# When monad preserves filtered colimits?

Given an adjunction pair $$(F, G)$$, the functor $$G \circ F$$ is a monad. My question is whether there exists certains conditions for which this monad perverses $$\lambda$$-filtered colimit?

Or maybe more concretely, when the adjunction pair are free functor and forgetful functor between the category of differential graded modules and the category of differential graded Lie algebras/ commutative differential graded algebras, does the monad preserves filtered colimits? If so, can you give me some hint to prove it?

Any ideas is welcome. Thank you in advance.

When $$(G,F)$$ is monadic, then $$G\circ F$$ preserves the same colimits as $$G$$ : in one direction it's easy to see, and in the other it's just checking that for any monad $$T$$, the forgetful functor from $$T$$-algebras preserves these colimits (as it is then "the same" as $$G$$), and that's done by building the colimits by hand.

For nonmonadic adjunctions, the question is not so easy; perhaps there is an answer (I'd like to see it) but I don't know it. If you look at e.g. the adjunction between the category of free abelian groups and the category of sets (the restriction of the usual one between abelian groups and sets) and the colimit of the directed diagram $$\mathbb{Z\overset{2}\to Z \overset{3}\to Z \overset{5}\to \dots}$$, its colimit in free abelian groups is the free group on no generators (this is the case because if we have a cocone to a free abelian group $$F(X)$$, then the image of the first $$\mathbb Z$$ is divisible by all integers, but $$F(X)$$ is residually finite so that's not possible unless that map is $$0$$, the same for the rest), whereas its colimit in abelian groups (and hence in sets) is $$\mathbb Q$$, so it's not preserved under the forgetful functor.

On the other hand, differential graded Lie or commutative algebras are (I think) monadic over differential graded modules, so it's equivalent to check that the forgetful functor preserves $$\lambda$$-filtered colimits; and this should be easier.

ADDED : Note first that the direction we're interested in here is the easy one : of $$G$$ preserves $$I$$-shaped colimits, then so does $$G\circ F$$: this simply follows as $$F$$ preserves all colimits. The point about monadic adjunctions was simply to say that it was in fact equivalent in our situation.

Here's a proof of that claim. It is clear that it suffices to show that for a monad $$T$$ on a category $$C$$, the forgetful functor $$U: C^T \to C$$ preserves all colimits that $$T$$ does (where $$C^T$$ is the category of algebras of $$T$$). For that we explicitly build colimits in $$C^T$$ : let $$I$$ be a category such that $$C$$ has $$I$$-shaped colimits and $$T$$ preserves them, and let $$D: I\to C^T$$ be a diagram.

Let $$L = \mathrm{colim}_I U\circ D$$ (which exists by assumption) and $$\iota_i D(i)\to L$$ the canonical maps (I'm abusing notation and writing $$D(i)$$ for the object of $$C$$ as well as for the $$T$$-algebra). The point is to try and construct a $$T$$-algebra structure on $$L$$ such that the $$\iota_i$$ are $$T$$-morphisms.

But now we also have a natural isomorphism $$\mathrm{colim}_I TD(i) \to TL$$ : the natural map is always defined, and it is an isomorphism because by assumption $$T$$ preserves colimits. So to define a map $$TL\to L$$ it suffices to define compatible maps $$TD(i) \to L$$. But that's easy : just composte $$TD(i)\to D(i)\to L$$ where the first map is the structure map for $$D(i)$$. These maps are compatible precisely because $$D$$ is a functor into $$T$$ -algebras, so the squares $$\require{AMScd}\begin{CD}TD(i) @>>> D(i) \\ @V{T(f)}VV @V{f}VV \\ TD(j) @>>> D(j)\end{CD}$$

commute for all $$f:i\to j$$ in $$I$$. Therefore this system of maps allows one to define $$\mathrm{colim}_I TD(i)\to L$$ so $$TL\to L$$.

Now the annoying bookkeeping comes : we have to check that this does define a $$T$$-algebra structure; and that this structure is in fact a colimit of $$D$$ in $$C^T$$. This is simply using uniqueness in universal properties, naturality of certain morphisms and preservation of $$I$$-colimits by $$T$$ a bunch of times; if you're not convinced you should do it as an exercise.

A full proof (no details skipped) should be in Borceux's Handbook of Categorical algebra (volume 1 or 2)

• Thank you very much, max. I think your answer is perfect for most people, but maybe I learn not well enough for filtered colimits, I have no idea how to building the colimit, and why the forgetful functor preserves filtered colimits? Could you please give me more details about the crucial technique or maybe a reference? Jul 19, 2019 at 3:23
• Sure, let me add a word about that. Note, however, that in the specific situation for differential graded things, you only care about the "easy" direction : if $U$ preserves these colimits then $U\circ F$ does too; but that's easy because $F$ preserves all colimits. I have nonetheless added a sketch of proof. Jul 19, 2019 at 10:23
• I really appreciate your help. So it suffices to prove the forgetful functor $F$ preserves filtered colimits. My idea is to equipped the colimit $L$ of the underlying chain complex diagram $F\circ D$ of $D: I\to\text{dgLie}$ with a lie structure by defining the value of $[\~x,\~y]$( where $\~x$ is the class of $x$) by $f([z_x,z_y])$. Here, $x\in D(i_x)$ and $y\in D(i_y)$ and there exists $i_z\in I$ and two maps in $I$, $p:i_x\to i_z$ and $i_y\to i_z$, let $z_x=D(p)(x)$ same for $z_y$. $f:F\circ D(i_z)\to L$ is canonical map. Is it true? Jul 20, 2019 at 11:02