dual algebra of a coalgebra Given a coalgebra $A$ over a field $F$ (for example, $H_*(X:F)$, i.e. the homology of a space $X$ equipped with $\Delta_*$, where $\Delta: X\to X\times X$, $x\to (x,x)$) how to obtain its dual $A^*$ as an algebra over $F$ (for example, the dual of $H_*(X;F)$ is the cohomology ring $H^*(X;F)$ equipped with cup product)? 
For deriving the product structure of $A^*$ from the coproduct of $A$, is there any standard algorithm to follow?
 A: If $A$ is any $F$-vector space and $(A,\Delta,\epsilon)$  is a coalgebra, then its dual vector space $A^*=Hom_{F}(V,F)$ i.e. the v.s. of linear maps from $V$ to the field $F$, becomes an $F$-algebra in the following way: 
First let me introduce some notation: If $ \ U$, $V$ are any two vector spaces over $F$, $u,v\in A$, $f,g\in A^*$, then the map 
$$
\lambda:U^*\otimes V^*\rightarrow(U\otimes V)^* \\
f\otimes g\mapsto\lambda(f\otimes g)
$$
defined by
$$\lambda(f\otimes g)(u\otimes v)=f(u)g(v)\in F$$
is a monomorphism of vector spaces (i.e. a linear map). Furthermore, if at least one out of $U,V$ is finite dimensional, then $\lambda$ is an isomorphism of vector spaces. We will also denote with $\psi$, the natural v.s. isomorphism 
$F\cong F^*$ defined by $\psi(\kappa)=f$ with $f(1)=\kappa$ and $\psi^{-1}(f)=f(1)$. 
Under these notations, the multiplication of $A^*$ is given by: $\star=\Delta^{\ast}\!\!\!\circ\!\!\lambda$:
$$ 
\begin{array}{ccl}
  A^{\ast} \otimes A^{\ast} \stackrel{\lambda}{\longrightarrow} & (A \otimes
  A)^{\ast}
  \stackrel{\Delta^{\ast}}{\longrightarrow} & A^{\ast} \\
  f \otimes g \;\;\;\;\;\longmapsto & \lambda(f \otimes g) \;\longmapsto & (\Delta^{\ast}\!\!\!\circ\!\!\lambda)
  (f \otimes g)\equiv f \star g \\
\end{array}
$$
and the unity map, embedding the field $F$ inside the center of $A^*$, will be  given by:  $\epsilon^{\ast}\!\circ\!\psi$
$$\begin{array}{ccl}
  F \stackrel{\psi}{\longrightarrow} & F^{\ast}\stackrel{\epsilon^{\ast}}{\longrightarrow} &A^{\ast}  \\
  \kappa \longmapsto & f \longmapsto &  (\epsilon^{\ast}\!\circ\!\psi)(\kappa)=\epsilon^{\ast}(f)=\kappa\,\epsilon \\
\end{array}$$
i.e. the map $(\epsilon^{\ast}\!\circ\!\psi)(1)= \epsilon\in A^{\ast}$ will be the identity element of the algebra $A^*$.
Now we can straightforwardly check the desired properties (associativity and unity) to conclude that $(A^*,\Delta^{\ast}\!\!\!\circ\!\!\lambda,\epsilon^{\ast}\!\circ\!\psi)$ is an algebra. 
