Lecture 2: Abelian Categories

Teacher: Benjamin Hennion

F is right adjoint to F

HomGrp(F(A),G)Hom(A,FG)

NB: A=FG

HomGrp(FFG,G)HomAb(FG,FG) {G1x

F(G) cannot be the center of G

F()F()=F()F()F()=F()

F does not preserve , so it does not preserve colimits, so it cannot have a right adjoint. F does not exist.


II. Abelian categories

1. Additive categories

A category 𝒞 is additive:

if

  1. x,y𝒞,Hom𝒞(x,y) is an abelian group
  2. composition is bilinear: x,y,z.Hom𝒞(x,y)×Hom𝒞(y,z)Hom𝒞(x,z)

  3. 𝒞 admits a 0 object: an object which is both initial and terminal

  4. the category 𝒞 admits finite coproducts and products (i.e. all binary co/products exist)

Prop: If 𝒞 satisfies (1), (2) and (3) and admits finite coproducts, then it is additive and x,y𝒞.xyx×y

Notation: If 𝒞 is additive, we denote xy and x×y by xy

Proof:

  • in any category, if exists, then x𝒞,xx
  • yP0:

    xyPx=idxPx0xPy:xyy

    We get a morphism xyx×y if x×y exists.

    We have to prove that xy satisfies the universal property of the product.

    ψ, induced by Px and Py, is a bijection:

    z𝒞,Hom𝒞(z,xy)ψHom𝒞(z,x)×Hom𝒞(z,y)

    Build ψ1:

    • take f:zx and g:zy

    • f¯:zxxy, g¯:zyxy

    in Hom𝒞(z,xy), I can take f¯+g¯=h=ψ1(f,g)

    We can check that

    • f=Pxh:zxyx

      • Indeed: ξ:zh=f¯+g¯xyPxxξ=Px(f¯+g¯)=Pxf¯+Pxg¯

        look at Pxf¯:

        xyxx0xx0idxzf

        (same for Pxg¯)

    • g=Pyh:zxyx

    We have ψ1ψ=id


NB: f,g:xy in 𝒞 additive.

f+g:xxxx×xfgyyyyηy {Hom(z,x)Hom(z,x)×Hom(z,x)f(f,f) Hom(y,y)id(id,id)Hom(y,y)×Hom(y,y)Hom(yy,y)η

Let 𝒞,𝒟 be additive categories, F:𝒞𝒟 a functor.

F is additive if:

x,y𝒞, the map Hom𝒞(x,y)Hom𝒟(Fx,Fy) is a morphism of abelian groups

Prop: if F:𝒞𝒟 is additive, then F(0)0 and x,y𝒞:

F(xy)F(x)F(y)

And reciprocally.

Proof:

: easy using the previous remark.

: Fix x,y𝒞.

idF(xy)=F(idxy)=F(idxidy) F(xy)idF(xy)F(x)F(y)

Use the universal property of F(x)F(y) to find that F(x)F(y)F(xy)F(x)F(y) is an isomorphism (actually the identity)

Examples:

  • 𝕂 field: Vect𝕂 is additive
  • R ring left (or right) R-modules form an additive category
  • Sheaves in an additive category are again an additive category
  • Any sort of topological vector spaces (Fréchet/Banach)

2. Abelian categories

Goal of this entire class: talk about exact sequences. In all previous examples it works, except the last one (sort of topological vector spaces (Fréchet/Banach)).

Let 𝒞 be an additive category. f:xy

Kerfx×y0

it means that for all g:zx st zxy is the 0 map, then the map g factors uniquely through Kerfx.

coKerfyx0(pushout)

We get:

ImfKer(ycoKerf)coImfcoKer(Kerfx) Kerf0xf0y0coKerfcoImfψfImf
An additive category 𝒞 is called abelian:

if

  • kernels and cokernels always exist
  • for all f:xy𝒞, the morphism ψf is an isomorphism

Examples:

  • Vect𝕂,ModR,Sh(X,R) are abelian
  • ACRing: ModAft (modules over A of finite type) is abelian if A is noetherian (if A is not noetherian, then kernels are not necessarily of finite type).
  • TopVect over k: additive but not abelian.

    • 0k[t]k[[t]]

      with the t-adic topology: i=0NaitiN+i=0aiti

      The kernel and cokernel are trivial, the image is k[t], the coimage k[[t]]

      0k[t]ψf=fk[[t]]0
  • TopAbGrp: p, similar situation

Lemma: 𝒞 abelian category, f:xy𝒞.

f is an iso Kerf=coKerf=0

Proof:

: ok : Kerf=coKerf=0Imf=y and coImf=x and ψf=f


f:xy in 𝒞 abelian

f is a monomorphism:

if Kerf=0

f is an epimorphism:

if coKerf=0

NB: there are notions of monomorphisms and epimorphisms in general categories.

