the regular category embedding theorem
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The regular category embedding theorem Michael Barr Math & - PowerPoint PPT Presentation

The regular category embedding theorem Michael Barr Math & Stats, McGill University Abstract I have given two apparently different the regular category embedding theorem. The first, gotten by adapting the Lubkins argument for the


  1. The regular category embedding theorem Michael Barr Math & Stats, McGill University

  2. Abstract I have given two apparently different the regular category embedding theorem. The first, gotten by adapting the Lubkin’s argument for the abelian category, is rather opaque. The second, gotten by adapting Mitchell’s proof is much more elegant. Mitchell used Grothendieck’s theorem that an AB5 category with a generator has an injective cogenerator. However, the analogous result for regular categories fails. It turns out that full injectivity is not needed. Surprisingly, it turns out that “under the hood” the two proofs are really doing much the same thing. It is using functors rather than representing diagrams that makes the difference. 2 / 16

  3. Regular and exact categories A category C is called regular if it has finite limits, coequalizers and if the regular epimorphisms are stable under pullback. It is called exact if, in addition, every equivalence relation is a kernel pair. These conditions can be weakened somewhat, but it is not worth the effort to do so. The regular category embedding theorem states that every small regular category has a full and faithful embedding into a set-valued functor category that preserves finite limits and regular epics. For exact categories, we can add the preservation of coequalizers of equivalence relations. That is an easy gloss on the regular embedding theorem so will concentrate in this talk on that result. 3 / 16

  4. Some properties of regular categories � B The most important property is that ever morphism f : A g � � C � h � B where g is regular epic and h is can be factored A monic. Among other things this implies that the composite of regular epics is regular epic, which is not true for general categories. In a regular category, a morphism is a regular epic iff it is extremal. This means that it is not possible to factor it through any proper subobject of its codomain. In general extremal epics (in a complete category) are composites, even transfinite composites, of regular epics. 4 / 16

  5. Finite limit preserving functors From now on, C is a small regular category, F = FL ( C , Set ) is the category of finite limit preserving functors from C to sets and X = F op . The functor that takes A ∈ C to Hom ( A , − ) gives a contravariant embedding of C into F and therefore a covariant embedding of C into X . For the most part, we will treat C as a subcategory of X so write C instead of Hom ( C , − ). We begin with X is complete and cocomplete. We know that every functor F is � F with A ∈ C and it is the colimit of all the arrows Hom ( A , − ) an easy exercise to show that this diagram is filtered iff F preserves finite limits. Thus in X , every F is a filtered limit of all arrows � A . F Proposition. X is regular. We will need some diagrams to show the regularity. 5 / 16

  6. But first, an important obeservation � A is an isomorphism. Then Suppose colim A i � Hom ( colim A i , − ) and therefore Hom ( A , − ) � lim Hom ( A i , − ) are isomorhisms in ( C , Set ) and Hom ( A , − ) hence also in F , the subcategory of finite limit preserving functors. � Hom ( A , − ) is an isomorphism in But then colim Hom ( A i , − ) � X preserves colimits as well, of X = F op . This means that C course, as finite limits. In particular, it preserves regular epics and the regular epic/monic factorization of morphisms, which is what we need. 6 / 16

  7. � � � Some diagrams Consider a pullback: K H G � � F 7 / 16

  8. � � � � � Some diagrams And some maps to representables: C ⑧ ⑧ ⑧ ⑧ ⑧ ⑧ ⑧ ⑧ ⑧ ⑧ ⑧ ⑧ ⑧ ⑧ ⑧ K H G � � F ❄ � ⑧⑧⑧⑧⑧⑧⑧⑧⑧⑧⑧⑧⑧⑧⑧ ❄ ❄ ❄ ❄ ❄ ❄ ❄ ❄ ❄ ❄ ❄ ❄ ❄ ❄ ❄ B A 7 / 16

  9. � � � � � � Some diagrams A factorization: C ⑧ ⑧ ⑧ ⑧ ⑧ ⑧ ⑧ ⑧ ⑧ ⑧ ⑧ ⑧ ⑧ ⑧ ⑧ K H G � � F ❄ ⑧ ❄ ❄ ⑧ ⑧ ❄ ⑧ ❄ ⑧ ❄ � ⑧ ❄ ❄ A × B ❄ ❄ ❳ ❳ ❳ ❄ ❳ ❳ ❳ ❄ ⑧ ❳ ❳ ❳ ⑧ ❳ ❄ ❳ ❳ ⑧ ❳ ❄ ❳ ❳ ⑧ ❳ ❄ ❳ ❳ ⑧ ❳ ❄ ❳ ❳ ❳ � ⑧ ❳ ❳ ❳ ❳ ❳ B A 7 / 16

  10. � � � � � � � Some diagrams � 1 � � supp B B � � A × supp B � A × 1 = A A × B � � F H H H F F � ✄ ✄ ✄ ✄ ✄ ✄ ✄ ✄ ✄ ✄ A × B A × B A × supp B A × supp B A × supp B A × supp B A A 7 / 16

  11. � � � � � � Some diagrams Continuing: C ⑧ ⑧ ⑧ ⑧ ⑧ ⑧ ⑧ ⑧ ⑧ ⑧ ⑧ ⑧ ⑧ ⑧ ⑧ K H G � � F ❄ ⑧ ❄ ❄ ⑧ ❄ ⑧ ❄ ⑧ ⑧ ❄ � ⑧ A × B � � A × supp B ❄ ⑧ ❄ ⑧ ❄ ⑧ ❄ ⑧ ❄ ⑧ ❄ � ⑧ B A 7 / 16

