Showing tensor product of coalgebras is a coalgebra. Let $(C, \Delta, \epsilon)$ and $(C',\Delta', \epsilon')$ be two coalgebras over the field $k$. I'm trying to show that $C \otimes C'$ is a coalgebra for the comultiplication
$$\overline{\Delta}:=(id_{C} \otimes \tau_{C,C'} \otimes id_{C'}) \circ (\Delta \otimes \Delta')$$
where $\tau_{C,C'}(c \otimes c') = c' \otimes c$ and the counit
$$\overline{\epsilon}:=\epsilon \otimes \epsilon'$$
Here is my attempt to show that we have a comultiplication:
It suffices to check that
$$(\overline{\Delta} \otimes id_{C \otimes C'}) \circ  \overline{\Delta}(c\otimes c')= ( id_{C \otimes C'}\otimes \overline{\Delta}) \circ  \overline{\Delta}(c\otimes c')$$
where $c \in C, c' \in C'$.
I'm unsure how to proceed next. I started calculating $\Delta \otimes \Delta'(c \otimes c')= \Delta (c) \otimes \Delta'(c')$ but then I wrote $\Delta(c) = \sum_c c_{(1)} \otimes c_{(2)}$ and $\Delta(c') = \sum_{c'} c'_{(1)} \otimes c'_{(2)}$ (I believe this is called Sweedler notation?) and tried to proceed.
Is this the right way to continue?
Any input is appreciated! Thanks.
 A: It is possible to do this with Sweedler notation and surely a good excercise to get used to it:
For $c \in C$, we write $\Delta(c) = \sum c_{(1)} \otimes c_{(2)}$ and coassociativity then means 
$$\sum (c_{(1)})_{(2)} \otimes (c_{(1)})_{(2)} \otimes c_{(2)} = \sum c_{(1)} \otimes (c_{(2)})_{(1)} \otimes (c_{(2)})_{(2)}$$
Note that one usually writes only one sum symbol even though we are actually having two nested sums here. 
With $\overline{\Delta}$ defined as you did we have
$\overline{\Delta}(c \otimes c') = \sum c_{(1)} \otimes c'_{(1)} \otimes c_{(2)} \otimes c'_{(2)} $ for $c \in C$ and $c' \in C$ and thus
$$ (\overline{\Delta} \otimes id_{C \otimes C'}) \circ \overline{\Delta}(c \otimes c') = \sum (c_{(1)})_{(1)} \otimes (c'_{(1)})_{(1)} \otimes (c_{(1)})_{(2)} \otimes (c'_{(1)})_{(2)} \otimes c_{(2)} \otimes c'_{(2)}$$
and 
$$ ( id_{C \otimes C'}\otimes \overline{\Delta} ) \circ \overline{\Delta}(c \otimes c') = \sum c_{(1)} \otimes c'_{(1)} \otimes (c_{(2)})_{(1)} \otimes (c'_{(2)})_{(1)} \otimes (c_{(2)})_{(2)} \otimes (c'_{(2)})_{(2)} $$
Again, we only write one sum symbol, even though we have essentially four sums to deal with. 
In any case, coassociativity for $\Delta$ and $\Delta'$ shows that the two expressions are identical.
Another way is by direct manipulation of the maps:
First of all, note that 
$$\begin{align} &(\Delta \otimes \Delta' \otimes id_{C \otimes C'})(id_C \otimes \tau_{C,C'} \otimes id_{C'}) \\ = &
(id_C \otimes id_{C'} \otimes \tau_{C,C'} \otimes id_{C \otimes C'}) (id_{C \otimes C} \otimes \tau_{C,C'} \otimes id_{C'} \otimes id_{C' \otimes C'}) (\Delta \otimes id_C \otimes \Delta' \otimes id_{C'}) \end{align}$$
which just translates to instead of first switching the two interior tensor factors and then apply $\Delta$ and $\Delta'$ on the first and second factor, you can apply $\Delta$ and $\Delta'$ to the first and third factor and then switch the resulting factors around.
This gives
$$\begin{align} 
&(\overline{\Delta} \otimes id_{C \otimes C'}) \overline{\Delta} \\
   =&  (id_C \otimes \tau_{C,C'} \otimes id_{C'} \otimes id_{C \otimes C'}) (\Delta \otimes \Delta' \otimes id_{C \otimes C'})(id_C \otimes \tau_{C,C'} \otimes id_{C'}) (\Delta \otimes \Delta') \\
 = &(id_C \otimes \tau_{C,C'} \otimes id_{C'} \otimes id_{C \otimes C'})  (id_{C \otimes C'} \otimes \tau_{C,C'} \otimes id_{C \otimes C'}) (id_{C \otimes C} \otimes \tau_{C,C'} \otimes id_{C'} \otimes id_{C' \otimes C'}) 
 \\&(\Delta \otimes id_C \otimes \Delta' \otimes id_{C'}) (\Delta \otimes \Delta') \end{align}$$
and a similar calculation gives 
$$ \begin{align}
&(id_{C \otimes C'} \overline{\Delta}) \overline{\Delta} \\
=& (id_{C \otimes C'} \otimes id_{C} \otimes \tau_{C,C'} \otimes C') 
(id_C \otimes \tau_{C,C'} \otimes id_C \otimes id_{C' \otimes C'})
(id_{C \otimes C} \otimes \tau_{C,C'} \otimes id_{C' \otimes C'})
\\&( id_C \otimes \Delta  \otimes id_{C'} \otimes \Delta') (\Delta \otimes \Delta')
\end{align} 
$$
Coassociativity of $\Delta$ and $\Delta'$ implies that 
$$ ( id_C \otimes \Delta  \otimes id_{C'} \otimes \Delta') (\Delta \otimes \Delta') = (\Delta \otimes id_C \otimes \Delta' \otimes id_{C'}) (\Delta \otimes \Delta')$$
and the identity 
$$(id_C \otimes \tau_{C,C'} \otimes id_{C'} \otimes id_{C \otimes C'})  (id_{C \otimes C'} \otimes \tau_{C,C'} \otimes id_{C \otimes C'}) (id_{C \otimes C} \otimes \tau_{C,C'} \otimes id_{C'} \otimes id_{C' \otimes C'}) 
= (id_{C \otimes C'} \otimes id_{C} \otimes \tau_{C,C'} \otimes C') 
(id_C \otimes \tau_{C,C'} \otimes id_C \otimes id_{C' \otimes C'})
(id_{C \otimes C} \otimes \tau_{C,C'} \otimes id_{C' \otimes C'}) $$
is easily verified as well. It corresponds to two different decompositions of the 'shuffle' 
$$ c_1 \otimes c_2 \otimes c_3 \otimes c_1' \otimes c_2' \otimes c_3' \mapsto c_1 \otimes c_1' \otimes c_2 \otimes c_2' \otimes c_3 \otimes c_3'.$$
