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 $T$?
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    $\begingroup$ 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. $\endgroup$ – Simon Henry Apr 12 '19 at 12:21
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    $\begingroup$ @SimonHenry Should $\mathcal{O}(n)$ be initial object of the category, actually? $\endgroup$ – Najib Idrissi Apr 12 '19 at 12:30
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    $\begingroup$ 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. $\endgroup$ – Simon Henry Apr 12 '19 at 12:42
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    $\begingroup$ @SimonHenry That's true, thanks. $\endgroup$ – 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 $n \neq 1$.

  • $\begingroup$ 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? $\endgroup$ – FKranhold Apr 12 '19 at 13:30
  • $\begingroup$ @FKranhold Yes, you're right. $\endgroup$ – Najib Idrissi Apr 12 '19 at 15:18

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