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 $T$?
  • 6
    $\begingroup$ 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. $\endgroup$ Apr 12, 2019 at 12:21
  • 3
    $\begingroup$ @SimonHenry Should $\mathcal{O}(n)$ be initial object of the category, actually? $\endgroup$ Apr 12, 2019 at 12:30
  • 2
    $\begingroup$ 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. $\endgroup$ Apr 12, 2019 at 12:42
  • 2
    $\begingroup$ @SimonHenry That's true, thanks. $\endgroup$ Apr 12, 2019 at 12:50

1 Answer 1


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, 2019 at 13:30
  • $\begingroup$ @FKranhold Yes, you're right. $\endgroup$ Apr 12, 2019 at 15:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.