Coradical of coalgebra

Question

I'm reading a book about Hopf algebras and one definition is confusing for me.

If $C$ is a coalgebra then we can define a coradical of $C$ as a sum of all simple subcoalgebras of $C$.

Is there sum direct ? If no, is there simple counterexample ?

Answer 1

Yes, the sum is direct; see Theorem 8 below. The following argument is an elementarization of an argument from Hans-Jürgen Schneider's class on Hopf algebras (see here for the notes in German):

We fix a field $\mathbf{k}$, once and for all. All coalgebras in the following are $\mathbf{k}$-coalgebras, and are assumed to be coassociative and counital. All tensor products are over $\mathbf{k}$. We shall always use the notations $\Delta$ and $\epsilon$ for the comultiplication and the counit of a coalgebra. The word "finite-dimensional" will always mean "finite-dimensional as a $\mathbf{k}$-vector space".

Proposition 1. Let $D$ and $E$ be two subcoalgebras of a coalgebra $C$. Then, $D\cap E$ is a subcoalgebra of $C$ as well.

Proof of Proposition 1. We have $\Delta\left( D\right) \subseteq D\otimes D$ (since $D$ is a subcoalgebra of $C$) and $\Delta\left( E\right) \subseteq E\otimes E$ (similarly).

Now, recall the following fact from linear algebra: If $V$ and $W$ are two $\mathbf{k}$-vector spaces, if $P$ and $P^{\prime}$ are two vector subspaces of $V$, and if $Q$ and $Q^{\prime}$ are two vector subspaces of $W$, then $\left( P\otimes Q\right) \cap\left( P^{\prime}\otimes Q^{\prime}\right) =\left( P\cap P^{\prime}\right) \otimes\left( Q\cap Q^{\prime}\right) $ (where we regard $P\otimes Q$, $P^{\prime}\otimes Q^{\prime}$ and $\left( P\cap P^{\prime}\right) \otimes\left( Q\cap Q^{\prime}\right) $ as vector subspaces of $V\otimes W$). Applying this to $V=C$, $W=C$, $P=D$, $Q=D$, $P^{\prime}=E$ and $Q^{\prime}=E$, we obtain $\left( D\otimes D\right) \cap\left( E\otimes E\right) =\left( D\cap E\right) \otimes\left( D\cap E\right) $.

Now, let $x\in D\cap E$. Then, $x\in D\cap E\subseteq D$ and therefore $\Delta\left( x\right) \in\Delta\left( D\right) \subseteq D\otimes D$. Combining this with the relation $\Delta\left( x\right) \in E\otimes E$ (which is obtained similarly), we obtain $\Delta\left( x\right) \in\left( D\otimes D\right) \cap\left( E\otimes E\right) =\left( D\cap E\right) \otimes\left( D\cap E\right) $.

Now, let us forget that we fixed $x$. We thus have shown that $\Delta\left( x\right) \in\left( D\cap E\right) \otimes\left( D\cap E\right) $ for every $x\in D\cap E$. In other words, $\Delta\left( D\cap E\right) \subseteq\left( D\cap E\right) \otimes\left( D\cap E\right) $. In other words, $D\cap E$ is a subcoalgebra of $C$. This proves Proposition 1.

Remark. Proposition 1 breaks if we drop the assumption that $\mathbf{k}$ be a field. When $\mathbf{k}$ is a commutative ring, already the notion of a "subcoalgebra" becomes tricky (because for $D\subseteq C$, the canonical map $D\otimes D\rightarrow C\otimes C$ is not always injective). Maybe some restrictions on $D$ and $E$ and $D\cap E$ (flatness? direct addends?) can be used to salvage Proposition 1, though.

Proposition 2. Let $C$ be a coalgebra. Let $\left( C_{i}\right) _{i\in I}$ be a family of subcoalgebras of $C$. Then, $\sum_{i\in I}C_{i}$ is a subcoalgebra of $C$.

