The following apparently simple exercise is driving me mad.

What I want to show is that, given a monad $T$ on $\bf C$, then every $F \colon \bf C \to D$ induces a monad $\text{Ran}_F (FT)$ on $\bf D$. I tried the direct path but I'm not sure this is the smartest choice.

1. Call $S=\text{Ran}_F(FT)$; then the universal property of $\text{Ran}_F$ gives $$\text{Nat}(SSF, FT) \cong \text{Nat}(SS,S)$$ so I'm led to show that the mate of the arrow $SSF\xrightarrow{S\sigma}SFT\xrightarrow{\sigma T} FTT \xrightarrow{F\mu}FT$ is the multiplication of a monad (and that the mate of $F\to FT$ is ist unit), if $\sigma\colon SF\to FT$ denotes the counit of the adjunction $F^*\dashv \text{Ran}_F$, and $\mu\colon TT\to T$ is the multiplication of the given monad. How do I have to treat such a monster as the diagram $$\begin{array}{ccc} SSS & \to & SS \\ \downarrow && \downarrow \\ SS &\to & S \end{array}$$ (I don't even dare to write the arrows explicitly!)
2. Looking for a more conceptual argument, one is led to show that the correspondence $\hat E=\text{Ran}_F(F-)\colon \bf[ C,C]\to [D,D]$ is a lax functor; if this is true, the functor $\hat E$ restricts to a functor between monoids in both categories, giving the result. This boils down to show that the mate of the arrow $$\hat E(P) \hat E(Q) E \to \bar E(P) E Q \to E PQ$$ enjoys the properties of the laxity cell of a lax monoidal functor. Same associativity, different context, so I'm left with the same .

I am sure there is a smart way to do these computations, but I'm not able to find it. A dry and conceptual argument to solve this problem is welcome, but in the end I'd rather see an explicit computation, or a clever way to sistematically avoid such boring checks.

• The right Kan extension doesn't have to exist (even if $C,D$ are complete, since the index categories may be large). So probably we just have to assume the existence here? I think the argument should be easy if we would have pointwise right Kan extensions. – Martin Brandenburg Jan 25 '14 at 20:59
• You can assume whatever you need. At this point I'm more curious to know how to finish the computation than to find the sharpest hypotheses! (btw, arguing pointwise to my eyes involves an extremely big, nested end: is it what you mean too?) – Fosco Jan 25 '14 at 21:12
• The pointwise right Kan extension has a very simple limit expression, and limits have an easy universal property. This is how I would test the commutativity of the diagrams: Compose with the limit projections, and use the given monad diagrams. I am pretty sure that nothing more happens here. – Martin Brandenburg Jan 25 '14 at 21:40
• This is by no means intended to bother you, but... I'm completely aware of what you say. And nevertheless I'm not able to conclude, simply because the diagrams I have to use are simply HUGE, whatever the universal property I use to reduce them. I'm convinced I'm missing a trick, or a conceptual general method (e.g.: "Whenever you have to check that the mate of an arrow fits into a c.d., do the following[...]"). – Fosco Jan 26 '14 at 13:36
• (...) This s16.postimg.org/ik5wtsmr9/CAM00046.jpg is the diagram I have to check the commutativity of, to ensure associativity of the multiplication. And it's not even in its final form! If you want to "compose it with the limit projections and use the given monad diagrams" you're welcome (and this is by no means an ironic or offensive tone), but honestly I'm not able to simplify it. – Fosco Jan 26 '14 at 13:37

I assume that $S$ exists pointwise, i.e. $$S(d) = \lim\limits_{d \to F(c)} F(T(c)).$$ For every $i : d \to F(c)$ we have the projection $$p(i) : S(d) \to F(T(c)).$$ Notice that we can view $p(i)$ as a "new $i$" and form $p(p(i)) : S(S(d)) \to F(T(T(c))$, and likewise $p(p(p(i))) : S(S(S(d))) \to F(T(T(T(c))))$.

We have a canonical morphism $\alpha : SF \to FT$ such that commutes, namely $\alpha_c := p(\mathrm{id}_{T(c)})$. We define $$\overline{\mu}_d : S(S(d)) \to S(d)$$ by the requirement that for all $i$ the diagram is commutative (one checks existence using the universal property of $S(d)$). Then $\overline{\mu} : SS \to S$ is a morphism. We claim that $\overline{\mu}$ is associative. Look at the following huge diagram: Now let us check commutativity:

• The two triangles commute because of $(\star)$ (applied to $p(i)$ resp. $p(p(i))$ in place of $i$).
• The four outer trapezoids commute by the very definition of $\overline{\mu}$.
• The middle square on the left commutes because $\alpha$ is natural.
• The middle square on the right commutes because $\mu$ is associative.

Since $(S(d) \xrightarrow{p(i)} F(T(c)))_i$ is a limit cone, a diagram chase shows that also the outer rectangle commutes. But this says that $\overline{\mu}$ is associative.

I haven't checked it, but probably the unit business will be quite similar: One takes the unit diagram for $S$ and decomposes it into smaller diagrams, including the unit diagram for $T$. Commutativity will follow in each case either from the definitions or from naturality.

• C'est parfait. Thank you. – Fosco Jan 28 '14 at 19:13

The functor you are considering is a lax monoidal functor between the (strict) monoidal categories $[\mathbf C, \mathbf C]$ and $[\mathbf D,\mathbf D]$ (proving that is not difficult but is extremely tedious, it's just proving that some equations holds).

Theorems about lax monoidal functor tells us that lax functors preserve monoids hence it follows.

I've already written elsewhere the computations (or at least a good part of that). If you're interested you can find in this link and this link. Unfortunately they are in italian, but I don't think it would really worth rewrite them here.

Hope this helps.