Lecture 2: Abelian Categories
Teacher: Benjamin Hennion
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\[\Hom_{Grp}(F(A), G) ≅ \Hom(A, \underline F G)\]NB: $A = \underline F G$
\[\Hom_{Grp}(F \underline F G, G) ≅ \Hom_{Ab}(\underline F G, \underline F G)\] \[\begin{cases} ℤ &⟶G \\ 1 &⟼ x \end{cases}\]$\underline F(G)$ cannot be the center of $G$
\[F(ℤ) \coprod F(ℤ) = F(ℤ) \ast F(ℤ) ⟶ F(ℤ ⊕ ℤ) = F(ℤ \coprod ℤ)\]$F$ does not preserve $\coprod$, so it does not preserve colimits, so it cannot have a right adjoint. $\underline F$ does not exist.
II. Abelian categories
1. Additive categories
- A category $𝒞$ is additive:
-
if
- $∀x, y ∈ 𝒞, \; \Hom_{𝒞}(x,y)$ is an abelian group
-
composition is bilinear: \(∀x,y,z. \; \Hom_{𝒞}(x,y) × \Hom_{𝒞}(y,z) → \Hom_{𝒞}(x,z)\)
-
$𝒞$ admits a $0$ object: an object which is both initial and terminal
- 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 \(∀x,y ∈ 𝒞. \; x \coprod y ≅ x × y\)
Notation: If $𝒞$ is additive, we denote $x \coprod y$ and $x × y$ by $x ⊕ y$
Proof:
- in any category, if $∅$ exists, then $∀x ∈ 𝒞, \; x \coprod ∅ ≅ x$
-
$y \xto P 0$:
\[x \coprod y \xto {Px \; = \; id_x \coprod P} x \coprod 0 ≅ x\\ Py: x \coprod y ⟶ y\]We get a morphism $x \coprod y ⟶ x × y$ if $x × y$ exists.
We have to prove that $x \coprod y$ satisfies the universal property of the product.
$ψ$, induced by $Px$ and $Py$, is a bijection:
\[∀ z ∈ 𝒞, \quad \Hom_{𝒞}(z, x \coprod y) \xto {ψ} \Hom_{𝒞}(z,x) × \Hom_{𝒞}(z,y)\]Build $ψ^{-1}$:
-
take $f: z ⟶ x$ and $g: z ⟶ y$
-
$\bar f: z ⟶ x ⟶ x \coprod y$, $\bar g: z ⟶ y ⟶ x \coprod y$
in $\Hom_{𝒞}(z, x \coprod y)$, I can take $\bar f + \bar g = h = ψ^{-1}(f,g)$
We can check that
-
$f = Px \circ h: z ⟶ x \coprod y ⟶ x$
-
Indeed: \(ξ: z \xto {h \, = \, \bar f + \bar g} x \coprod y \xto {Px} x\\ ξ = Px(\bar f + \bar g) = Px \bar f + Px \bar g\)
look at $Px \bar f$:
\[\begin{xy} \xymatrix{ x \coprod y \ar[r] \ar@{<-}[d] & x ≅ x \coprod 0 \\ x ≅ x \coprod 0 \ar@{->}[ru]_{ id_x } & \\ z \ar[u] \ar@{->}[ruu]_{ f } } \end{xy}\](same for $Px \bar g$)
-
-
$g = Py \circ h: z ⟶ x \coprod y ⟶ x$
We have $ψ^{-1} \circ ψ = id$
-
NB: $f, g: x ⟶ y$ in $𝒞$ additive.
\[f+g: x ⟶ x ⊕ x ≅ x × x \xto {f ⊕ g} y ⊕ y ≅ y \coprod y \xto {η} y\] \[\begin{cases} \Hom(z,x) &⟶ \Hom(z,x) × \Hom(z,x) \\ f &⟼ (f,f) \end{cases}\] \[\Hom(y,y) \ni id ⟼ (id,id) ∈ \Hom(y,y) × \Hom(y,y) ≅ \Hom(y \coprod y, y) \ni η\]Let $𝒞, 𝒟$ be additive categories, $F: 𝒞 ⟶ 𝒟$ a functor.
