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In answering this MO question, the issue was raised of characterizing when a given endofunctor $R:C\to C$ has the form $U\circ F$ where $F:C\to D$ is left adjoint to $U:D\to C$, i.e. which admit a monad structure. Is there an algebraic or purely categorical characterization of such $R$?

We know $F$ has to preserve colimits and $U$ has to preserve limits. I'd be happy with an answer saying $R$ is of the form $U\circ F$ iff $R$ preserves $\langle$ fill in the blank $\rangle$. This seems like it should be known classically (e.g. in Categories for the Working Mathematician), but I never learned it and feel like it would be useful to know.

From this MO question we know that in order to be a composition of adjoints $R$ must be a homotopy equivalence on the nerve $N(C)$. According to this MO question, the converse fails, so this is not a characterization.

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Straight out of Categories for the Working Mathematician, every monad is of this form, usually in more than one way. Wikipedia has a brief discussion: en.wikipedia.org/wiki/… –  Eric Wofsey Jan 13 '13 at 21:02
    
Agreed: $R$ is of this form iff $R$ admits a monad structure, and there are two canonical adjunctions that give rise to $R$, one from Kleisli and the other from Eilenberg-Moore. –  Todd Trimble Jan 13 '13 at 21:23
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@Eric, Todd: In response to your excellent comments, I've edited it to make clearer what I'm asking. Thanks for your help in turning this from a bad question into a (hopefully) interesting one. –  David White Jan 13 '13 at 22:43
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Somehow I feel the scope of the question is a still bit too general to admit good characterizations. A slightly less general question is: in a monoidal category, which objects admit a monoid structure? Even this feels too general. For specific monoidal categories, sometimes the answer is interesting (e.g., Eckmann-Hilton lemma), sometimes not so interesting IMO. –  Todd Trimble Jan 13 '13 at 23:09
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For example, for the monoidal category $(Set, \times)$, every nonempty set admits a monoid structure, for the following very silly reason: if $X$ has more than one element, then pick elements (denoted $0, 1$), declaring $1$ to be the identity, and defining $x \ast y = 0$ whenever neither $x$ nor $y$ is $1$. –  Todd Trimble Jan 14 '13 at 0:48
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1 Answer

up vote 5 down vote accepted

Characterizing endofunctors on various categories which admit a monad structure is the same as characterizing objects in various monoidal categories which have a monoid structure: "=>" If $(C,\otimes,\dotsc)$ is a monoidal category and $X \in C$, then $- \otimes X : C \to C$ has a monad structure iff $X$ has a monoid structure. "<=" If $C$ is a category and $T : C \to C$ is a functor, then monad structures on $T$ are precisely the monoid structures of $T$ in $(\mathrm{End}(C),\circ,\dotsc)$.

And as a topologist you probably know that it is hard to characterize monoid objects in $\mathrm{hTop}_*$ or $\mathrm{hTop}^{\mathrm{op}}_\*$ without any further requirements.

But I even doubt that there is any characterization of those endofunctors of $\mathsf{Set}$ which admit a monad structure. I only know of the following necessary condition: The endofunctor preserves epi-mono factorizations up to isomorphism (see Linton, Coequalizers in categories of algebras).

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Thanks for your answer. It does seem the question was too broad to admit a sharp characterization. Still, with your answer and the comments above there's a lot of useful information here. I'm satisfied. –  David White Jan 14 '13 at 16:21
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