In their paper Strongly homotopy Lie algebras, Lada and Markl first show, that there is a symmetrization functor $(-)_L:\mathcal{A}(m)\rightarrow \mathcal{L}(m)$ from the category of $A(m)$-algebras to the category of $L(m)$-algebras. Then there is a left adjoint $\mathcal{U}_L$for that functor. They define the **universal enveloping $A(m)$-algebra** for a $L(m)$-algebra $L$ to be $\mathcal{U}_L(L)$.

In addition the category of $A(m)$-algebras has a strict monoidal structure, which is given by taking the free presentations $\mathcal{F}_m(X_A)/R_A$ and $\mathcal{F}_m(X_B)/R_B$ for two $A(m)$-algebras $A\equiv(X_A,\left\lbrace \mu_i\right\rbrace),B\equiv(X_B,\left\lbrace \tilde{\mu}_i\right\rbrace)$ and define \begin{equation} A\Box B\equiv \mathcal{F}_m(X_A\oplus X_B)/(R_A,S_{AB},R_B) \end{equation}

where $S_{AB}$ are the relations generated by $\sum_\sigma \chi(\sigma) \lambda_i(x_{\sigma(1)},\dots,x_{\sigma(i)})=0$ ($\lambda_i$ are the i-ary operations for $A\Box B$). For two $L(m)$-algebras $L_1,L_2$ it holds that \begin{equation} \mathcal{U}_L(L_1\oplus L_2)\cong \mathcal{U}_L(L_1)\,\Box\, \mathcal{U}_L(L_2) \end{equation} and that the homomorphism $\Delta:\mathcal{U}_L(L)\rightarrow \mathcal{U}_L(L\oplus L)\cong \mathcal{U}_L(L)\,\Box\, \mathcal{U}_L(L)$ gives the universal enveloping $A(m)$ algebra the structure of coassociative cocommutative coalgebra in the category $(\mathcal{A}(m),\Box,1)$. Here $1$ is the unit $A\Box 1\cong 1\Box A\cong A$ obtained from taking the free $A(m)$ algebra for the trivial vector space.

My question is, if this is the same as having a $A_\infty$-coalgebra structure in the sense of operations $\left\lbrace \omega^i:X\rightarrow X^{\otimes i}\right\rbrace $ on the underlying vector space of $\mathcal{U}_L(L)$?

If this holds true, are the $A_\infty$-algebra and coalgebra structures automatically compatible, i.e. is $\mathcal{U}_L(L)$ a $A_\infty$-bialgebra?