Proof that the dual of a coalgebra is an algebra via commutative diagrams

We know that the algebra and coalgebra axioms are given via following commutative diagrams (algebra $A$ and coalgebra $C$ are over a field $\mathbb{K}$):

I am now trying to show that the dual of a coalgebra $C^*$ has an algebra structure given by $$\mu_{C^*} := \Delta^* \circ \lambda \quad \text{and} \quad \eta_{C^*} := \varepsilon^*,$$ where \begin{align*}\lambda : C^* \otimes C^* &\longrightarrow (C \otimes C)^* \\ p \otimes q &\longmapsto [x \otimes y \longmapsto p(x)q(y)].\end{align*}

My question is the following: I can verify the axioms by a painful direct calculation, but the book I am using simply says that the result follows from considering the diagrams above. How do I actually consider these diagrams? From my understanding, they simply encode the axioms in a neat way (e.g. we do not have to write out explicitly what the coproduct does to elements of $C$), but I do not see how from simply looking at the diagrams we can deduce the result in question.

• Note that the diagrams giving the axioms for an algebra are almost the same as those for a coalgebra. The difference is the direction of the arrows. Taking duals is a contravariant functor (so it reverses the arrows). – David Hill Feb 3 '16 at 23:01
• @DavidHill I am unfamiliar with those concepts, so is there, perhaps, some other way to explain this? – Vitaly B Feb 3 '16 at 23:08
• @DavidHill One also needs to know that dualisation is compatible with the tensor product... – Zhen Lin Feb 4 '16 at 7:28