$M$ comodule if and only if $N$ and $L$ comodules Let $k$ be a field, $C$ a $k$-coalgebra, and $M$ a left $C$-comodule. Then, for a short exact sequence
$$
0 \rightarrow N \rightarrow M \rightarrow L \rightarrow 0
$$
of vector spaces, we have that $N$ and $L$ are also $C$-comodules. Does the converse hold?
I've tried using the correspondence of left $C$-comodules and rational right $C^\ast$-modules, but it's not clear to me why the map $\varphi: M\rightarrow M \otimes C$ in the definition of rational modules exists in general. It's clear it exists if the short exact sequence splits.
 A: For any (coassociative, counital) coalgebra $C$ over a field $k$, there is a fully faithful exact functor from the category of $C$-comodules to the category of $C^*$-modules.  The essential image of this functor is the full subcategory of rational $C^*$-modules (hence one can identify $C$-comodules with rational $C^*$-modules).
The full subcategory of rational $C^*$-modules is closed under subobjects and quotient objects in the category of $C^*$-modules.  But it is not closed under extensions, generally speaking.
For a counterexample, consider the matrix coalgebra $C=\begin{pmatrix} k & V \\ 0 & k\end{pmatrix}$, where $V$ is a $k$-vector space.  Its dual algebra $C^*$ is the matrix algebra $C^*=\begin{pmatrix} k & V^* \\ 0 & k\end{pmatrix}$.  Then $C$-comodules $E$ are essentially pairs of vector spaces $(E_1,E_2)$ endowed with a linear map $\psi\colon E_1\longrightarrow V\otimes_k E_2$, while $C^*$-modules are pairs of vector spaces $(M_1,M_2)$ endowed with a linear map $g\colon V^*\otimes_k M_1\longrightarrow M_2$.  The functor from $C$-comodules to $C^*$-modules is constructed in the obvious way.
Now take $V$ to be an infinite-dimensional vector space, and let $g\colon V^*\longrightarrow k$ be a linear function not arising from any element of $V$ (i.e., $g$ belongs to $V^*{}^*\setminus V$).  Take $M_1=M_2=k$ to be one-dimensional vector spaces. Then $M=(k,k,g)$ is an irrational $C^*$-module.  Still it is an extension of two rational $C^*$-modules,
$$0\longrightarrow N=(0,k)\longrightarrow M=(k,k)\longrightarrow L=(k,0)\longrightarrow0.$$
