Why coproduct in the algebra of linear operators is actually a coproduct?

Question

Coproduct in a coalgebra $V$, where $V$ is also a vector space over a field $\mathbb{K}$, is defined as a $\mathbb{K}-$linear map $\Delta : V \rightarrow V\otimes V$ satisfying two diagrams obtained from the diagrams describing associativity and existence of unit in associative unital algebras by inverting the arrows. In particular, the coassociativity diagram is satisfied: Coassociativity of coproduct

which means $$(\Delta \otimes I) \circ \Delta = (I \otimes \Delta) \circ \Delta$$

It is suggested (for example, on the corresponding Wikipedia page) that the coproduct in the algebra of linear operators is $$\Delta : A \rightarrow A \otimes 1 + 1 \otimes A$$ where $\otimes$ is the "exterior tensor product" (what is its exact definition is unclear to me, but it appears to have properties similar to the regular tensor product). I assume $1$ here is just the unit operator $I$, because otherwise the map wouldn't be from $V$ to $V\otimes V$. Let us check the coassociativity: take a random operator $A$. $$\Delta A = A \otimes 1 + 1 \otimes A$$ $$\Delta \otimes I (A \otimes 1 + 1 \otimes A)=$$ $$(A \otimes 1 + 1 \otimes A)\otimes 1 + (1 \otimes 1 + 1 \otimes 1)\otimes A=$$ $$A \otimes 1 \otimes 1 + 1 \otimes A \otimes 1 + 2(1 \otimes 1 \otimes A)$$ Now let us calculate $(I \otimes \Delta) \circ \Delta$: $$(I \otimes \Delta)(A \otimes 1 + 1 \otimes A)=$$ $$= 2(A \otimes 1 \otimes 1)+ 1 \otimes A \otimes 1 + 1 \otimes 1 \otimes A$$

Clearly, $(\Delta \otimes I) \circ \Delta \ne (I \otimes \Delta) \circ \Delta$. So why is this a coproduct? What am I missing?

Edit: the Wikipedia article appears to refer to the fact (p.20) that the universal enveloping algebra over a Lie group $\mathfrak{g}$ has a coproduct defined as $\Delta: x \rightarrow 1 \otimes x + x \otimes 1$, where $1$ is the actual $1$ as a number, living in the 0-th degree of the universal enveloping algebra. The angular momentum operator used as example in the article is an element of $\mathfrak{so}_{3}$, so it makes sense. However, I still do not understand why coassociativity condition is satisfied. The universal enveloping algebra is the factor of the tensor algebra $T(\mathfrak{g})$ by the ideal generated by elements of the form $x \otimes y - y \otimes x - [x, y]$. How do I use that to prove the two expressions to be equivalent? If $[x, y]$ was defined for any pair of elements of $U(\mathfrak{g})$, I could change the order of elements in monomials of the form $A \otimes 1 \otimes 1$ as follows: $$A \otimes 1 \otimes 1 = 1 \otimes A \otimes 1 - 1 \otimes A \otimes 1 + A \otimes 1 \otimes 1$$ then, using the fact that $A \otimes 1 \otimes 1 - 1 \otimes A \otimes 1= (A \otimes 1 - 1 \otimes A)\otimes 1$ and $A \otimes 1 - 1 \otimes A - [A, 1] = 0$, I would rewrite $$A \otimes 1 \otimes 1 = 1 \otimes A \otimes 1 + [A, 1] \otimes 1$$ However, $[A, 1]$ is not defined (or so I think).

Answer 1

I have found the answer to my original question in the comments here. First, as was correctly indicated in the comments, I was originally mistaken about the algebra on which this coproduct is defined, and did not understand the definition itself. The coproduct $\Delta: x \rightarrow 1 \otimes x + x \otimes 1$ is defined on the universal enveloping algebra $U(\mathfrak{g})$ of a Lie algebra $\mathfrak{g}$, which is the factor of the tensor algebra $\mathbb{K}\oplus \bigoplus_{i \ge 1}\mathfrak{g}^{\otimes i}$ by the ideal generated by elements of the form $x \otimes y - y \otimes x - [x, y]$, where $x, y \in \mathfrak{g}$, $[,]$ is the bracket in the Lie algebra. It has such a form specifically for elements of $\mathfrak{g}$ embedded into $U(\mathfrak{g})$. For the element $1 \in \mathbb{K}$, the coproduct is defined as $\Delta: 1\rightarrow 1 \otimes 1$, and not as $\Delta: 2(1 \otimes 1)$. It is done in this way, because coproduct in a Hopf algebra must be a morphism of algebras, hence, it must map the unit of $\mathbb{K}$ to the unit of $U(\mathfrak{g})$ (also because unit and counit must compose in identity). Then it is easy to check explicitly that coproduct is coassociative.

Answer 2

You have misunderstood the Wikipedia page. It states that, for a certain coalgebra $C$ and a certain element $j$ of $C$ we have $\Delta(j) = j \otimes 1 + 1 \otimes j$. Your deduction that this equation defines a coalgebra structure for every vector space* and any element $1$ of it and all elements $j$ is not correct. Your proof confirms this.

For a more concrete example, consider the divided power bialgebra (in one variable), which is $K[X]$ with $\Delta(X) = X \otimes 1 + 1 \otimes X$ and $\varepsilon(X)=0$. Here, we have $\Delta(X^2)=X^2 \otimes 1 + 2 \cdot X \otimes X + 1 \otimes X^2$, and thus $\Delta(X^2) \neq X^2 \otimes 1 + 1 \otimes X^2$.

*Even though you have said this only for $\mathrm{End}(V)$, it is clear that you never use the algebra structure on this algebra, you only use the vector space structure and some random element $1$.