Axioms of a coalgebra restated using Sweedler's notation I'm struggling with understanding manipulation using Sweedler's notation at a very fundamental level. I don't understand the equivalence of the axioms of coalgebras in the standard notation
[Coproduct Map] $$\Delta:A\rightarrow A\otimes A$$ is a coassociative map;
$$(\Delta\otimes id_C)\circ\Delta=(id_C\otimes\Delta)\circ\Delta$$
in terms of Sweedler's notation;
Coassociativity $$\sum\Delta(a_{(1)})\otimes a_{(2)} =\sum a_{(1)}\otimes \Delta(a_{(2)})$$
I get upto being able to expand the left hand side of the above;
$$
\sum\sum(a_{(1)})_{(1)}\otimes (a_{(1)})_{(2)}\otimes(a_{(2)})
$$
but I don't know how to make it equivalent to $\sum\sum(a_{(1)})\otimes (a_{(2)})_{(1)}\otimes(a_{(2)})_{(2)}$. I'm assuming it would involve applying the coassociativity axiom.
For the counit axiom, I can't even scratch the surface and make a start.
$\eta:C\rightarrow K$ is the counit map that satisfies $(\eta\otimes id_C)\circ\Delta=(id_C\otimes \eta)\circ\Delta$ as
$$\sum\eta(c_1)c_2=\sum c_1\eta(c_2)=c, \forall c=\in C$$
Could someone help with a basic, step by step walk through of showing the equivalence of the two expressions?
 A: You say:

$\dots$ but I don't know how to make it equivalent to $\cdots$

You don't have to make it an equality, it is an equality by precisely the assumption that $\Delta$ is coassociative.
Let's step back a little. A comultiplication is a linear map $\Delta\colon C\to C\otimes C$ satisfying coassociativity, i.e.~$(\Delta\otimes 1)\circ \Delta=(1\otimes \Delta)\circ \Delta$. As every element of $C\otimes C$ can be written as a finite linear combination of elementary tensors (elements of the form $c\otimes c'$ with $c,c'\in C$), we can write $$\Delta(x)=\sum_{i=1}^nx_{1,i}\otimes x_{2,i}$$
with all $x_{j,i}\in C$. The above is not yet the Sweedler notation but is certainly correct notation. The annoying thing is that the above notation gets complicated really quickly when doing multiple operations. That's where the Sweedler notation comes in!
Instead of writing $$\Delta(x)=\sum_{i=1}^nx_{1,i}\otimes x_{2,i},$$
we surpress the index of the summation and obtain $$\Delta(x)=\sum x_{(1)}\otimes x_{(2)}.$$
By using these round squares, we remind ourselves that actually an index is omitted from the notation, but the labels $1$ and $2$ recall the order in which we 'blew up' the element $x$. Coassociativity precisely says that when doing repeated 'blow ups', the order doesn't matter.
So with this explanation, let's use this notation to understand the counit property $(\eta\otimes id_C)\circ\Delta=(id_C\otimes \eta)\circ\Delta$.
We find that $$\begin{aligned} (\eta\otimes id_C)\circ\Delta(x)&=(\eta\otimes id_C)(\Delta(x))\\ &= (\eta\otimes id_C)(\sum x_{(1)}\otimes x_{(2)}) \\ &= \sum \eta(x_{(1)})\otimes id_C(x_{(2)})\end{aligned}$$
Note that the above actually lives in $k\otimes C$ (where $k$ is the field we're working over). Clearly $k\otimes_k C\cong C$ by mapping $\lambda\otimes c$ to $\lambda c$. Hence we can identify $\sum \eta(x_{(1)})\otimes id_C(x_{(2)})$ with $\sum \eta(x_{(1)})x_{(2)}$. The other side can be computed similarly and yields the desired equality.
