Augmented coalgebras

Question

Let $C, \Delta$ be a coalgebra. Assume that it is coaugmented with coaugmentation $u\: : \: k\to C$ and co unit $\epsilon\: : \ C\to k$. Since $\epsilon\circ u=id $ we get $$ C=\text{Kern}(\epsilon )\oplus k $$ Set $\bar{C}:=\text{Kern}(\epsilon )$. Then $\bar{C}$ equipped with the reduced coproduct $\bar{\Delta}(x):=\Delta(x)-1\otimes x-x\otimes 1$ is a (non-coaugmented) coalgebra. Then we have a functor between the category of augmented coalgebras into the category of coalgebras.

Does this functor defines an equivalences of categories?

I ask this because a similar thing is true in the contest of algebras and aumented algebras. The idea is starting from a generaleal coalgebra $C',\Delta'$ choose a symbol 1 and define $C:=C'\oplus 1 k$, then define $$ \Delta(x):=\Delta'(x)+1\otimes x+x\otimes 1 $$ Do you know some reference about that? Does this process works in the contest of differential graded coalgebras?

Answer 1

Actually it's simpler (to follow the duality with the case of algebras) to consider the cokernel of the coaugmentation, instead of the kernel of the augmentation. More precisely define an equivalence: $$F : \{ \text{coaugmented counital dg-coalgebras} \} \leftrightarrows \{ \text{coalgebras (without counit)} \} : G$$ as follows. Let $(C,\Delta,\varepsilon,d,u)$ be a coaugmented dg-coalgebra (with the obvious notations, $\varepsilon$ the counit and $u$ the coaugmentation). Then $F(C,\Delta,\varepsilon,d,u) = (C', \Delta', d')$ where:

• $C' = \operatorname{coker}(u) = C / \operatorname{im}(u)$;
• since the coaugmentation is a morphism of coalgebras and $\Delta(1_k) = 1_k \otimes 1_k$, it follows that $\Delta(1) = 1 \otimes 1$ (where $1 = u(1_k)$), thus $\Delta : C \to C \otimes C$ induces a map on the quotient $\Delta' : C' \to C' \otimes C'$, which is again coassociative;
• similarly $d(1_k) = 0$ thus $d(1) = 0$ and so $d$ induces a map $d' : C' \to C'$, which is a coderivation that squares to zero (because $d$ was).

Conversely let $(C',\Delta',d')$ be a coalgebra (without counit), then you can define a coaugmented counital coalgebra $G(C',\Delta',d') = (C, \Delta, \varepsilon, d, u)$ by:

• $C = C' \oplus k1$ (where $1$ is some symbol);
• $\Delta(x) = \Delta'(x) + x \otimes 1 + 1 \otimes x$ for $x \in C$ and $\Delta(1) = 1 \otimes 1$, extended linearly;
• $\varepsilon(x) = 0$ for $x \in C'$ and $\varepsilon(1) = 1$;
• $d(x) = d'(x)$ for $x \in C'$ and $d(1) = 0$;
• $u(1_k) = 1$.

You get a coaugmented counital dg-coalgebra this way. Skip the next paragraph if you don't need convincing.

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Most axioms are easy to check, the most troublesome one (in terms of lines of computation needed) is perhaps coassociativity. Let me write down $\Delta'(x) = x_{(1)} \otimes x_{(2)}$ using Sweedler notation. By coassociativity of $\Delta'$ we know that $(\Delta' \otimes \operatorname{id}) \Delta'(x) = x_{(1)} \otimes x_{(2)} \otimes x_{(3)} = (\operatorname{id} \otimes \Delta') \Delta'(x)$. Then: $$\begin{align} (\Delta \otimes \operatorname{id}) (\Delta(x)) & = (\Delta \otimes \operatorname{id}) \bigl( x_{(1)} \otimes x_{(2)} + x \otimes 1 + 1 \otimes x \bigr) \\ & = \Delta(x_{(1)}) \otimes x_{(2)} + \Delta(x) \otimes 1 + \Delta(1) \otimes x \\ & = \bigl( \Delta'(x_{(1)}) \otimes x_{(2)} + x_{(1)} \otimes 1 \otimes x_{(2)} + 1 \otimes x_{(1)} \otimes x_{(2)} \bigr) \\ & + \bigl( \Delta'(x) \otimes 1 + x \otimes 1 \otimes 1 + 1 \otimes x \otimes 1 \bigr) + 1 \otimes 1 \otimes x \\ & = x_{(1)} \otimes x_{(2)} \otimes x_{(3)} + x_{(1)} \otimes 1 \otimes x_{(2)} + 1 \otimes x_{(1)} \otimes x_{(2)} \\ & + x_{(1)} \otimes x_{(2)} \otimes 1 + x \otimes 1 \otimes 1 + 1 \otimes x \otimes 1 + 1 \otimes 1 \otimes x \end{align}$$ and now clearly you can rearrange the terms in that last expression to get back to $(\operatorname{id} \otimes \Delta)\Delta(x)$ (whew! it took me some time to get it right).

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Now you need to prove that $F$ and $G$ are inverse equivalences. This comes from the fact that if $C$ is a coaugmented dg-coalgebra, the exact sequence $$0 \to k \xrightarrow{u} C \to \operatorname{coker}(u) \to 0$$ is split by the counit: $\varepsilon(u(1_k)) = 1_k$. So $C$ is isomorphic as a chain complex to $\operatorname{coker}(u) \oplus k$, which you can check is exactly $G(F(C))$. The converse $F(G(C')) \cong C'$ is almost immediate by definition. (All that's left is checking naturality of everything, which is a completely mechanical exercise.)