How to prove the reduced comultiplication of a coaugmented coalgebra is coassociative? a coalgebra over a field $k$ is a vector space $C$ over $k$ together with  $k$-linear maps
$$\text{comultiplication } \Delta: C \to C\otimes_k C \text{ and} $$
$$\text{counit } \epsilon: C \to k$$ such that

*

*$$(\mathrm{id}_C \otimes \Delta) \circ \Delta = (\Delta \otimes \mathrm{id}_C) \circ \Delta;$$


*$$(\mathrm{id}_C \otimes \epsilon) \circ \Delta = \mathrm{id}_C = (\epsilon \otimes \mathrm{id}_C) \circ \Delta.$$
A coalgebra $C$ is called coaugmented if there exists another $k$-linear map $u: k\to C$ such that $\epsilon\circ u=\mathrm{id}_k$.
For an coaugmented coalgebra $(C,k,\Delta,\epsilon, u)$, let $\bar{C}$ denote $\ker \epsilon\subset C$. It is clear that $C\cong \text{Im}(u)\oplus \ker \epsilon$ as a $k$-linear space. Moreover we can prove that for any $x\in \bar{C}$ we have
$$
\Delta(x)=x\otimes u(1_k)+u(1_k)\otimes x +\sum x^{\prime}\otimes x^{\prime\prime}
$$
where $x^{\prime}\otimes x^{\prime\prime}\in \bar{C}\otimes_k \bar{C}$. See Lemma 2.4 of this paper.
With the above result, it is natural to define the reduced comultipliciation $\bar{\Delta}:\bar{C}\to \bar{C}\otimes_k\bar{C}$ as
$$
\bar{\Delta}(x):=\Delta(x)-x\otimes u(1_k)-u(1_k)\otimes x.
$$
Now I want to prove that $\bar{\Delta}$ is coassiciative, i.e.
$$
(\mathrm{id}_\bar{C} \otimes \bar{\Delta}) \circ \bar{\Delta} = (\bar{\Delta} \otimes \mathrm{id}_\bar{C}) \circ \bar{\Delta}.
$$
By routine computation, it reduced to the identity
$$
\Delta(u (1_k))=u (1_k)\otimes u (1_k).
$$
Using Properties 1 and 2 above I can show that $\Delta(u (1_k))$ has no component in $\text{Im}(u)\otimes_k \bar{C}$ and $\bar{C}\otimes_k \text{Im}(u)$ and its component in $\text{Im}(u)\otimes_k \text{Im}(u)$ is $u (1_k)\otimes u (1_k)$. However I don't know how to prove $\Delta(u (1_k))$ has no component in $\bar{C}\otimes_k\bar{C}$, although it is true for the easies case: the tensor coalgebra.
It will be appreciated if you can let me know how to prove the coassociativity or if my argument is wrong.
 A: (Modified and Updated)
For your last concern, $u$ should be a coalgebra map. Guess this is why your computation does not follow. In standard references like Loday's Algebraic Operad and Peter May's More Concise Algebraic Topology, the coaugmentation map is always a coalgebra homomorphism. It follows $u(1)$ is also a group-like element, since $u$ is an coalgebra map, so $\Delta(u(1)) = (u \otimes u)(\Delta_k(1)) = u(1) \otimes u(1)$ and $\epsilon_C(u(1)) = \epsilon_k(1) = 1$). $u(1) \not\in \ker\epsilon$.
Back to your main question, this is straightforward computation without any trick and this has been more or less done in Augmented coalgebras.
The key in that computation is that $1$ is an group-like element. I thought the short comment that $u(1)$ is group-like should solve your question already. Looks like I need to give a detail answer in the answer section.
Here we assume $g$ is any group-like element and $\Delta$ is coassociative. Define $\bar\Delta(x) = \Delta(x) - g \otimes x - x \otimes g$. I assume the Sweddler notation.
$(\overline{\Delta} \otimes Id)\overline{\Delta}(x) = 
 x_{(11)} \otimes x_{(12)} \otimes x_{(2)} - x_{(1)} \otimes g \otimes x_{(2)} -  g \otimes x_{(1)}  \otimes x_{(2)}  - x_{(1)} \otimes x_{(2)} \otimes g + x \otimes g \otimes g + g \otimes x \otimes g + g \otimes g \otimes x$
$(Id \otimes \overline{\Delta})\overline{\Delta}(x) = 
x_{(1)} \otimes x_{(21)} \otimes x_{(22)} - x_{(1)} \otimes x_{(2)} \otimes g -  x_{(1)} \otimes g \otimes x_{(2)} + x \otimes g \otimes g -g \otimes x_{(1)} \otimes x_{(2)} + g \otimes x \otimes g + g \otimes g \otimes x $
$(\overline{\Delta} \otimes Id)\overline{\Delta}(x) - (Id \otimes \overline{\Delta})\overline{\Delta}(x) = x_{(11)} \otimes x_{(12)} \otimes x_{(2)} - x_{(1)} \otimes x_{(21)} \otimes x_{(22)} = 0$ since $\Delta$ coassociative.
Update: Here Well-definedness of reduced coproduct details on well-definedness of reduced comultiplication for your last concern.
