TITLE: How to understand the coidea of a colagebra?
QUESTION [5 upvotes]: Let $C$ be a coalgebra with coproduct $\Delta$ and counit $\epsilon$. Then a subset $I\subseteq C$ is a coideal if $\Delta(I)\subseteq I\otimes C+C\otimes I$ and $\epsilon(I)=0$.
My question is why the coideal should be defined by this way? Does it show the  absorptivity ?

REPLY [9 votes]: Supossing that we are speaking about coalgebras over a field $k$, the definition of the notion of a coideal of a $k$-coalgebra has been much inspired by the relation between ideals of a $k$-algebra and kernels of $k$-algebra morphisms: we are actually trying to "mimic" this relation and define the notion of a coideal in a way  which would allow for a similar relation between coideals of a $k$-coalgebra and kernels of $k$-coalgebra morphisms:
Recall that, given a field $k$ and the two $k$-coalgebras $C$, $D$, if the $k$-linear map $f:C\to D$ is a coalgebra map, then the following diagram is commutative: 
$$
\require{AMScd}
\begin{CD}
C@>f>> D  \\
 @V\Delta_C VV @VV \Delta_D V\\
C\otimes C @>{f\ \otimes\ f}>>D\otimes D 
\end{CD}
$$ 
Now, we would like to determine which $k$-subspaces of $C$ are in the kernel of the coalgebra map $f:C\to D$.
Since $\Delta_D\big(f(Ker f)\big)=0$, by the comm. diagram above, we get that $\big(f\otimes f\big)\big(\Delta_C(Ker f)\big)=0$ and thus
$$
\Delta_C(Ker f)\subseteq Ker(f\otimes f)
$$
Consequently, the description of the kernel of $f$ and its behaviour under $\Delta$ are related to the description of $Ker(f\otimes f)$.
Now, recall that from linear algebra: 
$$
Ker(f\otimes f)=Ker f\otimes C+C\otimes Ker f
$$ 
Since we want our notion of coideal to describe something which behaves similarly to kernels of coalgebra maps, the standard definition seems reasonable, in view of the relation
$$
\Delta_C(Ker f)\subseteq Ker f\otimes C+C\otimes Ker f
$$
On the other hand, the second relation which is implied by the fact that $f:C\to D$ is a coalgebra morphism is that 
$$
\varepsilon_D\circ f=\varepsilon_C
$$
thus: $\varepsilon_D\big(f(Ker f)\big)=0=\varepsilon_C(Ker f)$ and consequently:
$$
\varepsilon_C(Ker f)=0
$$
which justifies the second relation of the definition of the coideal. 
I hope the above description, will shed some light into the ... coidea behind  the definition of a coideal (for coalgebras over fields).