Dear Colleagues, let F:C -> D be a functor between two small categories. Its induced functor F^*:Fun(D,Set) -> Fun(C,Set) has both a left adjoint L and a right adjoint R. 1. Under which precise conditions on F is L F^* = Id. 2. Under which precise conditions on F is R F^* = Id. 3. Under which precise conditions on F is F^* R = Id. 4. Under which precise conditions on F is F^* L = Id. Does anybody know any answers to one of these queries? What happens if equality is replaced by natural equivalence? Does anybody know of good references to these or similar problems? Markus Pfenniger Andy Tonks School of Mathematics School of Mathematics Dean Street Dean Street Bangor, LL57 1UT Bangor, LL57 1UT UK UK ==============================================================================
The first thing to note is that your question, as stated, is meaningless. The proper statement is that the induced functor has left and rights adjoints. If for one of them, one of the composites you mention is naturally equivalent to the identity, then another could be chosen (using AC) for which the composite in question was equal to the identity. On the other hand, in the usual set theory, the usual construction that shows that adjoint exists will use more complicated sets than the functor and thus the composite will never be the identity. Thus the only question that can meaningfully raised is when one or the other of the composites is naturally equivalent to the identity. I think sufficient conditions are known, although I don't recall them offhand, but it seems awfully unlikely to me that useful necessary and sufficient conditions are known. Michael Barr ==============================================================================
Since L and R are only determined up to isomorphism, how can one ask for equalities like F^* R = Id? Ross ==============================================================================
>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> let F:C -> D be a functor between two small categories. Its induced functor F^*:Fun(D,Set) -> Fun(C,Set) has both a left adjoint L and a right adjoint R.
1. Under which precise conditions on F is L F^* = Id. 2. Under which precise conditions on F is R F^* = Id. 3. Under which precise conditions on F is F^* R = Id. 4. Under which precise conditions on F is F^* L = Id. Does anybody know any answers to one of these queries? What happens if equality is replaced by natural equivalence? Does anybody know of good references to these or similar problems? Markus Pfenniger Andy Tonks <<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<< I don't know the answers. However, I can describe an analogous context (ring theory) where there is a surprising answer to 2: the appropriate analogue holds iff F is an epimorphism. Perhaps someone already knows whether the corresponding conjecture is true in the category (or monoid) context, and anyway perhaps the different perspective will cast light on the problem. First, as Mike Barr points out, natural equivalence is the appropriate property to look for, and in fact I don't know how you might go beyond the question of when the units or counits of the adjunctions are natural isos. If R is a ring, then the category of (right, say) modules over R is very like a functor category. R _is_ a category (one object, lots of morphisms), and a module is a functor from R to Abelian Groups. Of course, it's more special than that, because we also require that the functor should preserve the additive structure. However, the idea that a functor F from a category C to Sets is a kind of module is quite a natural one. The "elements of the module" are the elements of the sets F(i) where i ranges over objects of C, and a morphism f: i -> j of C "acts on" the elements (at least, those in F(i)) by xf = F(f)(x). You'll see this more clearly in the case where C is a monoid (only one object). This perspective is explained in Popescu, "Abelian categories with applications to rings and modules" (LMS Monographs 3, Academic Press, 1973). It extends readily to "ringoids", i.e. categories enriched over Abelian groups (Barry Mitchell "Rings with several objects", Advances in Mathematics 37 (1972) 1-161). If F: R -> S is a ring homomorphism, then you get a functor F^*: Mod-S -> Mod-R ("restriction of scalars") with left adjoint L, M |-> M tensor_R S, and right adjoint R, M |-> S hom_R M. It is known that the counit of the adjunction L -| F^* is a natural iso if and only if F is an epimorphism in the category of rings. (See, e.g., Stenstro"m "Rings of Quotients", Springer 1975.) (Note that epimorphisms are not always surjective - e.g. the embedding of the integers in the rationals is epi.) This is also true for ringoids (I believe it's in Mitchell's paper). It's conceivable that the direct analogue holds for categories, i.e. (2) (in suitable form) iff F is an epimorphism in the category of (small) categories, though at first glance the proof in Stenstro"m doesn't seem to generalize. If you look into this, I'd suggest you study the monoid case first. Steve Vickers ==============================================================================
1. Under which precise conditions on F is L F^* = Id. 2. Under which precise conditions on F is R F^* = Id. 3. Under which precise conditions on F is F^* R = Id. 4. Under which precise conditions on F is F^* L = Id.
