In the definition of operads, if we restrict our attention to $\mathbb{S}$-modules where the action by the symmetric groups is free, then the *free operads* 
have still an underlying free $\mathbb{S}$-module? Even the colimits over this kind of operads have still an underling free $\mathbb{S}$-module?


And finally, in this kind of symmetric operads, the free operad construction (using trees) will be much simpler right? I mean if it is similar to the one for the non-symmetric case (where essentially we only need to label every vertex with the elements of the $\mathbb{S}$-module)? In the sense that we don't need all the technical combinatorial details about the behavior of $\mathbb{S}$-modules.


I'm not an expert in operads but these questions came to me when i was reading the book **Algebraic Operads** of Bruno Vallette and Jean-Louis Loday. In Section 5.5, the free operad construction is described.



Free Symmetric Operads and $\mathbb{S}$-modules
1. What you say about free operads on free $\mathbb{S}$-modules being themselves free $\mathbb{S}$-modules and being describable in terms of the free non-symmetric operad is correct.


In section 5.9.11 of Loday–Vallette, they construct adjoint functors between ns operads and (symmetric) operads. The free ns operad and free symmetric operad functors also admit right adjoint forgetful functors. Finally, the a free $S$-module functor on arity graded modules also has a forgetful adjoint.


The forgetful functors commute by inspection:
$$
\require{AMScd}
\begin{CD}
Op@>>>ns\ Op\\
@VVV @VVV\\
\mathbb{S}\text{-}mod@>>>\mathbb{N}\text{-}mod
\end{CD}
$$
and so the left adjoint free functors commute
$$
\require{AMScd}
\begin{CD}
Op@<{\otimes \mathbb{K}[\mathbb{S}\_n]}<<ns\ Op\\
@A{F}AA @A{F\_{ns}}AA\\
\mathbb{S}\text{-}mod@<{\otimes\mathbb{K}[\mathbb{S}\_n]}<<\mathbb{N}\text{-}mod
\end{CD}
$$
and thus the free operad $F(M)$ for $M\cong N\otimes \mathbb{K}[\mathbb{S}\_n]$ is canonically isomorphic to $F\_{ns}(N)\otimes \mathbb{K}[\mathbb{S}\_n]$.




---


2. As explained in my comment, this is no longer true for colimits of free operads on free $\mathbb{S}$-modules. Take a free $\mathbb{S}$-module $M$ and an endomorphism $f$ of $M$ whose cokernel are not free $\mathbb{S}$-modules. For instance, let $M$ be a free $\mathbb{S}\_2$-module and let $f$ be symmetrization or skew-symmetrization.


Then the pushout of the diagram
$$
\require{AMScd}
\begin{CD}
F(M)@>{F(f)}>>F(M)\\
@VVV \\
0
\end{CD}
$$
is the free operad on the cokernel of $f$ (in our example, the trivial or sign representation of $\mathbb{S}\_2$), and in particular is not free as an $\mathbb{S}$-module.