Proof of Proposition 2. This follows from

$\Delta\left( \sum_{i\in I}C_{i}\right) =\sum_{i\in I}\Delta\left( C_{i}\right) =\sum_{j\in I}\underbrace{\Delta\left( C_{j}\right) }_{\substack{\subseteq C_{j}\otimes C_{j}\\\text{(since }C_{j}\text{ is a}\\\text{subcoalgebra of }C\text{)}}}$

$\subseteq\sum_{j\in I}\underbrace{C_{j}}_{\subseteq\sum_{i\in I}C_{i}} \otimes\underbrace{C_{j}}_{\subseteq\sum_{i\in I}C_{i}}\subseteq\sum_{j\in I}\left( \sum_{i\in I}C_{i}\right) \otimes\left( \sum_{i\in I}C_{i}\right) $

$\subseteq\left( \sum_{i\in I}C_{i}\right) \otimes\left( \sum_{i\in I} C_{i}\right) $.

Definition. A coalgebra $C$ is said to be simple if and only if it is nonzero and the only subcoalgebras of $C$ are $0$ and $C$.

Lemma 3. Let $C$ be a coalgebra. Let $E$ be a simple subcoalgebra of $C$. Let $D$ be a subcoalgebra of $C$. Let $x\in E$ be such that $x\in D$ and $x\neq 0$. Then, $E\subseteq D$.

Proof of Lemma 3. We have $x\in D\cap E$ (since $x\in D$ and $x\in E$). Hence, $D\cap E$ contains the nonzero element $x$. Consequently, $D\cap E\neq0$.

Also, $D\cap E$ is a subcoalgebra of $C$ (by Proposition 1), and thus a subcoalgebra of $E$. Since the only subcoalgebras of $E$ are $0$ and $E$ (because $E$ is simple), this shows that $D\cap E$ is either $0$ or $E$. Since $D\cap E\neq0$, we thus obtain $D\cap E=E$, so that $E=D\cap E\subseteq D$. Lemma 3 is proven.

Lemma 4. Let $C$ be a coalgebra. Let $C_{1}$ and $C_{2}$ be two subcoalgebras of $C$. Let $E$ be a simple subcoalgebra of $C$ such that $E\subseteq C_{1}+C_{2}$. Then, either $E\subseteq C_{1}$ or $E\subseteq C_{2}$ (or both).

Proof of Lemma 4. Assume the contrary. Thus, neither $E\subseteq C_{1}$ nor $E\subseteq C_{2}$.

We don't have $E\subseteq C_{1}$. Thus, $E\cap C_{1}\neq E$. But $E\cap C_{1}$ is a subcoalgebra of $C$ (by Proposition 1, applied to $E$ and $C_{1}$ instead of $D$ and $E$), and thus a subcoalgebra of $E$. Since the only subcoalgebras of $E$ are $0$ and $E$ (because $E$ is simple), this shows that $E\cap C_{1}$ is either $0$ or $E$. Since $E\cap C_{1}\neq E$, this shows that $E\cap C_{1}=0$. Thus, the sum $E+C_{1}$ is a direct sum. Thus, there exists a $\mathbf{k}$-linear projection $\pi:E+C_{1}\rightarrow E$ which annihilates $C_{1}$. Consider this $\pi$. Hence, there exists a $\mathbf{k}$-linear map $g:C\rightarrow\mathbf{k}$ such that $g\mid_{E}=\epsilon\mid_{E}$ and $g\mid_{C_{1}}=0$. (Indeed, we can construct such a map $g$ by extending the map $\epsilon\circ\pi : E + C_1 \to \mathbf{k}$ to the whole $C$.) Consider this $g$. From $g\mid_{C_{1}}=0$, we obtain $g\left( C_{1}\right) =0$.

We have

(1) $\left( \operatorname*{id}\otimes g\right) \left( \alpha\right) =\left( \operatorname*{id}\otimes\epsilon\right) \left( \alpha\right) $ for every $\alpha\in E\otimes E$.

(Indeed, it is easy to verify (1) for every pure tensor $\alpha =e_{1}\otimes e_{2}\in E\otimes E$, using the fact that $g\mid_{E} =\epsilon\mid_{E}$.)

Let $x\in E$. Then, $x\in E\subseteq C_{1}+C_{2}$. Thus, there exist $y\in C_{1}$ and $z\in C_{2}$ such that $x=y+z$. Consider these $y$ and $z$. Since $y\in C_{1}$, we have $\Delta\left( y\right) \in C_{1}\otimes C_{1}$ (since $C_{1}$ is a subcoalgebra of $C$). Similarly, $\Delta\left( z\right) \in C_{2}\otimes C_{2}$. Now, $x\in E$ and thus $\Delta\left( x\right) \in E\otimes E$ (since $E$ is a subcoalgebra of $C$), so that

$\left( \operatorname*{id}\otimes g\right) \left( \Delta\left( x\right) \right) =\left( \operatorname*{id}\otimes\epsilon\right) \left( \Delta\left( x\right) \right) $ (by (1), applied to $\alpha =\Delta\left( x\right) $)

$=x\otimes1$ (by the axioms of a coalgebra).

