Thinking about whether or not there is a natural way to transform $L_\infty$-algebras into $A_\infty$-algebras, I wonder if there is a morphism of cooperads
$\mathcal{A}ss^i\to\mathcal{L}ie^i$
from the Koszul symmetric, associative cooperad $\mathcal{A}ss^i$ to the Koszul Lie cooperad $\mathcal{L}ie^i$. I think there is no such map and my argument is as follows:
Suppose there is such a map. Since $\mathcal{A}ss^i\simeq \mathcal{S}\otimes_H\mathcal{A}ss^*$ and $\mathcal{L}ie^i\simeq \mathcal{S}\otimes_H\mathcal{C}om^*$ (with $\otimes_H$ the tensor product of cooperads), this would imply that there is although a cooperad map
$id_\mathcal{S}\otimes_H \theta : \mathcal{S}\otimes_H\mathcal{A}ss^* \to \mathcal{S}\otimes_H\mathcal{C}om^*$, which in turn would mean that the map
$\theta : \mathcal{A}ss^* \to \mathcal{C}om^*$ exists, too. This however is not true, since both $\mathcal{A}ss$ as well as $\mathcal{C}om$ are finite dimensional in each arity and there is no morphism of operads $\mathcal{C}om\to\mathcal{A}ss$.
Is this a valid argument?
