}{\rightarrow}A^*\otimes A^*$$

And $\rho: V^*\otimes W^* \rightarrow (V\otimes W)^*$ is given by $<\rho(v^*, w^*), v\otimes w>=<v^*, v><w^*,w>$.

I have proven that $\rho$ is injective and since $A$ is finite dimensional $\rho$ is also bijective and we can take the inverse $\rho^{-1}$.

But I have problems understanding how does $\Delta$ work.

By definition we have $<M^*(c^*), a\otimes b>=<c^*, M(a\otimes b)>=c^*(ab)$. But I can't understand what is $\rho^{-1}(M^*(c^*))$, or in other words which element of $A^*\otimes A^*$ can act like $M^*(c^*)$ via $\rho$?

P.S. Please correct me if I have grammar mistakes. Thanks!

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One little comment: when doing those angled brackets , \langle and \rangle look nicer than < and >. –  Alex Petzke Jun 18 '12 at 12:59
This is the kind of question that makes me wish we had xymatrix on M.SE. –  darij grinberg Mar 10 '13 at 21:46

Given $M^*c^*=:d^* \in (A \otimes A)^*$, $\rho^{-1}(d^*)=d_1^* \otimes d_2^* \in A^* \otimes A^*$, where $d_1^*(a)=d^*(a \otimes 1)$ and $d_2^*(a)=d^*(1 \otimes a)$. Notice that $\rho (d_1^* \otimes d_2^*)=d^*$.
I am trying to check this $\rho (d_1^* \otimes d_2^*)=d^*$: $$<\rho (d_1^* \otimes d_2^*), a\otimes b>=<d_1^*, a><d_2^*,b>=d^*(a\otimes 1)\cdot d^*(1\otimes b)=c^*(a)\cdot c^*(b) \\ <d^*, a\otimes b>=c^*(ab)$$ and so $c^*(ab)$ must be equal to $c^*(a)\cdot c^*(b)$ what, I beleive, is not always right. Or I misunderstood something? –  grozhd Apr 9 '12 at 10:19