Hence,

$x\otimes1=\left( \operatorname*{id}\otimes g\right) \left( \Delta\left( \underbrace{x}_{=y+z}\right) \right) =\left( \operatorname*{id}\otimes g\right) \left( \Delta\left( y+z\right) \right) $

$=\left( \operatorname*{id}\otimes g\right) \left( \underbrace{\Delta \left( y\right) }_{\in C_{1}\otimes C_{1}}\right) +\left( \operatorname*{id}\otimes g\right) \left( \underbrace{\Delta\left( z\right) }_{\in C_{2}\otimes C_{2}}\right) $

$\in\underbrace{\left( \operatorname*{id}\otimes g\right) \left( C_{1}\otimes C_{1}\right) }_{\substack{\subseteq C_{1}\otimes g\left( C_{1}\right) =0\\\text{(since }g\left( C_{1}\right) =0\text{)} }}+\underbrace{\left( \operatorname*{id}\otimes g\right) \left( C_{2}\otimes C_{2}\right) }_{\subseteq C_{2}\otimes g\left( C_{2}\right) }$

$\subseteq0+C_{2}\otimes g\left( C_{2}\right) =C_{2}\otimes g\left( C_{2}\right) $.

Applying the canonical map $C\otimes\mathbf{k}\rightarrow C,\ c\otimes \lambda\mapsto\lambda c$ to both sides of this relation, we obtain

$1x\in g\left( C_{2}\right) C_{2}\subseteq C_{2}$,

so that $x=1x\in C_{2}$.

Now, let us forget that we fixed $x$. We thus have proven that $x\in C_{2}$ for every $x\in E$. In other words, $E \subseteq C_{2}$. This contradicts the fact that we do not have $E\subseteq C_{2}$; this contradiction concludes the proof of Lemma 4.

Lemma 5. Let $V$ be a $\mathbf{k}$-vector space. Let $\left( V_{i}\right) _{i\in I}$ be a family of vector subspaces of $V$. Let $E$ be a finite-dimensional vector subspace of $V$ such that $E\subseteq\sum_{i\in I}V_{i}$. Then, there exists a finite subset $J$ of $I$ such that $E\subseteq\sum_{i\in J}V_{i}$.

Proof of Lemma 5. Let $e\in E$ be arbitrary. Then, $e\in E\subseteq \sum_{i\in I}V_{i}$. Hence, there exists a family $\left( v_{i}\right) _{i\in I}\in\prod_{i\in I}V_{i}$ such that all but finitely many $i\in I$ satisfy $v_{i}=0$ and such that $e=\sum_{i\in I}v_{i}$. If we throw out the zero entries from this family $\left( v_{i}\right) _{i\in I}$, we obtain a subfamily $\left( v_{i}\right) _{i\in K}\in\prod_{i\in K}V_{i}$ for some finite subset $K$ of $I$ which still satisfies $e=\sum_{i\in K}v_{i}$. Hence, for this subset $K$, we have $e=\sum_{i\in K}\underbrace{v_{i}}_{\in V_{i}}\in\sum_{i\in K}V_{i}$.

Now, let us forget that we fixed $e$. We thus have shown that if $e\in E$, then

(2) there exists a finite subset $K$ of $I$ such that $e\in\sum_{i\in K}V_{i}$.

Now, the $\mathbf{k}$-vector space $E$ is finite-dimensional and thus has a finite basis $\left( e_{1},e_{2},\ldots,e_{k}\right) $. For every $p\in\left\{ 1,2,\ldots,k\right\} $, we can apply (2) to $e=e_{p}$ (since $e_{p}\in E$), and thus conclude that there exists a finite subset $K$ of $I$ such that $e_{p}\in\sum_{i\in K}V_{i}$. We denote this subset $K$ by $K_{p}$; thus, $e_{p}\in\sum_{i\in K_{p}}V_{i}$. We set $J=K_{1}\cup K_{2} \cup\cdots\cup K_{k}$. Then, $J$ is a finite set (since it is defined as a union of finitely many finite sets). For every $p\in\left\{ 1,2,\ldots ,k\right\} $, we have $e_{p}\in\sum_{i\in K_{p}}V_{i}\subseteq\sum_{i\in J}V_i$ (since $K_{p}\subseteq K_{1}\cup K_{2}\cup\cdots\cup K_{k}=J$). Hence, the basis $\left( e_{1},e_{2},\ldots,e_{k}\right) $ of $E$ is contained in the vector subspace $\sum_{i\in J}V$. As a consequence, the whole subspace $E$ is contained in the vector subspace $\sum_{i\in J}V$. Since $J$ is finite, this proves Lemma 5.

