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.
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'
$$ c_1 \otimes c_2 \otimes c_3 \otimes c_1' \otimes c_2' \otimes c_3' \mapsto c_1 \otimes c_1' \otimes c_2 \otimes c_2' \otimes c_3 \otimes c_3'.$$