Well-definedness of reduced coproduct I am studying Algebraic Operads with the book Algebraic Operads, by Jean-Louis Loday and Bruno Vallette. The authors present the following definitions of coaugmented coalgebra and reduced coproduct:

My first problem is:  Who is $1$? Is it an element that we can choose?
My first idea of choice was $1=u(1_{\mathbb{K}})$.  Since $u$ is a coalgebra morphism,   this gives us that $\Delta(1) = 1\otimes 1$ and maybe I could use this for to solve my last question: How I prove that $\bar{\Delta}(x) \in \bar{C}\otimes \bar{C}$? 
Thanks! ;)
 A: Yes, $1 = u(1_{\mathbb{K}})$. You have a short exact sequence
$$0 \to \ker(\epsilon) \to C \xrightarrow{\epsilon} \mathbb{K} \to 0$$
and it is split by the coaugmentation $u : \mathbb{K} \to C$, so by the splitting lemma, $C \cong \ker(\epsilon) \oplus \mathbb{K}$, and $\mathbb{K}$ is identified as a submodule of $C$ as the image of $u$.
You simply need to compute $(\epsilon \otimes \operatorname{id})(\bar{\Delta}(x))$ and $(\operatorname{id} \otimes \epsilon)(\bar{\Delta}(x))$ to check that $\bar{\Delta(x)}$ is in $\bar{C} \otimes \bar{C}$. But for example
$$(\epsilon \otimes \operatorname{id})(\bar{\Delta}(x)) = (\epsilon \otimes \operatorname{id})(\Delta(x)) - \epsilon(x) \otimes 1 - \epsilon(1) \otimes x;$$
since $x \in \bar{C} = \ker(\epsilon)$, $\epsilon(x) \otimes 1 = 0$; since $\epsilon \circ u = \operatorname{id}$, $\epsilon(1) = 1_K$; and by the counit relations,  $(\epsilon \otimes \operatorname{id})(\Delta(x)) = x$. So you're left with $x - 0 - x = 0$, and so $\bar{\Delta(x)} \in \bar{C} \otimes C$. Similarly $(\operatorname{id} \otimes \epsilon)(\bar{\Delta}(x)) = 0$ so $\bar{\Delta}(x) \in \bar{C} \otimes \bar{C}$.
A: You are correct about the identity $1=u(1_{\mathbb{K}})$.  For the second part, this should follow from the distributivity of tensor product over direct sum and the fact that, as an element of $\overline{C}$, $x$ has no nonzero part in $\mathbb{K}1$.