Lemma 6. Let $C$ be a coalgebra. Let $I$ be a finite set. Let $\left( C_{i}\right) _{i\in I}$ be a family of subcoalgebras of $C$. Let $E$ be a simple subcoalgebra of $C$ such that $E\subseteq\sum_{i\in I}C_{i}$. Then, there exists an $i\in I$ such that $E\subseteq C_{i}$.

Proof of Lemma 6. We shall prove Lemma 6 by induction on $\left\vert I\right\vert $:

Induction base: We cannot have $\left\vert I\right\vert =0$ under the assumptions of Lemma 6 (because if $\left\vert I\right\vert =0$ under the assumptions of Lemma 6, then $I=\varnothing$ and thus $E\subseteq\sum_{i\in I}C_{i}=0$ (since $I=\varnothing$), so that $E=0$, which contradicts the assumption that $E$ be simple). Hence, Lemma 6 is vacuously true in the case when $\left\vert I\right\vert =0$. This completes the induction base.

Induction step: Let $N\in\mathbb{N}$. Assume (as the induction hypothesis) that Lemma 6 is proven in the case when $\left\vert I\right\vert =N$. Now, we must prove Lemma 6 in the case when $\left\vert I\right\vert =N+1$.

Assume that we are in the setting of Lemma 6, and we have $\left\vert I\right\vert =N+1$. We must show that there exists an $i\in I$ such that $E\subseteq C_{i}$.

The set $I$ is nonempty (since $\left\vert I\right\vert =N+1\geq1$). Hence, there exists an $i_{0}\in I$. Pick such an $i_{0}$, and let $I^{\prime} =I\setminus\left\{ i_{0}\right\} $; then, $\left\vert I^{\prime}\right\vert =\left\vert I\right\vert -1=N$ (since $\left\vert I\right\vert =N+1$).

Our goal is to show that there exists an $i\in I$ such that $E\subseteq C_{i}$. If $E\subseteq C_{i_{0}}$, then we are done (because we can just take $i=i_{0}$). Thus, we WLOG assume that we don't have $E\subseteq C_{i_{0}}$.

But $\sum_{i\in I^{\prime}}C_{i}$ is a subcoalgebra of $C$ (by Proposition 2, applied to $I^{\prime}$ instead of $I$).

Now, $E\subseteq\sum_{i\in I}C_{i}=C_{i_{0}}+\sum_{i\in I\setminus\left\{i_0\right\}}C_{i} = C_{i_{0}}+\sum_{i\in I^{\prime}}C_{i}$ (since $I\setminus\left\{ i_{0}\right\} = I^{\prime}$). Since $C_{i_{0}}$ and $\sum_{i\in I^{\prime}}C_{i}$ are subcoalgebras of $C$, we thus have either $E\subseteq C_{i_{0}}$ or $E\subseteq\sum_{i\in I^{\prime}}C_{i}$ (or both) (by Lemma 4, applied to $C_{i_{0}}$ and $\sum_{i\in I^{\prime}}C_{i}$ instead of $C_{1}$ and $C_{2}$). Hence, we have $E\subseteq\sum_{i\in I^{\prime}} C_{i}$ (since we don't have $E\subseteq C_{i_{0}}$). But the induction hypothesis tells us that we can apply Lemma 6 to $I^{\prime}$ and $\left( C_{i}\right) _{i\in I^{\prime}}$ instead of $I$ and $\left( C_{i}\right) _{i\in I}$ (since $\left\vert I^{\prime}\right\vert =N$). Hence, we conclude that there exists an $i\in I^{\prime}$ such that $E\subseteq C_{i}$ (since $E\subseteq\sum_{i\in I^{\prime}}C_{i}$). Thus, there exists an $i\in I$ such that $E\subseteq C_{i}$ (since $I^{\prime}\subseteq I$). Thus, Lemma 6 holds for our $I$ and $\left( C_{i}\right) _{i\in I}$. This completes the induction step, and thus Lemma 6 is proven by induction.

