# Free operad over a monoid object

Let $$\mathcal{O}$$ be an operad in the monoidal category $$M$$. Then $$\mathcal{O}(1)$$ together with the morphisms $$\mathcal{O}(1)\otimes \mathcal{O}(1)\to \mathcal{O}(1)$$ and the unit $$\eta:1\to \mathcal{O}(1)$$ is a monoid object. Moreover, a morphism $$\varphi:\mathcal{O}\to \mathcal{O}'$$ of operads induces a morphism $$\mathcal{O}(1)\to \mathcal{O}(1)$$ of monoid objects. Therefore, we get a forgetful functor $$\mathrm{Operads}(M)\to \mathrm{Monoids}(M).$$ If we work with coloured operads, we get a functor from coloured $$M$$-operads to $$M$$-enriched categories. Conversely, if $$T$$ is a monoid object, we can build an operad by $$\mathcal{O}_T(r) := T^{\otimes r}.$$ and the structure maps $$T^{\otimes r}\otimes \bigotimes_{i=1}^r T^{\otimes k_i}\to T^{\otimes (k_1+\dotsb+k_r)}$$ as follows: Let $$\Delta:T\to T^{\otimes k}$$ be the diagonal (existence is clear if $$\otimes$$ is the categorical product). Then $$T\otimes T^{\otimes k}\stackrel{\Delta\otimes \mathrm{id}}{\to}T^{\otimes k}\otimes T^{\otimes k} \cong (T^{\otimes 2})^{\otimes k} \to T^{\otimes k}.$$ In $$\mathbf{Set}$$, this just means $$t(t_1,\dotsc,t_k)=(tt_1,\dotsc,tt_k)$$. It should be clear that this construction gives us an operad. Now two problems/questions:

1. Does the morphism $$\Delta$$ always exist in the general setting? It is obviously not the same as $$T\cong T\otimes 1^{\otimes (k-1)}\stackrel{\mathrm{id}\otimes \eta^{\otimes (k-1)}}{\to} T^{\otimes k}.$$
2. Obviously $$\mathcal{O}_T(1)\cong T$$, but it seems to be not true that $$\mathcal{O}_T$$ is the free operad over the monoid object $$T$$. Is there another construction for the “free” operad over $$T$$?
• The Free operads over $T$ is alot simpler than this: it has $\mathcal{O}(1)=T$ and all the $\mathcal{O}(n)$ for $n>1$ are the unit. – Simon Henry Apr 12 '19 at 12:21
• @SimonHenry Should $\mathcal{O}(n)$ be initial object of the category, actually? – Najib Idrissi Apr 12 '19 at 12:30
• Yes, you're right, sorry ! And one needs the tensor product to preserves the initial object in each variable, I'm not sure what happen otherwise. – Simon Henry Apr 12 '19 at 12:42
• @SimonHenry That's true, thanks. – Najib Idrissi Apr 12 '19 at 12:50

Let me mention that this is related to this earlier question of mine (which is unanswered :-( ) and more generally to semi-direct products of operads by bialgebras. You construction is the semi-direct product $$\mathtt{Com} \rtimes T$$ of the commutative operad $$\mathtt{Com}$$ by $$T$$.
1. No, there is no diagonal in general. You need a cocommutative bimonoid object, i.e. an object equipped with a multiplication $$\mu : T \otimes T \to T$$ and a comultiplication $$\Delta : T \to T \otimes T$$ such that $$\mu$$ is associative and unital, $$\Delta$$ is coassociative, cocommutative and counital, and they satisfy a compatibility relation $$\Delta \circ \mu = \mu \circ (\Delta \otimes \Delta)$$.
2. I'll assume that by "free operad" you mean a left adjoint to the forgetful functor. Let $$\varnothing$$ be the initial object of your category and suppose that $$\varnothing \otimes X = \varnothing = X \otimes \varnothing$$ for all $$X$$. Then (as Simon Henry mentions in the comments), the free operad on $$T$$ is simply given by $$\mathtt{O}(1) = T$$ and $$\mathtt{O}(n) = \varnothing$$ for $$n \neq 1$$.
• Thank you! Am I correct that you mean the compatibility $\Delta\circ \mu = (\mu\otimes \mu)\circ (\mathrm{id}\otimes \beta\otimes\mathrm{id})\circ (\Delta\otimes\Delta)$ where $\beta:T^{\otimes 2}\to T^{\otimes 2}$ is the symmetric braiding? – FKranhold Apr 12 '19 at 13:30