In an abelian category, everything behaves like it does in ModR, for R a ring.

Actually: any abelian category can be “locally” embedded in a category of the ModR (Freyd-Mitchell)

Exact sequences

𝒞 an abelian category.

  • xfygz𝒞 is exact in y if KergImf

  • A short exact sequence is a sequence

    0xyz0

    exact in x,y,z

  • A sequence is

    0xfygz0

    split (scindée in fr) if one of the following equivalent assertion holds:

    • g admits a section (s:zy st gs=idz)
    • f admits a retract (r:yx st rf=idx)
    • the sequence is isomorphic to 0xxzz0

Exercise: Prove the assertions are equivalent

Lemma: a split sequence is exact

Examples:

  • in Vectk, any short exact sequence splits
  • 0n/n0 does not split in Ab

  • 0k[t]×tnk[t]k[t]/tn0 is k[t]-mod exact, k-mod split

NB: F:𝒞𝒟 additive functor between abelian categories. F maps split exact sequences to split exact sequences, but F does not necessarily preserve exact sequences.

3. Exact functors

M an A-module, MA is exact iff M is flat.

F:𝒞𝒟 is an additive functor, 𝒞,𝒟 abelian.

We say that F is exact (resp. left/right exact) if for any exact sequence

0xyz0

the sequence

0F(x)F(y)F(z)0

is exact (resp. 0F(x)F(y)F(z) / F(x)F(y)F(z)0 is exact).

Lemma:

  • F is left exact iff it preserves kernels: for all f:xy, F(Kerf)Ker(Ff)

  • F is right exact iff it preserves cokernels

Proof: Up to changing y by Imf, we can assume f to be an epimorphism.

0Kerfxy00F(Kerf)FxFfFy0

Then

0F(Kerf)φF(x)F(y)0Ker(Ff)F(x)F(y)

is exact iff φ is an isomorphism.

Examples:

  • MA-Mod: MA:A-ModA-Mod is always right exact

    It’s left exact iff M is flat (definition)

  • HomA(,Q):A-ModopA-Mod left exact (if A is commutative)

  • G group, A ring

    {RepA(G)ModAMMG{mMgG,gm=m}

    is left exact.

III. Complexes

Fix an abelian category.

Definitions

A complex in 𝒞:

is a sequence x1d1x0d0x1d1x2 in 𝒞 st d2=0 (n,dn+1dn=0)

d is called the differential

We will denote it by x

A morphism f:xy is the datum of n, a morphism fn:xnyn in 𝒞 st

dynfn=fn+1dxn

All the rectangles

xndnfnxn+1fn+1yndnyn+1

commute

Denote by C(𝒞) the category of complexes in 𝒞.

We denote by C(𝒞),C+(𝒞),Cb(𝒞) the full subcategories of C(𝒞) whose objects are complexes x with

  • xn=0 if n»0 for C(𝒞)
  • xn=0 if n«0 for C+(𝒞)
  • xn=0 if n»0 or n«0 for Cb(𝒞)

NB:

{𝒞C(𝒞)x0x0xn=0 if n0

is a fully faithful functor.

Prop: C(𝒞),C(𝒞),C+(𝒞),Cb(𝒞) are abelian categories.

Proof:

  • Abelian group structure on them: ok
  • 0: ok
  • coprod, prod = because they are computed degree-wise: (xy)n=xnyn

  • kernels and cokernels are computed degree-wise

Cohomology

Let x𝒞.

The cohomology of x is:

the family Hn(x)=Kerdn/Imdn1𝒞

NB: Quotient = cokernel of the map

  • If 𝒞=ModA,

    • then we call cocycles the elements of Zn(x)=Kerdn

    • we call coboundaries the elements of Bn(x)=Imdn1

  • A complex x is called acyclic/exact if n,Hn(x)=0

Examples:

  • M manifold

    0𝒞(M,)ddRΩM1ddRΩM2

    de Rham complex.

    H() is called the de Rham cohomology of M

  • Cech cohomology

  • Singular cohomology

  • Dolbeault complex

NB: Cohomology vs homology:

  • difference = natural complexes:

    • increasing the degree: cochain complexes ⟶ cohomology
    • decreasing the degree: chain complexes ⟶ homology

    ⟹ equivalent categories, no big difference

  • usually: for historical reasons, some theories are called homology while others are called cohomology: this has to do with the variance of the functor

Group cohomology

Fix a group G and MRep(G).

