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Let $\mathcal{C}$ and $\mathcal{D}$ be categories, and suppose $F\colon\mathcal{C}\to\mathcal{D}$ is a functor. It induces two adjoint pairs between $Set~^{\mathcal{C}}$ and $Set~^{\mathcal{D}}$; one is denoted $(F^\star,F_\star)$ and one is denoted $(F_!,F^\star)$. One proves easily that the counit to $(F^\star,F_\star)$ is a natural isomorphism of functors $\mathcal{C}\to Set$ if and only if $F$ is fully faithful.

I am interested in the counit of the other adjunction $F_!:Set~^{\mathcal{C}}\Longleftrightarrow Set^{\mathcal{D}}:F^*$. Lets denote it by $$\epsilon_F\colon F_!F^*\to \operatorname {id}_{Set^{\mathcal{D}}}.$$

Question: Under what conditions on $F$ is $\epsilon_F^~$ a natural isomorphism?

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  • $\begingroup$ Fixed a LaTeX problem. $\endgroup$
    – Harry Gindi
    Jun 28, 2010 at 22:29
  • $\begingroup$ There is still a TeX problem, and also an unfortunate need to read between the lines to understand the question. Reading between the lines, $F^*$ is composition with $F$, "counit to" is "counit of", and $F_*$ and $F_!$ are respectively the right and the left adjoint of $F^*$, yes? $\endgroup$
    – Tom Goodwillie
    Jun 28, 2010 at 22:37
  • $\begingroup$ Tom: yes, yes, and yes (respectively). $\endgroup$
    – David Spivak
    Jun 28, 2010 at 23:06
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    $\begingroup$ The question sounds extremely unnatural (thinking about the case when $F$ corresponds to a map of sheaf categories associated to a morphism of schemes, say using etale topologies). Is there a reason to expect an interesting example to satisfy such a conclusion? Basically, what is the reason for posing this question? $\endgroup$
    – BCnrd
    Jun 29, 2010 at 3:30
  • $\begingroup$ @BCnrd: I disagree. You should keep in mind, although topoi may have been invented in algebraic geometry, they live and breathe outside of it. This counit being an isomorphism is equivalent to the induced geometric morphism being a connected morphism of topoi. These generalize maps of topological spaces which have connected fibers to the world of topoi. $\endgroup$
    – David Carchedi
    Jun 29, 2010 at 18:15

2 Answers 2

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It appears to me that the condition on $F:\cal C\to\cal D$ would be:

For any morphism $s: a\to b$ of $\cal D$, the following category is connected:

An object consists of a $\cal C$-object $c$ and a factorization $a\to F(c)\to b$ of $s$.

A morphism $c_1\to c_2$ is a $\cal C$-morphism such that the induced map $F(c_1)\to F(c_2)$ is compatible with the maps from $a$ and to $b$.

I don't recall ever having run into this sort of 'two-sided comma category' before, but it seems to be the answer.

I got this by choosing $G$ from $\cal D$ to Set to be represented by the object $a$ and thinking about the fiber of the map $\epsilon: (F_!F^*G)(b)\to G(b)$ over the element $s$.

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  • $\begingroup$ Incidentally, this condition is exactly the same as the condition on a functor $F$ discussed here. Such functors $F$ have been called "lax epimorphisms" (though I'm not sold on the name) because they are exactly those functors $F: C \to D$ such that the functor $Fun(D,E) \to Fun(C,E)$, induced by precomposition with $F$, is fully faithful for any category $E$. They are also exactly those functors which are "cofinal for coends" as discussed at the link, along with other characterizations. $\endgroup$
    – Tim Campion ♦
    Jul 17, 2020 at 21:51
  • $\begingroup$ Note that the counit $F_! F^\ast \Rightarrow 1$ is an isomorphism iff $F^\ast$ is fully faithful, iff the unit $1 \Rightarrow F_\ast F^\ast$ is an isomorphism, so the question here was equally asking about when $F^\ast$ is fully faithful. $\endgroup$
    – Tim Campion ♦
    Jul 17, 2020 at 21:57
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In general, the counit of an adjunction is an isomorphism if and only if the right-adjoint is fully faithful (dually the unit is an iso iff the left-adjoint is fully-faithful). So, your question is easily seen to be equivalent to asking "When is $F^{*}$ fully-faithful? In topos-theory lingo, when is the induced geometric morphism $\mathbf{F}:Set^{C^{op}} \to Set^{D^{op}}$ satisfies $F^*$ is faithful, then $\mathbf{F}$ is said to be a SURJECTION of topoi. In this setting, this is equivalent to every object in $D$ being a retract of an object of the form $F(C)$.

Ok, so how about asking for $F^*$ to also be full? $F^*$ being faithful AND full means you are looking at what is called a CONNECTED geometric morphism of topoi. What properties $F$ do we need to ensure this? This is in general a hard problem. However, there are at least sufficient conditions. Given $F$, you first construct the category $Ext_{F}$ of "F-extracts"- these are quadruples $(U,V,r,i)$ with $U \in C$, $V \in D$, $r:FU \to B$, and $i:V \to FU$ such that $ri=1$, with the evident morphisms. There is a canonical functor $\tilde F:Ext_{F} \to D$ which sends $(U,V,r,i) \mapsto V$. Denote by $Ext_F(V)$ the fiber over $V$ of this functor. Then if $\tilde F$ is full and each $Ext_F(V)$ is a connected category, then $\mathbf{F}$ is a connected morphism.

This is in "Sketches of an Elephant" C.3.3.

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  • $\begingroup$ Thanks David. This is a really helpful answer. In particular, it answers another question I was preparing to think about, namely "when is the unit of the (F^*,F_*) adjunction an isomorphism?" You point out that the answers to this new question and the question above agree. $\endgroup$
    – David Spivak
    Jun 29, 2010 at 19:04
  • $\begingroup$ No problem :-). I'm curious, what is your intended application? Perhaps there's more that can be said. $\endgroup$
    – David Carchedi
    Jun 29, 2010 at 19:39
  • $\begingroup$ Hi David, the intended application is to "categorical information theory." One can consider a small category $C$ as a "database schema" and a $C$-set as data of that specification. A morphism of schemas is just a functor $f\colon C\to D$ between categories. It induces the two aforementioned adjunctions, each of which has real-world meaning. A user of a database $(D, x\colon D\to Set)$ might have access to a small part $C$ and he may update $f^\star x$. These updates in $C$ can be transported to $D$ using $f_!$ and $f_\star$. I'm looking at the effects of various kinds of updates. $\endgroup$
    – David Spivak
    Jun 29, 2010 at 20:51

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