Dual of an algebra is a coalgebra Let $A$ be an algebra over the commutative unital ring $k$ that is finitely generated and projective as a $k$-module. Let $A^*= \operatorname{Hom}_k(A,k). $ Then the natural map
$$i: A^* \otimes A^* \to (A \otimes A)^ *$$ is bijective and this allows us to define
$$\Delta:= i^{-1}\circ m_A^*$$
where $m_A: A \otimes A \to A$ is the multiplication map on $A$. I am trying to verify that $\Delta$ is a comultiplication on $A^*$.
If $\{(e_i, f_i)\}_{i=1}^n$ is a dual base for $A^*$, then it is easy to see (please ask if you want me to add details) that
$$\Delta(g) = \sum_{i,j=1}^n g(e_i e_j) f_i \otimes f_j$$
With this, I calculated $$(\Delta \otimes \operatorname{id})\Delta(g) = \sum_{i,j,r,s} g(e_i e_j) f_i(e_r e_s) f_r \otimes f_s \otimes f_j$$ and
$$(\operatorname{id}\otimes \Delta)\Delta(g) = \sum_{i,j,r,s} g(e_i e_j) f_j(e_r e_s) f_i \otimes f_r \otimes f_s$$
These two expressions should be equal (if I did not make a calculation error), but I can't see why. I tried changing the summation indices but could not get there.
 A: May be I'm going to say something silly, but I believe that you will need the structure constants to complete your argument. That is, write
$$
e_ie_j = \sum_{k=1}^n c_{ij}^k e_k,
$$
and
$$
g = \sum_{\ell=1}^n g(e_\ell) f_\ell,
$$
and after a tedious caculation, you will obtain the result.
I don't believe that this is the best way to prove this result. This is a less effort argument: Let
$$
i\colon A^* \otimes A^* \to (A\otimes A)^* \quad \text{and} \quad j\colon A^*\otimes A^*\otimes A^* \to (A\otimes A\otimes A)^*
$$
the natural isomorphisms. Show by a direct calculation that
$$
j^{-1}\circ (\mbox{id}\otimes m_A)^*\circ i = \mbox{id}\otimes \Delta \quad \text{and} \quad j^{-1}\circ (m_A \otimes \mbox{id})^*\circ i = \Delta\otimes \mbox{id}.
$$
We know that
$$
m_A \circ (\mbox{id}\otimes m_A) = m_A \circ (m_A\otimes \mbox{id}),
$$
so by taking the dual and composing with $j^{-1}$ we obtain
$$
j^{-1}\circ(\mbox{id}\otimes m_A)^*\circ m_A^* = j^{-1}\circ(m_A\otimes \mbox{id})^*\circ m_A^*.
$$
Now note that
$$
j^{-1}\circ(\mbox{id}\otimes m_A)^*\circ m_A^* = j^{-1}\circ(\mbox{id}\otimes m_A)^*\circ i\circ i^{-1}\circ  m_A^* = j^{-1}\circ(\mbox{id}\otimes m_A)^*\circ i\circ \Delta
$$
and by using the above relation
$$
j^{-1}\circ(\mbox{id}\otimes m_A)^*\circ m_A^* = (\mbox{id}\otimes\Delta) \circ \Delta.
$$
Similarly
$$
j^{-1}\circ(m_A\otimes \mbox{id})^*\circ m_A^* = (\Delta\otimes \mbox{id}) \circ \Delta,
$$
which completes the proof.
