Cocommutativity, comultiplication and coalgebra maps Given a coalgebra $(C,\Delta,\varepsilon)$, over a field, the following is a well-known property: 

the comultiplication $\Delta:C\to C\otimes C$ is a coalgebra map if and only if $C$ is cocommutative

My question is how we can prove the above with an explicit computation? (The main problem to be handled here seems to be the treatment of the emerging elements $c_{(1)(2)}$ and $c_{(2)(1)}$). 
To the best of my knowledge, I have not found the details of the computation neither in some text nor in some article or some other question at this site. So, I have tried to devise the computation myself and I am posting the answer below. Is there some other approach ?  
P.S. This is exercise 3, p.66, Ch. III, from Moss E.Sweedler's book on Hopf algebras. 
 A: Given a (coassociative and counital) 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 
$$
(\Delta\otimes\Delta)=(Id\otimes\Delta\otimes Id)\circ(Id\otimes\Delta)= \\
=(Id\otimes\tau\circ\Delta\otimes Id)\circ(Id\otimes\Delta) = \\ =(Id\otimes\tau\otimes Id)\circ(Id\otimes\Delta\otimes Id)\circ(Id\otimes\Delta)=  (Id\otimes\tau\otimes Id)\circ(\Delta\otimes\Delta)=\Delta_{C\otimes C}\Rightarrow \\ 
\Rightarrow (\Delta\otimes\Delta)\circ\Delta=\Delta_{C\otimes C}\circ\Delta 
$$
A: Not an answer, but a clarification of the context and ``bording counterexamples''. In short, this property is true only in the context of counital coalgebras and NOT otherwise.
If it were true in general, it would be true in the subcategory $k-coalg^{fin}$ 
of finite dimensional coalgebras (we are over a field $k$) and the 
dual property (i.e. the multiplication is an algebra map iff the algebra is commutative) true in the dual category towards which we have 
an isofunctor (the vector dual). To see that this is not the case, take any finite dimensional non-commutative associative algebra $\mathcal{A}$ such that $\mathcal{A}^{(4)}$ (the space generated by the products of 4 elements) is zero.
Take, for example, a two-letter alphabet (i.e. a set $A=\{x,y\}$ of two variables). And consider the algebra $k[A^+]$ of the free semigroup $A^+$ (it is the algebra of non-commutative polynomials without constant term). Now, let $(A^+)_{\geq 4}$ be the set of words of length $\geq 4$. It generates the two-sided ideal $\mathcal{I}_{\geq 4}$ and the non-commutative algebra $\mathcal{A}=k[A^+]/\mathcal{I}_{\geq 4}$ is a counterexample. If one likes to stick to coalgebra, on can consider the dual coalgebra ($\mathcal{A}$ is finite dimensional). 
Remark : It is transparent in terms of identities that the property is true for coassociative coalgebras with counits. The dual property is that the identity 
$$
xy=yx\qquad (1)
$$
(commutativity) is true iff the identity 
$$
(x_1y_1)(x_2y_2)=(x_1x_2)(y_1y_2)\qquad (2)
$$
is true, which is false in general and straightforward if you suppose associativity and unit. 
Statement (after dualization) It seems that the full statement is: 
Let $\mathcal{C}=(C,\Delta)$ be a coassociative coalgebra, then


*

* if $\mathcal{C}$ is cocommutative, $\Delta$ is a coalgebra map

* if $\Delta$ is a coalgebra map and if $\mathcal{C}$ admits a counit then $\mathcal{C}$ is cocommutative.