Theorem 8. Let $C$ be a coalgebra. Let $\left( C_{i}\right) _{i\in I}$ be a family of pairwise distinct simple subcoalgebras of $C$. Then, the sum $\sum_{i\in I}C_{i}$ is a direct sum.

Proof of Theorem 8. Let $\left( c_{i}\right) _{i\in I}\in\prod_{i\in I}C_{i}$ be a family such that all but finitely many $i\in I$ satisfy $c_{i}=0$, and such that $\sum_{i\in I}c_{i}=0$. We must prove that $\left( c_{i}\right) _{i\in I}=\left( 0\right) _{i\in I}$.

Indeed, assume the contrary. Thus, $\left( c_{i}\right) _{i\in I}\neq\left( 0\right) _{i\in I}$. Hence, there exists an $i_{0}\in I$ such that $c_{i_{0} }\neq0$. Consider this $i_{0}$.

The subcoalgebra $C_{i_{0}}$ of $C$ is simple, and thus satisfies $C_{i_{0} }\neq0$.

Recall that the family $\left( C_{i}\right) _{i\in I}$ consists of pairwise distinct subcoalgebras of $C$. Thus,

(3) $C_{i}\neq C_{j}$ whenever $i$ and $j$ are two distinct elements of $I$.

Let $J$ be the set of all $i\in I$ satisfying $c_{i}\neq0$. Then, $J$ is finite (since all but finitely many $i\in I$ satisfy $c_{i}=0$), and we have $i_{0}\in J$ (since $c_{i_{0}}\neq0$). As a consequence, the set $J \setminus \left\{i_0\right\}$ is finite as well. Every $i\in I\setminus J$ satisfies $c_{i}=0$ (because $J$ is the set of all $i\in I$ satisfying $c_{i}\neq0$). Hence, $\sum_{i\in I\setminus J}\underbrace{c_{i}}_{=0}=\sum_{i\in I\setminus J}0=0$. Now,

$0=\sum_{i\in I}c_{i}=\sum_{i\in J}c_{i}+\underbrace{\sum_{i\in I\setminus J}c_{i}}_{=0}$ (since $J\subseteq I$)

$=\sum_{i\in J}c_{i}=c_{i_{0}}+\sum_{i\in J\setminus\left\{ i_{0}\right\} }c_{i}$ (since $i_{0}\in J$),

so that

$c_{i_{0}}=-\sum_{i\in J\setminus\left\{ i_{0}\right\} }\underbrace{c_{i} }_{\in C_{i}}\in-\sum_{i\in J\setminus\left\{ i_{0}\right\} }C_{i} =\sum_{i\in J\setminus\left\{ i_{0}\right\} }C_{i}$

(since $\sum_{i\in J\setminus\left\{ i_{0}\right\} }C_{i}$ is a $\mathbf{k}$-vector space).

But $\sum_{i\in J\setminus\left\{ i_{0}\right\} }C_{i}$ is a subcoalgebra of $C$ (by Proposition 2, applied to $J\setminus\left\{ i_{0}\right\} $ instead of $I$). Now, Lemma 3 (applied to $E=C_{i_{0}}$, $D=\sum_{i\in J\setminus \left\{ i_{0}\right\} }C_{i}$ and $x=c_{i_{0}}$) shows that $C_{i_{0} }\subseteq\sum_{i\in J\setminus\left\{ i_{0}\right\} }C_{i}$ (since $C_{i_{0}}$ is a simple subcoalgebra of $C$ and since $c_{i_{0}}\in C_{i_{0}} $). Lemma 6 (applied to $J\setminus\left\{ i_{0}\right\} $ and $C_{i_{0}}$ instead of $I$ and $E$) thus shows that there exists an $i\in J\setminus \left\{ i_{0}\right\} $ such that $C_{i_{0}}\subseteq C_{i}$ (since $C_{i_{0}}$ is a simple subcoalgebra of $C$). Consider this $i$.

The coalgebra $C_{i}$ is simple. Thus, the only subcoalgebras of $C_{i}$ are $0$ and $C_{i}$. Since $C_{i_{0}}$ is a subcoalgebra of $C_{i}$ (since $C_{i_{0}}\subseteq C_{i}$), we thus conclude that $C_{i_{0}}$ is either $0$ or $C_{i}$. Since $C_{i_{0}}\neq0$, this shows that $C_{i_{0}}=C_{i}$.