- $F$ is additive if:
-
$∀x,y∈ 𝒞$, the map \(\Hom_{𝒞}(x,y) ⟶ \Hom_{𝒟}(Fx,Fy)\) is a morphism of abelian groups
Prop: if $F: 𝒞 ⟶ 𝒟$ is additive, then $F(0) ≅ 0$ and $∀x,y ∈ 𝒞$:
\[F(x ⊕ y) ≅ F(x) ⊕ F(y)\]And reciprocally.
Proof:
$⟸$: easy using the previous remark.
$⟹$: Fix $x,y∈ 𝒞$.
\[id_{F(x ⊕ y)} = F(id_{x ⊕ y}) = F(id_{x} ⊕ id_{y})\] \[\begin{xy} \xymatrix{ F(x ⊕ y) \ar[r]^{ id } \ar[d] & F(x ⊕ y) \\ F(x) ⊕ F(y) \ar@{->}[ru] & } \end{xy}\]Use the universal property of $F(x) ⊕ F(y)$ to find that \(F(x) ⊕ F(y) ⟶ F(x ⊕ y) ⟶ F(x) ⊕ F(y)\) is an isomorphism (actually the identity)
Examples:
- $𝕂$ field: $Vect_𝕂$ 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. $f: x ⟶ y$
\[\Ker f \; ≝ \; x ×_y 0\]it means that for all $g: z ⟶ x$ st $z ⟶ x ⟶ y$ is the $0$ map, then the map $g$ factors uniquely through $\Ker f ⟶ x$.
\[\coKer f ≝ y \coprod_x 0 \qquad \text{(pushout)}\]We get:
\[\Im f ≝ \Ker (y ⟶ \coKer f)\\ \coIm f ≝ \coKer (\Ker f ⟶ x)\\\] \[\begin{xy} \xymatrix{ \Ker f \ar@<+2pt>[r] \ar@<-2pt>[r]_{ 0 } & x \ar@<+2pt>[r]^{ f } \ar@<-2pt>[r]_{ 0 } \ar[d] & y \ar[d] \ar@<+2pt>[r] \ar@<-2pt>[r]_-{ 0 } & \coKer f \\ & \coIm f \ar@{.>}[ru] \ar@{.>}[r]^{ψ_f} \ar@{.>}@/_3pc/[rru] & \Im f } \end{xy}\]- An additive category $𝒞$ is called abelian:
-
if
- kernels and cokernels always exist
- for all $f: x ⟶ y ∈ 𝒞$, the morphism $ψ_f$ is an isomorphism
Examples:
- $Vect_𝕂, \, Mod_R, \, Sh(X, R)$ are abelian
- $A ∈ CRing$: $Mod_A^{ft}$ (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.
-
\[0 ⟶ k[t] ⟶ k[[t]]\]
with the $t$-adic topology: \(\sum\limits_{ i = 0}^N a_i t^i \xto {N → +∞} \sum\limits_{ i = 0}^∞ a_i t^i\)
The kernel and cokernel are trivial, the image is $k[t]$, the coimage $k[[t]]$
\[0 ⟶ k[t] \xto{ψ_f = f} k[[t]] ⟶ 0\]
-
\[0 ⟶ k[t] ⟶ k[[t]]\]
- $Top Ab Grp$: $ℤ ⟶ ℤ_p$, similar situation
Lemma: $𝒞$ abelian category, $f: x ⟶ y ∈ 𝒞$.
\[f \text{ is an iso } ⟺ \Ker f = \coKer f = 0\]Proof:
$⟹$: ok $⟸$: $\Ker f = \coKer f = 0$ ⟹ $\Im f = y$ and $\coIm f = x$ and $ψ_f = f$
$f: x ⟶ y$ in $𝒞$ abelian
- $f$ is a monomorphism:
-
if $\Ker f = 0$
- $f$ is an epimorphism:
-
if $\coKer f = 0$
NB: there are notions of monomorphisms and epimorphisms in general categories.
In an abelian category, everything behaves like it does in $Mod_R$, for $R$ a ring.
