I have a question about comultiplication for coalgebras:
Suppose $C$ is a coalgebra over the field $k$. How does one show that the comultiplication map $\Delta:C\to C\otimes C$ is a coalgebra map if and only if $C$ is cocommutative?
The main problem that I encounter is that when I tried to do it by definition, I was struggling to find a relation between $c_{(1)(2)}$ and $c_{(2)(1)}$. In that case, is there any other way for which I could tackle this question?
Given a coalgebra $(C,\Delta,\varepsilon)$, over a field $k$, we can form the tensor product coalgebra $(C\otimes C,\Delta_{C\otimes C},\varepsilon_{C\otimes C})$ through: $$ \Delta_{C\otimes C}=(Id\otimes\tau\otimes Id)\circ(\Delta\otimes\Delta):C\otimes C\rightarrow C\otimes C \otimes C \otimes C , \\ \\ \\ \varepsilon_{C\otimes C}=\phi \circ (\varepsilon\otimes\varepsilon):C\otimes C\rightarrow k $$ where $Id$ is the identity map and $\phi:k\otimes k\stackrel{\cong}{\rightarrow} k$ the natural isomorphism.
The comultiplication $\Delta:C\rightarrow C\otimes C$ being a morphism of coalgebras, or a coalgebra map, by definition means that for an arbitrary $c\in C$ we have: $\varepsilon_C(c)=\varepsilon_{C\otimes C}\circ\Delta(c)=\varepsilon(c_1)\varepsilon(c_2)$ and $$ \Delta_{C\otimes C}\circ\Delta(c)=(\Delta\otimes\Delta)\circ\Delta(c) \Leftrightarrow \\ \\ \\ \Leftrightarrow \Delta_{C\otimes C}\big(\sum c_1\otimes c_2\big)=(\Delta\otimes\Delta)\big(\sum c_1\otimes c_2\big)\Leftrightarrow \\ \\ \\ \Leftrightarrow (Id\otimes\tau\otimes Id)\circ(\Delta\otimes\Delta)\big(\sum c_1\otimes c_2\big)=(\Delta\otimes\Delta)\big(\sum c_1\otimes c_2\big)\Leftrightarrow \\ \\ \\ \Leftrightarrow \sum c_1\otimes c_3\otimes c_2\otimes c_4=\sum c_1\otimes c_2\otimes c_3\otimes c_4 $$ In the last line of the above, we have made use of generalized coassociativity, expressing both sides in Sweedler's notation. Given that $\Delta(c)=\sum c_1\otimes c_2$, the last line of the above could have been written (without using generalized coassociativity) alternatively as: $$ \sum c_{1_1}\otimes c_{2_1}\otimes c_{1_2}\otimes c_{2_2}=\sum c_{1_1}\otimes c_{1_2}\otimes c_{2_1}\otimes c_{2_2} $$ Now, applying to both sides of the last line of the above, the map $(\varepsilon\otimes Id\otimes Id\otimes\varepsilon)$, we get $$ \sum \varepsilon(c_{1_1})\otimes c_{2_1}\otimes c_{1_2}\otimes \varepsilon(c_{2_2})=\sum \varepsilon(c_{1_1})\otimes c_{1_2}\otimes c_{2_1}\otimes \varepsilon(c_{2_2})\Leftrightarrow \\ \\ \\ \Leftrightarrow \sum\varepsilon(c_{2_2}) c_{2_1}\otimes \varepsilon(c_{1_1})c_{1_2}=\sum \varepsilon(c_{1_1})c_{1_2}\otimes \varepsilon(c_{2_2})c_{2_1} \Leftrightarrow \\ \\ \\ \Leftrightarrow \sum c_2\otimes c_1=\sum c_1\otimes c_2 $$ for any $c\in C$. In the last line, use has been made of the defining property of the counity map: $\sum\varepsilon(c_1)c_2=\sum c_1\varepsilon(c_2)=c$, for any $c\in C$.
Thus, we have shown that: If the comultiplication $\Delta:C\rightarrow C\otimes C$ is a morphism of coalgebras, or a coalgebra map then this implies the cocommutativity of $C$.
The converse implication, i.e. cocommutativity of $C$ implies that the comultiplication is a coalgebra map, comes from the fact that generalized coassociativity permits us to write
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\Delta\otimes\Delta)=(Id\otimes\Delta\otimes Id)\circ(Id\otimes\Delta)$
and thus, together with cocommutativity of $C$, they imply
$$
(\Delta\otimes\Delta)=(Id\otimes\Delta\otimes Id)\circ(Id\otimes\Delta)= \\
=(Id\otimes\tau\circ\Delta\otimes Id)\circ(Id\otimes\Delta) = \\ =(Id\otimes\tau\otimes Id)\circ(Id\otimes\Delta\otimes Id)\circ(Id\otimes\Delta)= (Id\otimes\tau\otimes Id)\circ(\Delta\otimes\Delta)=\Delta_{C\otimes C}\Rightarrow \\
\Rightarrow (\Delta\otimes\Delta)\circ\Delta=\Delta_{C\otimes C}\circ\Delta
$$
P.S. This is exercise 3, p.66, Ch. III, from Moss E.Sweedler's book on Hopf algebras.