MModAb

Linear map:

gG:MgM
C(G,M):

is defined as the complex where

Cn(G,M)Maps(Gn,M)

(all maps: no structure preserved)

dn:{Cn(G,M)Cn+1(G,M)(φ:GnM)dnφ:{Gn+1M(g1,,gn+1)g1φ(g2,,gn+1)+i=1n(1)iφ(g1,,gigi+1,,gn+1)+(1)n+1φ(g1,,gn)

Exercise: Show that dn+1dn=0

C(G,M)C+(Ab)

The group cohomology of G with coefficients in M:

is the cohomology of this complex C(G,M)

It’s denoted by H(G,M)

NB: H0(G,M)=Z0(G,M)pick an mM=kerd0

And what is kerd0?

d0:{C0(G,M)=MC1(G,M)=Map(G,M)mgmm

So kerd0=MG

Assume, for the sake of simplicity, that G acts trivially on M:

gm=mm,g Z2(G,M)=kerd2={φ:G2Mg1,g2,g3,φ(g2,g3)φ(g1g2,g3)+φ(g1,g2g3)φ(g1,g2)=0}

Pick such a φ, build a group structure on G×M:

(g1,m1)×(g2,m2)(g1g2,g1m2+g2m1=m2+m1 here+φ(g1,g2))

We have an exact sequence of groups:

1MG×φMG1

Theorem: H2(G,M){1MMG1 exact sequence}/iso

NB: you can compute group extensions thanks to this theorem (this is part of the reason why cohomology matters)

Sketch of the proof:

Start with an exact sequence

1MHπG1

Choose a splitting GsH as a set-theoretic map.

Define

φ(g1,g2)=s(g1)s(g2)s(g1g2)H

Actually: φ(g1,g2)M, as πφ=0 and M=kerπ.

Check that φZ2(G,M).

1MidHπ(π,ρ)Gid11MG×φMG1

where

ρ:{HMhhsπ(h1)

If you had chosen another section s~, then φφ~ lives in Imd1:

d1:C1(G,M)C2(G,M)

Other examples of complexes

Shift

Shift (décalage in fr):

xC(𝒞),p:

(x[p])nxn+pd[p](1)pdx

NB: x[p][q]=x[p+q],x[0]=x

Mapping cone

f:xyC(𝒞).

Mc(f)n=xn+1dxfyndyMc(f)n+1=xn+2yn+1

we have dn+1dn=0.

0yMc(f)x[1]0

exact sequence in C(𝒞)

Complex of maps

x,yC(𝒞).

Define Hom(x,y)C(Mod)C(Ab)

Homn(x,y)={(fp)pfp:xpyp+n} d:{Homn(x,y)Homn+1(x,y)(fp)p(dyp+nfp+(1)n+1fp+1dxp)p

What is Z0 of Hom(x,y)?

  • fp:xpyp
  • dypfpfp+1dxp=0
  • so Z0=HomC(𝒞)(x,y)

Functors, homotopies

F:𝒞𝒟 additive, with 𝒞,𝒟 abelian induces

F:C(𝒞)C(𝒟)

Problem: in general, F does not preserve the cohomology:

F(Hn(x))Hn(F(x))

f:xy morphism of complexes.

We say that f is nulhomotopic/homotopic to 0 if there are morphisms hn:xnyn1 st

fn=dyn1hn+hn+1dxn

Two morphisms f,g:xy are homotopic if fg is nulhomotopic.

We write fg (this is an equivalence relation).

A complex x is homotopic to 0 if its identity is.

NB: if x0x1x0x10, then homotopy to zero amounts to a splitting.

Exercises: f:xy

  • 0yMc(f)x[1]0 is an exact sequence

    Compare the data of splitting this sequence with the data of a homotopy f0.

  • Recall that Z0(Hom(x,y))=HomC(𝒞)(x,y)Z0(Hom(x,y))Imd1={f st f0}HomC(𝒞)(x,y)


Lemma: If two morphisms f,g:xy are homotopic, then they induce the same morphisms.

Hn(x)Hn(y)

Proof: Enough to tackle the case g=0.

Corollary: F:𝒞𝒟 additive, 𝒞,𝒟 abelian.

xC(𝒞) if x is homotopic to 0, then:

  • x is acyclic: Hn(x)=0n
  • F(x)0, hence Hn(F(x))=0n

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