I'm following Loday-Vallette's book Algebraic Operads, and I'm having some trouble understanding the quadratic cooperad associated to a quadratic data $(E, R)$.
They define the quadratic cooperad $\mathcal{C}(E,R)$ associated to the quadratic data $(E, R)$ with the following universal property: it is the universal cooperad $\mathcal{C}$ among the sub-cooperads of $\mathcal{T}^c(E)$ such that the composite $\mathcal{C} \to \mathcal{T}^c(E) \to \mathcal{T}^c(E)^{(2)}/(R)$ is zero.
What are the elements in $\mathcal{C}(E,R)$? My guess is that $\mathcal{C}(E,R) = \ker(\mathcal{T}^c(E) \to \mathcal{T}^c(E)^{(2)}/(R))$, am I right? I haven't been able to find any explicit description of $\mathcal{C}(E,R)$ -except from the weight zero, one and two components...
Moreover, I guess the decomposition $\Delta$ in $\mathcal{C}(E,R)$ is just the restriction of $\Delta_{\mathcal{T}^c(E)}$ (i.e. the sum of all admissible cuttings) to $\mathcal{C}(E,R)$, is this correct? Does it restrict, or we need to compose with a projection?
It's been more than two months since I posted the question, and even though nobody has answered it I believe I've figured it out on my own. Here I post my answer to my own question, hoping it can help someone in the same situation.
Recall the definition. If $E$ is an $\mathbb{S}$-module and $(R) \subseteq \mathcal{T}^c(E)$ is a cooperadic ideal, one defines $\mathcal{C}(E,R)$ as the cooperad satisfying the following universal property: For every subcooperad $\mathcal{C}' \subseteq \mathcal{T}^c(E)$ such that the projection to $\mathcal{T}^c(E)^{(2)}/(R)$ vanishes, there exists a unique morphism of cooperads $\mathcal{C}' \to \mathcal{C}(E,R)$ such that the composite $\mathcal{C}' \to \mathcal{C}(E,R) \subseteq \mathcal{T}^c(E)$ equals $\mathcal{C}' \subseteq \mathcal{T}^c(E)$.
It's not very difficult to see that this implies that the morphism $\mathcal{C}' \to \mathcal{C}(E,R)$ is injective. This means that $\mathcal{C}(E,R)$ is the biggest subcooperad of $\mathcal{T}^c(E)$ such that the projection to $\mathcal{T}^c(E)^{(2)}/(R)$ vanishes.
Finally, let us unwind the condition that the projection vanishes. That the projection $\mathcal{C}' \to \mathcal{T}^c(E)^{(2)}/(R)$ vanishes means that $\pi(\mathcal{C}') \subseteq (R)$, where $\pi: \mathcal{C}' \to \mathcal{T}^c(E)^{(2)}$. But $\pi(\mathcal{C}')$ is precisely the elements of $\mathcal{C}'$ of weight $2$, so we're asking that all the elements of weight $2$ in $\mathcal{C}'$ belong to the id al $(R)$.
We conclude: $\mathcal{C}(E,R)$ is the biggest subcooperad of $\mathcal{T}^c(E)$ such that every element of weight $2$ belongs to $(R)$.
Is $\mathcal{C}(E,R)$ the kernel of $\mathcal{T}^c(E) \to \mathcal{T}^c(E)^{(2)}/(R)$?
In general this question makes no sense, since the kernel needn't be a cooperad. What we do know (by definition) is that $\mathcal{C}(E,R) \subseteq \ker$. If the kernel turned out to be a cooperad (which is not always true), it would be the biggest subcooperad of $\mathcal{T}^c(E)$ satisfying the required property, so it would indeed coincide with $\mathcal{C}(E,R)$.