Confusion about dual of comodule I am confused about something.  Please help! :)   All objects are vector spaces over a fixed field $k$.
Let $C$ be a coalgebra with comultiplication $m:C\to C\otimes C$.  Let $M$ be a left comodule over $C$, so we have a map $a:M\to C\otimes M$ such that 
$$(\text{id}_{C}\otimes a)\circ a=(m\otimes\text{id}_M)\circ a$$
Let $M^*$ denote the dual vector space of $M$ over $k$.  Define
$b:M^*\to C\otimes M^*$ by $b(\varphi)(v)=(\text{id}_C\otimes \varphi)(a(v))$ where $\varphi\in M^*$ and $v\in M$.  I have computed several times that $b$ defines a left $C$-comodule structure on $M^*$.  Here is my work; let $\varphi\in M^*$, $v\in M$:
$$\begin{eqnarray*} 
(\text{id}_C\otimes b)\circ b(\varphi)(v)& = &(\text{id}_C\otimes b(\varphi))\circ a(v)\\
                           & = &(\text{id}_C\otimes \text{id}_C\otimes \varphi)\circ(\text{id}_C\otimes a)\circ a(v)\\
                           & = &(\text{id}_C\otimes \text{id}_C\otimes \varphi)\circ(m\otimes \text{id}_{M^*})\circ a(V)\\
                           & = &(m\otimes \text{id}_{M^*})\circ (\text{id}_C\otimes\varphi)\circ a(v)\\
                           & = & (m\otimes \text{id}_{M^*})\circ b(\varphi)(v)
\end{eqnarray*}$$
I haven't been able to find a mistake in this computation... however something tells me that we need more structure (i.e. an antipode or something like that) on $C$ in order to get dual comodules.  Where is the error?
Thank you!!
 A: This could not be a comment, so it became an answer.

I will omit the $\otimes$ between objects, objects concatenations instead.
For (natural) isomorphisms i will use an equality sign. I will write a $\Delta:C\to CC$ instead of the $m$.

Short answer: The evaluation morphism $E:M^*\otimes M\to k$, $E(\phi\otimes v)=\phi(v)$, also plays a role in the computations. My feeling is that also trying to insert it explicitly in the computations breaks the chain of equalities.
This is formulated as a "feeling"... To have something explicitly in the discussion, i need to pass to the...

Long(er) answer: 
We have to show an equality of maps
$$
\begin{aligned}
&M^*\to CM^* \to C(CM^*)=CCM^* 
&&\phi\to b\phi \to(\operatorname{id_C}\otimes b)b(\phi)
\ ,\\
&M^*\to CM^* \to (CC)M^*=CCM^*
&&\phi\to b\phi \to(\Delta\otimes\operatorname{id_M})b(\phi)
\ .
\end{aligned}
$$ 
To show it we "test it" on some element $\phi\in M^*$ as above. The two maps, evaluated in $\phi$ give rise to two elements in $CCM^*$. 
We may and do identify this object with $\operatorname {Hom}(M,CC)$. 
To show the new equality of objects $M\to CC$, we "test it" again against elements $
v\in M$. So we have to show an equality in $CC$. Let us isolate now the equalities at the level of maps. The two maps that have to be equal are mapping an element $\phi\otimes v\in M^*M$ to an element in $CC$,
$$
\begin{aligned}
M^*M\to CM^*M 
\overset
{\operatorname {id}_C\otimes b\otimes \operatorname {id}_M}
\longrightarrow 
&C(CM^*)M=CCM^*M\to CCk=CC
\ ,\qquad(1)
\\
M^*M\to CM^*M
\overset
{\Delta\otimes \operatorname {id}_{M^*}\otimes \operatorname {id}_M}
\longrightarrow 
&(CC)M^*M=CCM^*M\to CCk=CC
\ .
\qquad(2)
\end{aligned}
$$
The above maps do not involve any $a$.
The definition of $b$ is so that the following compositions are equal, and we pass from the "$b$-world" to the "$a$-world":
$$
\begin{aligned}
&M^*M\overset{b\otimes\operatorname {id}_M }\longrightarrow CM^*M \overset {\color{red}E}\to C
\ ,\\
&M^*M = MM^*\overset{a \otimes\operatorname {id}_{M^*}}\longrightarrow CMM^*
\overset {\color{red}E}\to C
\ ,\text{ or also}\\
&M^*M\overset{\operatorname {id}_{M^*}\otimes a}\longrightarrow M^*(CM)= CM^*M
\overset {\color{red}E}\to C
\ .
\end{aligned}
$$ 
Above, i have invented a letter ${\color{red}E}$ for the evaluation map $M^*M\to k$, and the corresponding map $CM^*M\to Ck=C$ was also denoted by ${\color{red}E}$.
In order to pass from the $b$-world to the $a$-world it is necessary to have the further composition with $CM^*M\overset{\color{red}E}\to C$ (corresponding to the identity in the $C$-tensor factor).
Now i usually draw path diagrams for all compositions of morphisms as in $(1)$ and $(2)$ above. Here it is not possible. But i will write the maps (horizontally, although vertically is a better shape), please insert the corresponding paths. We start with $(1)$. There are two occurrences of $b$, only the second one is available with a factorization through $CM^*M\overset{\color{red}E}\to C$, so we work at this level first, and consider the following compositions...
$$
\begin{aligned}
M^*M
\overset
{b\otimes \operatorname {id}_M}
\longrightarrow 
C\color{green}{M^*M} 
\overset
{\operatorname {id}_C\otimes b\otimes \operatorname {id}_M}
\longrightarrow 
&C\color{green}{(CM^*)M}=C\color{green}{CM^*M}\to C\color{green}{Ck}=C\color{green}{C}
\ ,\qquad(1)
\\
&\qquad\text{ which is}
\\
M^*M
\overset
{b\otimes \operatorname {id}_M}
\longrightarrow 
C\color{green}{M^*M} 
\overset
{\operatorname {id}_C\otimes \operatorname {id}_{M^*}\otimes a}
\longrightarrow 
&C\color{green}{M^*(CM)}=C\color{green}{CM^*M}\to C\color{green}{Ck}=C\color{green}{C}
\\
&\qquad\text{ but why should the above factorize as a map}
\\
M^*M
\overset
{b\otimes \operatorname {id}_M}
\longrightarrow 
C\color{green}{M^*M} 
\overset
{\color{red}E}
\longrightarrow 
&C\color{green}{k}=C\ ?
\\[8mm]
&\qquad\text{ For the other map}
\\
M^*M
\overset
{b\otimes \operatorname {id}_M}
\longrightarrow 
C\color{green}{M^*M}
\overset
{\Delta\otimes \operatorname {id}_{M^*}\otimes \operatorname {id}_M}
\longrightarrow 
&(C\color{green}{C)M^*M}=C\color{green}{CM^*M}\to C\color{green}{Ck}
=C\color{green}{C}
\ ,
\qquad(2)
\\
&\qquad\text{ why should the above factorize as a map}
\\
M^*M
\overset
{b\otimes \operatorname {id}_M}
\longrightarrow 
C\color{green}{M^*M} 
\overset
{\color{red}E}
\longrightarrow 
&C\color{green}{k}=C\ ?
\end{aligned}
$$
If we do not have the factorizations i have no argument for applying the definition of $b$. 