Actually: any abelian category can be “locally” embedded in a category of the $Mod_R$ (Freyd-Mitchell)
Exact sequences
$𝒞$ an abelian category.
-
$x \xto f y \xto g z ∈ 𝒞$ is exact in $y$ if $\Ker g ≅ \Im f$
-
A short exact sequence is a sequence
\[0 ⟶ x ⟶ y ⟶ z ⟶ 0\]exact in $x, y, z$
-
A sequence is
\[0 → x \xto f y \xto g z → 0\]split (scindée in fr) if one of the following equivalent assertion holds:
- $g$ admits a section ($s: z ⟶ y$ st $gs = id_z$)
- $f$ admits a retract ($r: y ⟶ x$ st $rf = id_x$)
- the sequence is isomorphic to \(0 ⟶ x ⟶ x ⊕ z ⟶ z ⟶ 0\)
Exercise: Prove the assertions are equivalent
Lemma: a split sequence is exact
Examples:
- in $Vect_k$, any short exact sequence splits
-
\(0 ⟶ ℤ \overset{n}{⟶} ℤ ⟶ ℤ/n ⟶ 0\) does not split in $Ab$
- \(0 ⟶ k[t] \overset{× t^n}{⟶} k[t] ⟶ k[t]/t^n ⟶ 0\) is $k[t]$-mod exact, $k$-mod split
NB: $F: 𝒞 ⟶ 𝒟$ 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, $M ⊗_A \bullet$ is exact iff $M$ is flat.
$F: 𝒞 ⟶ 𝒟$ is an additive functor, $𝒞, 𝒟$ abelian.
We say that $F$ is exact (resp. left/right exact) if for any exact sequence
\[0 ⟶ x ⟶ y ⟶ z ⟶ 0\]the sequence
\[0 ⟶ F(x) ⟶ F(y) ⟶ F(z) ⟶ 0\]is exact (resp. $0 ⟶ F(x) ⟶ F(y) ⟶ F(z)$ / $F(x) ⟶ F(y) ⟶ F(z) ⟶ 0$ is exact).
Lemma:
$F$ is left exact iff it preserves kernels: for all $f: x ⟶ y$, \(F(\Ker f) ≅ \Ker (F f)\)
$F$ is right exact iff it preserves cokernels
Proof: Up to changing $y$ by $\Im f$, we can assume $f$ to be an epimorphism.
\[0 ⟶ \Ker f ⟶ x ⟶ y ⟶ 0\\ 0 ⟶ F(\Ker f) ⟶ Fx \xto {Ff} Fy ⟶ 0\\\]Then
\[\begin{xy} \xymatrix{ 0 \ar[r] & F(\Ker f) \ar[d]^{ φ }\ar[r] & F(x) \ar[r] \ar[d] & F(y) \ar[d] \\ 0 \ar[r] & \Ker (Ff)\ar[r] & F(x)\ar[r] & F(y) } \end{xy}\]is exact iff $φ$ is an isomorphism.
Examples:
-
$M ∈ A \text{-} Mod$: \(M ⊗_A \bullet: A \text{-} Mod ⟶ A \text{-} Mod\) is always right exact
It’s left exact iff $M$ is flat (definition)
-
\(\Hom_{A}(\bullet, Q): A \text{-} Mod^{op} ⟶ A \text{-} Mod\) left exact (if $A$ is commutative)
-
$G$ group, $A$ ring
\[\begin{cases} Rep_A(G) &⟶ Mod_A \\ M &⟼ M^G ≝ \lbrace m ∈ M \; \mid \; ∀ g ∈ G, gm = m \rbrace \end{cases}\]is left exact.
III. Complexes
Fix an abelian category.