  12. � � � � � � � Some diagrams Continuing: C ⑧ ⑧ ⑧ ⑧ ⑧ ⑧ A × supp B × C ⑧ ⑧ ⑧ ⑧ ⑧ K H G � � F ❄ ⑧ ❄ ❄ ⑧ ❄ ⑧ ❄ ⑧ ⑧ ❄ � ⑧ A × B � � A × supp B ❄ ⑧ ❄ ⑧ ❄ ⑧ ❄ ⑧ ❄ ⑧ ❄ � ⑧ B A 7 / 16

  13. � � � � � � � � � Some diagrams Pullback: C ⑧ ⑧ ⑧ ⑧ ⑧ ⑧ A × B × C � � A × supp B × C � ❄ ❄ ❄ ⑧ ⑧ ⑧ ⑧ ⑧ K H G � � F ❄ ⑧ ❄ ❄ ⑧ ❄ ⑧ ❄ ⑧ ⑧ ❄ � ⑧ A × B � � A × supp B ❄ ⑧ ❄ ⑧ ❄ ⑧ ❄ ⑧ ❄ ⑧ ❄ � ⑧ B A 7 / 16

  14. X is regular, finish � A × supp B × C The sequence A × B × B × C �� A × B × C is a coequalizer/kernel pair. As A , B , and C range over all the maps from F , G , and H , resp. to an object of C , A × supp B , A × B , and A × supp B × C , resp., range over coinitial subsets of those cofiltered families and hence their limits are, resp. F , G , and H . Since a limit of pullback diagrams is a pullback, we conclude that K = lim( A × B × C ) and similarly K × H K is the colimit of the A × B × B × C . Since filtered limits preserve finite � � H is a regular epic. colimits, we further conclude that K The argument actually shows that the opposite of the category of finite product preserving functors in ( C , Set ) is also regular. In order to do this, you need to know that such a functor preserves supports, for which it suffices that it preserve the property of being � 1 is a monomorphism iff the diagonal a subobject of 1. But E � E × E is an isomorphism. E 8 / 16

  15. ✤ � ✤ � ✤ ✤ ✤ Lemma � � F such that every For every F ∈ X , there is a regular epic F # diagram F # F # � � F F B B � � A A can be filled in as shown. 9 / 16

  16. � � � � Proof of lemma. Fix F and well order all possible diagrams of the form F F B B � � A A � A � Let F 0 = F and having chosen F α , choose the next F B in the well-ordering and let � � F F α +1 F α +1 F B B � � A A be a pullback. At a limit ordinal α , let F α = lim β<α F β . 10 / 16

  17. Theorem Every F can be covered by a projective. Proof. Let F (0) = F , F ( n +1) = F ( n )# , and F ∗ = lim F ( n ) . We � A factors through an F ∗ � F ( n ) . In claim that any map F ∗ fact, for any functor F Hom X ( F , Hom ( − , A )) ∼ = Hom F ( Hom ( − , A ) , F ) ∼ = FA (Yoneda) and then Hom X ( F ∗ , Hom ( − , A )) = Hom X (lim F ( n ) , Hom ( − , A )) = Hom F ( Hom ( − , A ) , colim F ( n ) ) = ( colim F ( n ) )( A ) 11 / 16

  18. � � � � Proof, continued But colimits in the functor category F are computed “pointwise”, meaning ( colim F ( n ) )( A ) = colim ( F ( n ) ( A )) so that each element of � A in X . ( colim F ( n ) )( A ) is represented by a morphism F ( n ) This gives � ✈✈✈✈✈✈✈✈✈✈✈✈✈✈ F ∗ F ∗ F ( n +1) F ( n +1) F ( n +1) F ( n +1) � F ( n ) F ( n ) F ( n ) F ( n ) B B � � A A so that F ∗ is a C -projective cover of F . 12 / 16

  19. � The subcategory P We let P denote the full subcategory of X consisting of projective cover P A of every A ∈ C as well as, for each such P A , a projective cover P ′ A of the kernel pair P A × A P A . The result is a coequalizer diagram in X d 0 d � A P ′ � P A A d 1 for each A ∈ C . Note that the functors in P take regular epics in C to surjections in Set . 13 / 16

  20. � � � The main theorem � ( P op , Set ) that takes A to evaluation at A is full The functor C and faithful. Proof. The embedding takes A ∈ C to � Set . A natural transformation Φ( A ) = Hom P ( − , A ) : P op � Φ( B ) assigns to each f : P � A , a map ν : Φ( A ) � B such that for all g : Q � P the square ν ( f ) : P Hom ( g , A ) � Hom ( Q , A ) Hom ( P , A ) Hom ( P , A ) Hom ( Q , A ) ν ν Hom ( P , B ) Hom ( P , B ) Hom ( Q , B ) Hom ( Q , B ) Hom ( g , B ) which means that ν ( fg ) = ν ( f ) g . 14 / 16

  21. ✤ � � ✤ ✤ ✤ � � ✤ Proof continued Apply this to d 0 d P ′ P ′ � P P P P A A ❉ d 1 ❉ ❉ ❉ ❉ ❉ ❉ ❉ f ❉ ❉ ν ( d ) ❉ ❉ ❉ ❉ B B From ν ( d ) d 0 = ν ( dd 0 ) = ν ( dd 1 ) = ν ( d ) d 1 � B such that fd = ν ( d ). Thus we see there is a unique f : A � ( P op , Set ) is full and faithful. C 15 / 16

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