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!!
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 $b$.