Every element of a comodule lies in a finite-dimensional subcomodule

(Coefficient criterion.) If $h_1,\dots ,h_n\in A$ are linearly independent and $x_1,\dots ,x_n$ lie in a $k$-vector space $X$ with ${\sum }_i{x}_i\otimes h_i=0$ in $X{\otimes }_{k}A$, then $x_1=\cdots =x_n=0$. Indeed, by [F2] the element ${\sum }_i{e}_{(x_i,h_i)}$ of the free module on $X\times A$ is a finite $k$-linear combination of finitely many bilinearity generators; the spans $X_0\subseteq X$ and $A_0\subseteq A$ of the initial $x_i,h_i$ together with all vectors occurring in that finite witness are finite-dimensional and the same combination exhibits ${\sum }_i{x}_i\otimes h_i=0$ in $X_0{\otimes }_k{A}_0$. Extend $h_1,\dots ,h_n$ to a finite basis $v_1,\dots ,v_N$ of $A_0$ with $v_i=h_i$ for $i\le n$, choose a finite basis $u_1,\dots ,u_M$ of $X_0$, and use the universal property [F2] to identify $X_0{\otimes }_k{A}_0$ with the matrices $k^{M\times N}$ by $u_a\otimes v_b\mapsto E_{ab}$; the element ${\sum }_i{x}_i\otimes h_i$ becomes the matrix whose $i$-th column is the coordinate vector of $x_i$ for $i\le n$ and whose remaining columns vanish, so this matrix is zero and every $x_i$ is zero. The same argument shows that $U{\otimes }_{k}H\to X{\otimes }_{k}H$ is injective for any inclusion $U\subseteq X$: take a finite witness of a zero relation, extend a basis of the span of its original first factors in $U$ to a basis of the finite ambient first-factor space, and compare the resulting tensor coordinates. Thus the subspace notation $U{\otimes }_{k}H\subseteq X{\otimes }_{k}H$ is legitimate.

Let $m\in M$ and write $\rho (m)={\sum }_{i=1}^n{m}_i\otimes a_i$ with $a_1,\dots ,a_n$ linearly independent; such a representation exists by deleting redundant terms from a finite tensor expression, and then $m=({\mathrm{id}}_M\otimes \epsilon )\rho (m)={\sum }_i\epsilon (a_i)m_i$ by [F1]. Put $N=\mathrm{span}(m,m_1,\dots ,m_n)$, a finite-dimensional subspace of $M$ containing $m$.

In the situation of step 1.2, let $q\text{\hspace{0.17em}⁣}:M\to M\mathrm{/}N$ be the quotient map. Applying $q\otimes {\mathrm{id}}_A\otimes {\mathrm{id}}_A$ to the coassociativity identity [F1] for $m$ gives ${\sum }_i(q\otimes {\mathrm{id}}_A)\rho (m_i)\otimes a_i={\sum }_i(q{m}_i)\otimes \mathrm{\Delta }(a_i)=0$, because $m_i\in N$; since the $a_i$ are linearly independent, step 1.1 yields $(q\otimes {\mathrm{id}}_A)\rho (m_i)=0$ for every $i$. The kernel of $q\otimes {\mathrm{id}}_A$ is $N{\otimes }_{k}A$: one inclusion is clear, and if a finite sum ${\sum }_j{x}_j\otimes b_j$ with linearly independent $b_j$ lies in that kernel, then ${\sum }_j(q{x}_j)\otimes b_j=0$, so step 1.1 gives $q{x}_j=0$ and $x_j\in N$ for every $j$. Hence $\rho (m_i)\in N{\otimes }_{k}A$ for all $i$, and since $m_i\in N$ one also has $\rho (m)={\sum }_i{m}_i\otimes a_i\in N{\otimes }_{k}A$; therefore $\rho (N)\subseteq N{\otimes }_{k}A$ and $N$ is a finite-dimensional subcomodule containing $m$.

For a finite set $S=\{s_1,\dots ,s_r\}$, apply step 2.1 to each $s_j$ to obtain finite-dimensional subcomodules $N_j\ni s_j$. The sum $N=N_1+\cdots +N_r$ is finite-dimensional and contains $S$, and it is a subcomodule because $\rho (N_j)\subseteq N_j{\otimes }_{k}A$ for each $j$ gives $\rho (N)\subseteq {\sum }_j{N}_j{\otimes }_{k}A\subseteq N{\otimes }_{k}A$.

Consequently every element of $M$ lies in a finite-dimensional subcomodule, and for two such subcomodules $N_1,N_2$ their sum contains both, so the finite-dimensional subcomodules of $M$ form a directed family under inclusion whose union is $M$.