(I guess some people know the precise answers at once, but don't bother to tell us. Here is what I figured.) Suppose, as M.Barr and R.Street promptly suggested, that the equalities in the query denote the natural isos. Of course, * 1 and 2 are equivalent: they mean that F^* is fully faithful; and * 3 and 4 are equivalent: they say that the essential morphism of toposes R: (D,Set)-->(C,Set) is an injection. A sufficient condition for 1 and 2 is that F:C-->D is a stably initial functor. (Initial functors are orthogonal to discrete opfibrations: cf. e.g. "The comprehensive factorisation of a functor" by Street and Walters, early seventies. Now F^* is pulling back of discrete opfibrations along F. Use orthogonality to show that it is fully faithful.) A sufficient condition for 3 and 4 is that F is fully faithful. (Just write down the pointwise Kan-extension formula for L (or R): to calculate F^*L(G) at X from C, we use the diagram obtained by projecting the comma category F/FX to C; we apply G on this diagram and calculate the colimit. When F is fully faithful, this diagram is just a cocone to X - so the colimit is GX.) Now what are the necessary conditions? Best regards, Dusko Pavlovic ==============================================================================
There is a need to say a few more elementary things on this question. My initial reply was short because I was busy teaching an intensive on-campus distance course, and I hoped someone else would respond in more detail. The equality question is meaningless. Also, the term "natural equivalence" is a term used "classically" by category theorists, but not a very good one; I suggest we should try to avoid it since "equivalence" means something else. Let F : C --> D be a functor and let L --| F* --| R as in the question. Let e : L F* --> 1, n : 1 --> F* L, e' : F* R --> 1, n' : 1 --> R F* be the counits and units for the adjunctions. Fact 1: n invertible iff L fully faithful iff R fully faithful iff e' invertible. Fact 2: e invertible iff F* fully faithful iff n' invertible. So there are really only two problems (as I just notice Dusko Pavlovic has pointed out): (a) when is L fully faithful? (b) when is F* fully faithful? Answer to (a): L fully faithful iff F is fully faithful. Proof: Since we are taking Kan extensions along F of functors into Set, there are formulas for L (pointwise left Kan extension). Any formula can be used to show F fully faithful implies n invertible, so L is fully faithful. This must be in all the textbooks (eg Mac Lane). Conversely, there is a square which commutes up to isomorphism (or equality if we choose L suitably on representables) involving L, two Yoneda embeddings, and F^op : C^op --> D^op. If L is fully faithful, so is F^op (since the Yon embs are), and so, so is F.///// Now some comments on (b): (i) Let's call F a "localization" when there exists a set S of arrows in C for which F is the universal functor out of C inverting the arrows of S. Given F, if there is an S, the set of arrows inverted by F is the largest such S. Localizations are bijective on objects. Localizations are coinverters of natural transformations between functors into C. If F is a localization, it follows that F* is an inverter of some natural transformation and hence is fully faithful. (ii) F* fully faithful does NOT imply F localization. For, if F induces an equivalence of categories on the cauchy (idempotent splitting) completions of C, D it will still have F* fully faithful, but need not have F bijective on objects. (iii) F* conservative (= reflects isos) iff e epic iff each object d of D is a retract of an object Fc for c in C. (iv) These things suggest to me that the answer to (b) could be: "F* fully faithful iff each object of D is a retract of an object in the image of F, and F = G H with G fully faithful and H a localization". However, localizations in Cat are notoriously difficult to characterize (easier in Lex, Rex, Pb, . . .). The above goes over to enriched categories with appropriate change in the notion of cauchy completion (and epic in (iii) becomes extremal epic); eg, additive categories for the context of Steve Vickers' response. Regards, Ross ==============================================================================
