Let $(C,\Delta)$ be a coalgebra over a field $K$. Denote by $()^{\ast}=Hom(-,K)$. Then we know that $(C^{\ast}, \Delta^{\ast} \circ \rho)$ is an algebra, where $\rho: C^\ast \otimes C^\ast \rightarrow (C \otimes C)^\ast$. Then we get $C^{\ast \ast}$ with a map $\Delta^{\ast \ast} :=(\Delta^{\ast} \circ \rho)^{\ast}: C^{\ast \ast} \rightarrow (C^\ast \otimes C^\ast)^\ast$. We also have a natural injection $\chi$ of $C$ into its double dual $C^{\ast \ast}$.
My quesiton is whether $\Delta^{\ast \ast}$ preserves the coalgebras structure of $C$? In other words, does the following equation hold$$\Delta^{\ast \ast} \circ \chi=\rho' \circ (\chi \otimes \chi) \circ \Delta?$$ Here $\rho':C^{\ast \ast} \otimes C^{\ast \ast} \rightarrow (C^{\ast} \otimes C^{\ast})^\ast$ is a natural injection.
Yes, this is true. The proof is a formal computation just using the definitions.
More generally, let $\Delta : X \to A \otimes B$ be any morphism of vector spaces (the same works in every symmetric monoidal closed category). Let $$e_X : X \to X^{**}, \quad \rho_{A,B} : A^* \otimes B^* \to (A \otimes B)^*$$ be the canonical morphisms defined by $$\begin{align*} e_C(x)(\omega) & :=\omega(x), \\ \rho_{A,B}(\omega \otimes \eta)(a \otimes b) & := \omega(a) \cdot \eta(b). \end{align*}$$We claim that the diagram $$\require{AMScd} \begin{CD} X @>{\large \Delta}>> A \otimes B @>{\large e_A \otimes e_B}>> A^{**} \otimes B^{**} \\ @V{\large e_X}VV @. @VV{\large \rho_{A^*,B^*}}V \\ X^{**} @>>{\large \Delta^{**}}> (A \otimes B)^{**} @>>{\large \rho_{A,B}^*}> (A^* \otimes B^*)^* \\ \end{CD}$$ is commutative. Let us complete it right away as follows:
$$\require{AMScd} \begin{CD} X @>{\large \Delta}>> A \otimes B @>{\large e_A \otimes e_B}>> A^{**} \otimes B^{**} \\ @V{\large e_X}VV @VV{\large e_{A \otimes B}}V @VV{\large \rho_{A^*,B^*}}V \\ X^{**} @>>{\large \Delta^{**}}> (A \otimes B)^{**} @>>{\large \rho_{A,B}^*}> (A^* \otimes B^*)^* \\ \end{CD}$$
The square on the left is commutative because, as is well-known, $e$ is natural. So we only need to look at the square on the right. So we have completely abstracted away $\Delta$. Nice, right?
Both sides are morphisms $f : A \otimes B \to (A^* \otimes B^*)^*$, and these are determined by the values $f(a \otimes b)(\omega \otimes \eta) \in K$ for $a \in A$, $b \in B$, $\omega \in A^*$, $\eta \in B^*$. So let us calculate these in each case. For the computation, also recall that the dual map $u^*$ is defined by $u^*(\omega) := \omega \circ u$.
The top arrow computes to:
$$\begin{align*} \rho_{A^*,B^*}(e_A \otimes e_B)(a \otimes b)(\omega \otimes \eta) &= \rho_{A^*,B^*}(e_A(a) \otimes e_B(b))(\omega \otimes \eta) \qquad (\color{green}{\text{Def. } e_A \otimes e_B})\\ & = e_A(a)(\omega) \cdot e_B(b)(\eta) \qquad (\color{green}{\text{Def. } \rho}) \\ & = \omega(a) \cdot \eta(b) \qquad (\color{green}{\text{Def. } e}) \end{align*}$$
The bottom arrow computes to:
$$\begin{align*} \rho_{A,B}^*(e_{A \otimes B}(a \otimes b))(\omega \otimes \eta) & = (e_{A \otimes B}(a \otimes b) \circ \rho_{A,B})(\omega \otimes \eta) \qquad (\color{green}{\text{Def. } u^*}) \\ & = e_{A \otimes B}(a \otimes b)(\rho_{A,B}(\omega \otimes \eta)) \qquad (\color{green}{\text{Def. } \circ})\\ & = \rho_{A,B}(\omega \otimes \eta)(a \otimes b) \qquad (\color{green}{\text{Def. } e})\\ & = \omega(a) \cdot \eta(b) \qquad (\color{green}{\text{Def. } \rho}) \end{align*}$$
These are the same, and we are done.
At least if we restrict finite-dimensional vector spaces (dualizable objects in the case of symmetric monoidal categories), the square can be interpreted in conceptual terms as follows. The assumption implies that the $\rho$'s are isomorphisms. Then the square can be rewritten as
$$\require{AMScd} \begin{CD} A \otimes B @>{\large e_A \otimes e_B}>> A^{**} \otimes B^{**} \\ @| @AA{\large \rho_{A^*,B^*}^{-1}}A \\ A \otimes B @. (A^* \otimes B^*)^* \\ @| @AA{\large \rho_{A,B}^*}A \\ A \otimes B @>>{\large e_{A \otimes B}}> (A \otimes B)^{**} \end{CD}$$
We see that the double dual functor $(-)^{**}$ can be equipped with the structure of a strong monoidal functor, and that $e$ is a monoidal transformation $\mathrm{id} \longrightarrow (-)^{**}$. (I leave out the diagram for the unit, which is easy.) In fact, $e$ is an isomorphism of strong monoidal functors. In other words, the identification of a finite-dimensional vector space with its double dual is compatible with tensor products.
This is not too surprising, but it is interesting that the square above also exists (and commutes) in the infinite-dimensional setting, and unfortunately I cannot interpret it as a form of monoidal naturality in this case.