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Suppose we have $inc:C \rightarrow D$ a full subcategory and an adjunction $F:D\leftrightarrow C: inc$ where

  1. $C$ and $D$ are complete and cocomplete categories.
  2. $F$ is a left adjoint such that $F\circ F=F$
  3. $inc$ commutes with colimits.

Does $F$ commute with finite limits in general and with pullback and product in particular?

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  • $\begingroup$ Consider the full inclusion of sets into directed graphs (aka quivers) given by taking the discrete graph on a set. This functor $\Delta$ is full and faithful, hence exhibits Set as a full subcategory of Graph. It has a right adjoint given by the forgetful functor $\hom(1, -): \text{Graph} \to \text{Set}$, and a left adjoint $\pi_0$ which assigns to a graph its set of connected components. This $\pi_0$ does not preserve pullbacks. $\endgroup$
    – Todd Trimble
    Commented Oct 15, 2016 at 12:02
  • $\begingroup$ @ToddTrimble As far as I understand $\pi_{0}$ preserves products, right ? is it a general fact that $F$ in my question has to preserve the product ? $\endgroup$ Commented Oct 15, 2016 at 12:14
  • $\begingroup$ What does $F \circ F = F$ mean? Anyway, I think the question is a duplicate of mathoverflow.net/questions/243514 $\endgroup$
    – HeinrichD
    Commented Oct 15, 2016 at 12:29
  • $\begingroup$ It's clear that "$F \circ F = F$" isn't literally correct; rather, the counit $F \circ i \to 1$ is an isomorphism since $i$ is full and faithful, and it follows that $F i F \cong F$ (or that we have an idempotent monad, etc.). $\endgroup$
    – Todd Trimble
    Commented Oct 15, 2016 at 17:08

1 Answer 1

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In general $F$ preserves neither pullbacks nor even products. In a comment I mentioned that the "discrete graph" functor $\text{Set} \to \text{Set}^{\bullet \rightrightarrows \bullet}$ is full and faithful and has a left adjoint $\pi_0$ which sends a graph to its set of connected components and a right adjoint which sends a graph to its set of vertices. The functor $F = \pi_0$ preserves neither finite products nor equalizers. For products, consider the product of two copies of $0 \to 1$, which has four vertices but only one edge. For equalizers, consider the equalizer of an obvious pair of maps from $0 \to 1 \to 2$ to the "diamond" consisting of edges $S \to W \to N$ and $S \to E \to N$; the equalizing object is the discrete graph on vertices $0, 2$).

(In an earlier version of this answer, I had mistakenly said that $\pi_0$ preserves products, but I was conflating the case of (directed) graphs = quivers with the case of reflexive graphs, where the $\pi_0$ does preserve products.)

Similarly, let $\text{Set}^\mathbb{Z}$ denote the category of $\mathbb{Z}$-sets, i.e., sets equipped with an automorphism. The functor $\text{Set} \to \text{Set}^\mathbb{Z}$ which assigns to a set $S$ the trivial $\mathbb{Z}$-action is full and faithful, and has a right adjoint which assigns to a $\mathbb{Z}$-set $X$ its set of fixed points $X^\mathbb{Z} = \hom(1, X)$ under the action, and a left adjoint which assigns to $X$ its set of orbits = connected components $\pi_0(X) = X \otimes_\mathbb{Z} 1$. The functor $\pi_0$ does not preserve products, as we easily see looking at $X = \mathbb{Z} \times \mathbb{Z}$: here $\mathbb{Z} \times \mathbb{Z}$ has infinitely many orbits under the action by the automorphism $(m, n) \mapsto (m+1, n+1)$, but $\mathbb{Z}$ has only one.

In general, we can say that $F$ preserves the terminal object: if $1$ denotes the terminal in $D$, then we have a unit map $1 \to i F 1$ which in fact is an isomorphism (because any algebra for an idempotent monad such as $i F$ has this property). In that case, for any object $c$ of $C$ there is a unique map $i c \to i F 1$, but then by full faithfulness of $i$, there is a unique map $c \to F 1$.

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  • $\begingroup$ This result in the last paragraph is quite interesting and sort of unexpected. Any functor with a fully faithful right adjoint preserves the terminal object. Is there any high-level explanation why this should be true? In your proof, $1$ probably denotes a terminal in $D$. And that $1 \to iF1$ is an isomorphism is simply because it is inverse to the unique map $iF1 \to 1$. $\endgroup$
    – HeinrichD
    Commented Oct 16, 2016 at 7:32
  • $\begingroup$ @HeinrichD Thanks; I got the $C$ and $D$ mixed up (fixed now). I'm not sure I have anything high-level to add, except for my parenthetical remark on algebras of idempotent monads (they are characterized by a property, that the unit is an isomorphism). The high-level explanation might reside within the general yoga of stuff, structure, and property (ncatlab.org/nlab/show/stuff,+structure,+property), where idempotent monads govern property-like structure due to full faithfulness of $i$. Then combine that with the general observation that the terminal object is an algebra of any monad. $\endgroup$
    – Todd Trimble
    Commented Oct 16, 2016 at 13:45
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    $\begingroup$ Here is an intuitive way of putting it: think of your typical idempotent monad as being a "completion monad"; examples include the field of fractions monad on integral domains, or the Cauchy completion of metric spaces, or the Stone-Cech compactification of Tychonoff spaces. (In each of these cases, the unit embeds an object into its completion, so this wouldn't apply to the abelianization monad on groups. Never mind that for now.) Thus $F$ is a completion functor; an algebra of the completion monad is something already complete, so $F$ doesn't "add anything". Then observe $1$ is complete. $\endgroup$
    – Todd Trimble
    Commented Oct 16, 2016 at 14:44

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