In the Comparison between Functor Categories question from Bangor Sept 23, we had a functor F:C->D, but now instead of Set, consider an arbitrary category A and the induced functor A^F:A^D->A^C. Left and right adjoints, when they exist, are patched together from Kan extensions. For any particular T in A^C, we could have Ran_F(T), Lan_F(T) in A^D with n.t.(S,Ran_F(T)) iso n.t.(A^F(S),T) or n.t(Lan_F(T),S) iso n.t.(T,A^F(S)) which occurrences we call right or left Kan extensions of T along F. These are approximations of the variable object T by images of A^F, which I think of heuristically as a sort of Dedekind cut situation, in my own naive way, as follows: think of those S in A^D for which SF<=T and call the Lower also think of those S for which T<=SF and call them Upper here SF=A^F(S) of course, and X<=Y just means there is some n.t. X->Y. Then A^F images of Lowers approximate T from below, and Uppers from above. Of course we proceed to look for best approximations. Among Lowers, the closest image to T (w.r.t. <=) would be the sup, and among Uppers it's the inf, corresponding to pointwise Kan ext. as limits. So T gets caught in a squeeze play (like a real number), but still might have lots of room to bounce around. Back to the Bangor question (as least partly), we could ask what happened if T got hit on the nose by A^F with some Z in A^D? ie. ZF->T is identity n.t. Surely Z is then the best approx. to T from above and below since it is bang on. But does Z have to be (iso to) a Kan ext. of T along F? (which it needs to be if it is going to participate in any adjoint for A^F). Consider left extensions (inf of Uppers): from ZF=T<=SF we need to produce a n.t. Z->S. Again heuristically, the temptation is to cross off F on the right in a kind of epi maneuver, which relates to Steve Vickers' comment for rings. By the way, the very nice perspective on modules SJV mentioned also appears as Exercise 3(b) Chap VII p.415 in Mac Lane/Moerdijk '92 "Sheaves..". But epi for F seems too strong to be necessary, whereas the nice suggestion of Ross Street that each object of D should be a retract of an object in the image of F with F factoring in a certain friendly way looks really neat. Still, not clear it should be necessary. Anyway, might help to think about it in different ways. Cheers ........................................Al Vilcius, Toronto ==============================================================================
Some not very exciting developments: If F: C -> D is a functor, then the question was about when F*: S^D -> S^C was full and faithful. (Incidentally, it took a lot of headache before I convinced myself that "fully faithful" just meant "full and faithful". I couldn't find the phrase defined in any of the standard texts. Is its mellifluousness really enough to justify its use?) My conjecture was that that this happens iff F is an epimorphism of categories. This can't be correct. An epimorphism must be surjective on objects (otherwise you can map the objects not in the image to distinct isomorphic copies), but if F is any equivalence then F* is as well, and hence full and faithful. (So my account of Mitchell's results for ringoids as opposed to rings was probably oversimplified.) I assume that "epimorphism", with its demands of on-the-nose equality, is simply not a good notion in the 2-categorical context of categories. In the monoid case, we do have an implication in at least one direction (adapted from the argument for rings): Suppose C and D are monoids. If F* is full and faithful, then F is an epimorphism (of monoids). Proof: Let G,H: D => E with F;G = F;H. E is a D-set, with action ed = e.H(d) Consider the function G: D -> E. This is a homomorphism of C-sets, for G(dc) = G(d).G(F(c)) = G(d).H(F(c)) = G(d)F(c) = G(d)c Hence by fullness of F* (note that faithfulness is automatic in this context), it is a homomorphism of D-sets, so for any d in D we have G(d) = G(1d) = G(1)d = 1.H(d) = H(d) Remaining questions: * Is the converse true in the monoid case? * Is there a 2-categorical generalization of epi that (includes equivalences and) restores the original conjecture? Steve Vickers. ==============================================================================
Just a quick comment on Steve Vickers' post. This ought to be well known, but either it isn't as well known as I thought, or people aren't thinking of it in this connection. If C --> D induces an equivalence between idempotent completions (for example, if it is inclusion of C into its idempotent completion), then the induced Set^D --> Set^C is an equivalence. In fact, the analagous statement is true for any base that is itself idempotent complete. Now of course an inclusion of monoids that had that property would already be an equivalence since the effect of idempotent completion is to add more objects, but (effectively) no more arrows. On the other hand, for a monoid with many objects (aka a category), the situation is quite different. On the other hand, it might be useful to confine the discussion to idempotent complete categories to avoid this particular problem. Michael ==============================================================================
participants (6)
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Al Vilcius -
barr＠triples.Math.McGill.CA -
MAS034＠BANGOR.AC.UK -
pavlovic＠triples.Math.McGill.CA -
sjv＠doc.ic.ac.uk -
street＠macadam.mpce.mq.edu.au