Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

A functor $C \to D$ between categories induces a morphism of presheaf categories $Pre(D) \to Pre(C)$. This functor has a left adjoint given by left Kan extension and I am interested in knowing when this left adjoint preserves pull-back squares.

I'm interested in any conditions that make this happen, but I am particularly interested in a special case. Let me say a little more about the context I am working in, and why I am interested. In the situation that this came up $C$ is a "lluf" subcategory of $D$, that is $C$ has the same objects as $D$ and the functor $C \to D$ is faithful. In that case it is good to call the functor $U:Pre(D) \to Pre(C)$ the forgetful functor. It automatically preserves limits and colimits. Let L be its left-adjoint.

Since C has the same objects as D, this forgetful functor is also conservative, meaning that it reflects isomorphisms. So by general non-sense (specifically Beck's monadicity theorem) this is a monadic adjunction. This means that $Pre(D) = T-alg$ is the category of T-algebras in $Pre(C)$ where T is the monad $T= UL$.

I'm trying to understand conditions under which this monad is cartesian in the sense described at the n-lab. This means, among other things, that the monad T is supposed to send fiber products to fiber products. This is equivalent to having L send fiber products to fiber products. I want to understand when this happens. Does it always happen? Are there reasonable conditions on C or D that ensure that this happens?

Notice that I am not asking for L to be "left-exact", i.e. to commute with all finite limits. This property is generally much too strong. In particular L will not usually preserve terminal objects. This means it doesn't preserve products, but should instead send products to fiber products over $L(1)$.

Here is an example. Let $C = pt$ be the singleton category and $D = G$ be the one object category with morphisms a group G. There is a unique inclusion $C \to D$ which is obviously faithful. The forgetful functor $$U:Pre(D) \to Pre(C)$$ sends a G-set to its underlying set. The left adjoint L sends a set S to the free G-set $S \times G$. This doesn't preserve terminal objects, but it does commute with fiber products. What is more, the monad $T=UL$ is a classic example of a cartesian monad in the n-lab sense.

I've played around with this, but can't seem to get it to work. I feel like this is going to be a well known result or there is going to be a counter example which sheds light on the situation.

Question: In the context I described above (where $C \to D$ is lluf), does the left adjoint $$L: Pre(C) \to Pre(D)$$ always commute with fiber products? If not what is a counter example, and are there conditions one can place on C and D to ensure that L does commute with fiber products?

share|improve this question
    
Here's an idea for an approach; I don't have time to work it through now, but maybe someone else can? First: "preserves pullbacks"="preserves finite connected limits". Now, the left adjoint $L = f \otimes -$ preserves finite limits iff the original functor $f$ is flat, i.e. "has (co?)filtered (co?)commas". Now, "filtered"="has a cocone under every finite diagram". What if we replace this with "…every finite connected diagram"? Will this condition be equivalent to $f \otimes -$ preserving finite connected limits? I suspect this won't quite work, but that something similar will. –  Peter LeFanu Lumsdaine Aug 3 '10 at 22:36

2 Answers 2

up vote 3 down vote accepted

$\newcommand{\C}{\mathbf{C}} \newcommand{\D}{\mathbf{D}} \newcommand{\Lan}{\mathrm{Lan}} \newcommand{\yon}{\mathbf{y}} \newcommand{\CC}{[\C^\mathrm{op},\mathbf{Set}]} \newcommand{\DD}{[\D^\mathrm{op},\mathbf{Set}]}$ Expanding on my comment above:

Define: a category is semi-filtered iff every pair of arrows $x \leftarrow z \rightarrow y$ can be completed to a commutative square, and every parallel pair of arrows $f,g \colon x \to y$ have some $h : y \to w$ with $hf = hg$; equivalently, if every finite connected diagram has some co-cone under it. (Afaik, this isn't standard terminology; I don't recall seeing this property discussed, but it almost certainly has been.)

It's filtered if moreover it's non-empty and every pair of objects $x,y$ is connected by some $x \to w \rightarrow y$; equivalently, if every finite diagram has some co-cone under it.

Answer: for $f \colon \C \to \D$, the left Kan extension $f_* \colon \CC \to \DD$ will preserve pullbacks (equivalently, all connected finite limits) exactly if the opposite of its every comma category $(f \downarrow d)$ is semi-filtered.

This is a close variant of the standard lemma (see e.g. Mac Lane and Moerdijk Sheaves in Geometry and Logic) that $f_*$ preserves pullbacks and the terminal object (equivalently, all finite limits) iff the opposite of every $(f \downarrow d)$ is filtered, i.e. if $f$ is flat.

Proof: the values of $f_*$ can be computed as colimits over the opposites of comma categories (see this answer). But in $\mathbf{Set}$, finite limits commute with filtered colimits; and similarly, pullbacks commute with semi-filtered colimits.

(The first of these facts is standard. The second follows because in a semi-filtered category, each connected component is filtered; so a semi-filtered colimit is a coproduct of filtered colimits; and pullbacks commute with both coproducts and filtered colimits.)

share|improve this answer
    
Mac Lane (CWM exercise IX.2.2) calls your property 'pseudo-filteredness' and attributes it to Verdier. –  Finn Lawler Aug 6 '10 at 20:16

Here's another way of getting to the same answer as Peter's. A functor $F\colon A\to B$ preserves pullbacks if and only if the induced functor $F/1 \colon A \to B/F1$ preserves all finite limits, where 1 is the terminal object of A. When F is left Kan extension $L\colon Psh(C) \to Psh(D)$ along a functor $f\colon C\to D$, it's not hard to check that Psh(D)/L1 is equivalent to presheaves on the opposite of the category el(L1) of elements of L1 (this is true with any presheaf replacing L1), and that L/1 is left Kan extension along the induced functor $f'\colon C\to el(L1)^{op}$. Thus, we need to know when that functor is flat.

Now an object of $el(L1)^{op}$ is an object $d\in D$ together with a connected component, call it X, of the comma category $(d\downarrow f)$. And a morphism in $el(L1)^{op}$ is a morphism $d_1\to d_2$ such that the induced functor $(d_2\downarrow f) \to (d_1\downarrow f)$ maps $X_2$ to $X_1$. You can then check that the comma category $((d,X)\downarrow f')$ is precisely the connected component X of the comma category $(d\downarrow f)$. Therefore, since $f'$ is flat just when all categories $((d,X)\downarrow f')$ are cofiltered, we conclude that left Kan extension along f preserves pullbacks iff all connected components of all comma categories $(d\downarrow f)$ are cofiltered, i.e. if all $(d\downarrow f)$ are "semi-filtered" in Peter's terminology.

Edit: you also asked for a specific counterexample when $f\colon C\to D$ is the inclusion of a lluf subcategory. Let D be the walking commutative square generated by arrows $a\to b$, $a\to c$, $b\to d$, and $c\to d$, and let C be its lluf subcategory containing the identities and the arrows $b\to d$ and $c\to d$. Then the comma category $(a\downarrow f)$ has two connected components, one of which is not semi-cofiltered.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.