I think this is probably elementary, but some searching (and asking on the chatroom) hasn't turned up a result. Could anyone point me to a reference for (or counterexample to) the following statement?

Given categories $\mathcal{C}, \mathcal{D}, \mathcal{E}$ and comonadic adjunctions $\mathcal{C} \rightleftarrows \mathcal{D}, \mathcal{D} \rightleftarrows \mathcal{E}$, the composite adjunction is comonadic.

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This is not true in general. Note that by passage to the opposite category, your question is equivalent to asking that a composite of monadic functors is monadic. A counterexample for that is the following:

The category $\mathbf{Cat}$ of small categories is monadic over the category $\mathbf{Grph}$ of (directed) graphs (with loops), and $\mathbf{Grph}$ is monadic over $\mathbf{Set}$. But there is no monadic functor from $\mathbf{Cat}$ to $\mathbf{Set}$. Details can be found on page 107 of "Toposes, Triples and Theories" by Barr and Wells.

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Are graphs really monadic over $\mathbf{Set}$? I would have expected $\mathbf{Set} \times \mathbf{Set}$. – Zhen Lin Jan 7 '14 at 16:41
Well, it is also monadic over $\mathbf{Set}\times \mathbf{Set}$, but the functor which sends a graph to its set of arrows is also monadic. In fact, there is a three element monoid in $\mathbf{Set}$ whose category of actions is equivalent to graphs (see exercise (GRMN) on page 107 of "TTT"). – Daniel Schäppi Jan 7 '14 at 16:58
Sorry, the functor in question sends a graph to the disjoint union of its set of arrows and its set of vertices, not just its set of arrows. The two nontrivial elements $s,t$ of the monoid act trivially on the set of objects, and send arrows to their source and target respectively. The usual relations that source of the target is target etc. have to be imposed. – Daniel Schäppi Jan 7 '14 at 17:14
Thank you for clearing up a misconception of mine! – Akhil Mathew Jan 7 '14 at 18:25
Right, the usual example is that $\mathbf{Cat}$ is monadic over reflexive graphs, and reflexive graphs are monadic over $\mathbf{Set}$ by sending a reflexive graph to its set of arrows; indeed the category of reflexive graphs is equivalent to the category of actions of the monoid with two generators $s, t$ subject to the relations $s s = s = t s$ and $s t = t = t t$. – Todd Trimble Mar 6 at 16:54

As Daniel Schäppi pointed out, this is the same as asking whether the composite of two monadic functors is monadic. The answer, unfortunately, is no.

Consider locally presentable categories. By the classification theorem, every locally presentable category can be embedded as a reflective subcategory of some presheaf topos. It is not hard to see that a fully faithful functor with a left adjoint is automatically monadic. On the other hand, $[\mathcal{A}^\mathrm{op}, \mathbf{Set}]$ is monadic over $\mathbf{Set}^{\operatorname{ob} \mathcal{A}}$, as one might expect for a multi-sorted algebraic theory. Thus, for any locally presentable category $\mathcal{C}$, there exist a small category $\mathcal{A}$ and monadic functors $$\mathcal{C} \to [\mathcal{A}^\mathrm{op}, \mathbf{Set}] \to \mathbf{Set}^{\operatorname{ob} \mathcal{A}}$$ but in general the composite $\mathcal{C} \to \mathbf{Set}^{\operatorname{ob} \mathcal{A}}$ is not monadic.

Indeed, it can be shown that any category monadic over $\mathbf{Set}^I$ (for some set $I$) must be an effective regular (= Barr-exact) category. But this is not true for a general locally presentable category, such as $\mathbf{Cat}$ or $\mathbf{Poset}$.

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Thank you for another interesting answer! – Akhil Mathew Jan 7 '14 at 18:24

Here is a positive result: Let $\mathcal{C} \to \mathcal{D}$ and $\mathcal{D} \to \mathcal{E}$ be two monadic functors. Assume that the underlying functor of the first monad $\mathcal{D} \to \mathcal{D}$ preserves reflexive coequalizers (this happens quite often!). Then, the composition $\mathcal{C} \to \mathcal{D} \to \mathcal{E}$ is monadic, too. This is sketched here. (Does anybody know a canonical reference for this result?) There you can also find another counterexample by Jim Dolan.

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Don't know where this precise result appears, but it's closely akin to the crude monadicity theorem. – Todd Trimble Mar 6 at 16:57