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I'd like to know what's the name (if any) of the following categorical structure, and also references where it has been considered.

Given a category $C$, let $O=\{O(n)\}_{n\geq 0}$ be a sequence of functors $O(n)\colon C\times\stackrel{n}\cdots\times C\rightarrow C$ equipped with natural transformations $O(n)(O(p_1),\dots,O(p_n))\Rightarrow O(p_1+\cdots + p_n)$ satisfying the usual relations (as for operads). (A unit natural transformation $id_C\Rightarrow O(1)$ may be added.)

If $C$ is symmetric monoidal and $P$ is an operad, then the sequence of functors $(X_1,\dots,X_n)\mapsto P(n)\otimes X_1\otimes\cdots\otimes X_n$ fits in the previous setting.

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    $\begingroup$ I've heard these being called "lax monoidal categories", "skew-monoidal categories" and "corepresentable non-symmetric (colored) operads". (But don't quote me on that without checking that the definitions exactly match.) Of those names, my favorite is "lax monoidal categories". $\endgroup$ Jun 1, 2014 at 14:54
  • $\begingroup$ @OmarAntolín-Camarena Thanks, Omar! I'm still interested in the name "corepresentable non-symmetric (colored) operads". Google is not helping much. Do you remember the place where you've seen this (or any similar) name for this structure? I'm interested in considering different such structures on a fixed category $C$, and I suspect that people thinking of this as generalized operads may have studied this before. $\endgroup$ Jun 1, 2014 at 18:52
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    $\begingroup$ One correction: I only said "non-symmetric" because I guessed incorrectly which operad axioms you meant. If you meant to include the symmetric group actions, then the names I've heard are "lax symmetric monoidal categories" and "corepresentable operads". The latter is Lurie's terminology, see Higher Algebra 7.2.4.3, for instance. The idea is simple: you can characterize which colored operads (also called symmetric multicategories) arise from the sort of construction you mention in the last line of your question; a slight generalization thereof gives you corepresentable operads. (to be cont'd) $\endgroup$ Jun 1, 2014 at 20:28
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    $\begingroup$ I was about to write what the definition through operads was when I remembered this is nicely explained in Tom Leinster's book that Eduardo already mentioned, see theorem 3.3.4. In the terminology used there: a monoidal category is a multicategory where every sequence of object is the domain of a universal arrow, and a lax monoidal category is a multicategory where every sequence of objects is the domain of a pre-universal arrow. $\endgroup$ Jun 1, 2014 at 20:43
  • $\begingroup$ @OmarAntolín-Camarena You're right in your guess that I was thinking of nonsym ones. Thanks for the reference to Lurie. I'll have to unwrap that definition though. $\endgroup$ Jun 2, 2014 at 10:35

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If I'm getting right what you mean by "satisfying the usual relations (as for operads)" then your structure is what Tom Leinster calls in his Higher Operads, Higher Categories book an unbiased lax monoidal category.

They can be seen as (lax) algebras for a $\mathbf{Cat}$-enriched operad:

First, start with the terminal object $1$ in $\mathbf{Set}^{\mathbb{N}}$ and build the free $\mathbf{Set}$-operad on it $F1$; this is generated by one operation per arity, but you get an awful lot of freely derived operations: $F1(n)$ is the set of planar trees with $n$ leaves. Now, make all the sets $F1(n)$ categories by taking the indiscrete category on them $I(F1(n))$. $I\colon \mathbf{Set} \to \mathbf{Cat}$ preserves products so this gives a $\mathbf{Cat}$-operad; lax algebras for it are unbiased lax monoidal categories.

All this is in Leinster's book, in the chapter about algebraic definitions of monoidal categories.

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    $\begingroup$ ¡Gracias Eduardo! $\endgroup$ Jun 1, 2014 at 18:50
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    $\begingroup$ De nada @FernandoMuro :) There's also Brian Day and Ross Street paper Lax monoids, pseudo-operads and convolution. There they call them lax monoidal categories, as an example of lax monoids in a Gray monoid (Cat in this case) $\endgroup$ Jun 1, 2014 at 19:33
  • $\begingroup$ por cierto, it seems I'm in the 'biased' case. $\endgroup$ Jun 2, 2014 at 15:13

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