But $i\in J\setminus\left\{ i_{0}\right\} $, so that $i_{0}\neq i$. Therefore, (3) (applied to $i_{0}$ and $i$ instead of $i$ and $j$) shows that $C_{i_{0}}\neq C_{i}$. This contradicts $C_{i_{0}}=C_{i}$. This contradiction concludes the proof of Theorem 8.

We notice that we have not used the coassociativity of our coalgebras anywhere in the above proofs! Actually we have not used half of the counitality axiom either; we have merely used the axiom that $\left( \operatorname*{id}\otimes\epsilon\right) \left( \Delta\left( x\right) \right) = x \otimes 1$ for every $x$ in a coalgebra.

---- OUTTAKES ----

I have written the following expecting it to be used for the proof, but it turned out to not be necessary:

The next proposition is the generalization of Lemma 6 to arbitrary (not necessarily finite) sets $I$:

Proposition 7. Let $C$ be a coalgebra. Let $\left( C_{i}\right) _{i\in I}$ be a family of subcoalgebras of $C$. Let $E$ be a simple subcoalgebra of $C$ such that $E\subseteq\sum_{i\in I}C_{i}$. Then, there exists an $i\in I$ such that $E\subseteq C_{i}$.

Proof of Proposition 7. The coalgebra $E$ is simple, and thus satisfies $E\neq0$. Hence, there exists an $x\in E$ such that $x\neq0$. Consider such an $x$.

Let $\left\langle x\right\rangle $ denote the $\mathbf{k}$-linear span of $x$. Thus, $\left\langle x\right\rangle \subseteq E$ (since $x\in E$), so that $\left\langle x\right\rangle \subseteq E\subseteq\sum_{i\in I}C_{i}$. Lemma 5 (applied to $C$, $C_{i}$ and $\left\langle x\right\rangle $ instead of $V$, $V_{i}$ and $E$) thus shows that there exists a finite subset $J$ of $I$ such that $\left\langle x\right\rangle \subseteq\sum_{i\in J}C_{i}$. Consider this $J$. We have $x\in\left\langle x\right\rangle \subseteq\sum_{i\in J}C_{i}$. We know that $\sum_{i\in J}C_{i}$ is a subcoalgebra of $C$ (by Proposition 2, applied to $J$ instead of $I$). Thus, Lemma 3 (applied to $D=\sum_{i\in J}C_{i}$) shows that $E\subseteq\sum_{i\in J}C_{i}$. Therefore, we can apply Lemma 6 to $J$ and $\left( C_{i}\right) _{i\in J}$ instead of $I$ and $\left( C_{i}\right) _{i\in I}$ (since $J$ is finite). As a consequence, we see that there exists an $i\in J$ such that $E\subseteq C_{i}$. Hence, there exists an $i\in I$ such that $E\subseteq C_{i}$ (since $J\subseteq I$). This proves Proposition 7.

Proposition A. Every coalgebra is the union of its finite-dimensional subcoalgebras.

Proposition A is the famous Sweedler finiteness theorem. For a proof, combine Corollary 4.6 in Chapter VIII of James Milne's notes "Basic Theory of Affine Group Schemes" (version 1.00) (this corollary states that every coalgebra $C$ is the union of its subcoalgebras $C_{V}$, where $V$ runs over the finite-dimensional subcomodules of $C$, and where $C_{V}$ is defined in Remark 4.3 before) with Remark 4.3 (a) in these same notes (which states that the subcoalgebras $C_{V}$ just mentioned are finite-dimensional).

Proposition B. Every simple coalgebra is finite-dimensional.

Proof of Proposition B. Let $C$ be a simple coalgebra. We need to prove that $C$ is finite-dimensional.

Proposition A shows that the coalgebra $C$ is the union of its finite-dimensional subcoalgebras. Each of these subcoalgebras must be either $0$ or $C$ (since $C$ is simple); but if they are all $0$, then $C$ must itself be $0$ (since $C$ is their union), which would contradict the fact that $C$ is simple. Hence, at least one of these subcoalgebras is $C$. Thus, $C$ is a finite-dimensional subcoalgebra of $C$. In particular, $C$ is finite-dimensional. This proves Proposition B.