Definitions
- A complex in $𝒞$:
-
is a sequence \(⋯ ⟶ x^{-1} \overset{d^{-1}}{⟶} x^{0} \overset{d^{0}}{⟶} x^{1} \overset{d^{1}}{⟶} x^2 ⟶ ⋯\) in $𝒞$ st $d^2 = 0$ ($∀n ∈ ℤ, \; d^{n+1} d^n = 0$)
$d$ is called the differential
We will denote it by $x^\bullet$
A morphism $f: x^\bullet ⟶ y^\bullet$ is the datum of $∀ n ∈ ℤ$, a morphism $f^n: x^n ⟶ y^n$ in $𝒞$ st
\[d_y^n f^n = f^{n+1} d^n_x\]All the rectangles
\[\begin{xy} \xymatrix{ ⋯ \ar[r] & x^n \ar[r]^{ d^n } \ar[d]_{ f^{n}} & x^{n+1} \ar[r] \ar[d]^{ f^{n+1}} & ⋯ \\ ⋯ \ar[r] & y^n \ar[r]_{ d^n } & y^{n+1} \ar[r] & ⋯ } \end{xy}\]commute
Denote by $C(𝒞)$ the category of complexes in $𝒞$.
We denote by $C^-(𝒞), \, C^+(𝒞), \, C^b(𝒞)$ the full subcategories of $C(𝒞)$ whose objects are complexes $x^\bullet$ with
- $x^n = 0$ if $n » 0$ for $C^-(𝒞)$
- $x^n = 0$ if $n « 0$ for $C^+(𝒞)$
- $x^n = 0$ if $n » 0$ or $n « 0$ for $C^b(𝒞)$
NB:
\[\begin{cases} 𝒞 &⟶ C(𝒞) \\ x &⟼ ⋯ ⟶ 0 ⟶ x ⟶ 0 ⟶ ⋯ && x^n = 0 \text{ if } n ≠0 \end{cases}\]is a fully faithful functor.
Prop: $C(𝒞), \, C^-(𝒞), \, C^+(𝒞), \, C^b(𝒞)$ are abelian categories.
Proof:
- Abelian group structure on them: ok
- $0$: ok
-
coprod, prod = $⊕$ because they are computed degree-wise: $(x^\bullet \coprod y^\bullet)^n = x^n \coprod y^n$
- kernels and cokernels are computed degree-wise
Cohomology
Let $x^\bullet ∈ 𝒞$.
- The cohomology of $x^\bullet$ is:
-
the family \(H^n(x^\bullet) = \Ker d^n / \Im d^{n-1} ∈ 𝒞\)
NB: Quotient = cokernel of the map
-
If $𝒞 = Mod_A$,
-
then we call cocycles the elements of $Z^n(x^\bullet) = \Ker d^n$
-
we call coboundaries the elements of $B^n(x^\bullet) = \Im d^{n-1}$
-
-
A complex $x^\bullet$ is called acyclic/exact if $∀n,\, H^n(x^\bullet) = 0$
Examples:
-
$M$ manifold
\[0 ⟶ 𝒞^∞(M, ℝ) \overset{d_{dR}}{⟶} Ω^1_M \overset{d_{dR}}{⟶} Ω^2_M ⟶ ⋯\]de Rham complex.
$H^\bullet(-)$ 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 $M ∈ Rep_ℤ(G)$.
$M ∈ Mod_ℤ ≃ Ab$
Linear map:
\[g ∈ G: M \xto {g} M\]- $C^\bullet(G, M)$:
-
is defined as the complex where
\[C^n(G, M) \; ≝ \; Maps(G^n, M)\](all maps: no structure preserved)
Exercise: Show that $d^{n+1}d^n = 0$
$C^\bullet(G, M) ∈ C^+(Ab)$
- The group cohomology of $G$ with coefficients in $M$:
-
is the cohomology of this complex $C^\bullet(G, M)$
It’s denoted by $H^\bullet(G, M)$
NB: \(H^0(G, M) = \underbrace{Z^0(G, M)}_{\text{pick an } m ∈ M} = \ker d^0\)
And what is $\ker d^0$?
\[d^0 : \begin{cases} C^0(G, M) = M &⟶ C^1(G, M) = Map(G, M) \\ m &⟼ g \cdot m - m \end{cases}\]So $\ker d^0 = M^G$
Assume, for the sake of simplicity, that $G$ acts trivially on $M$:
\[g \cdot m = m \qquad ∀ m, ∀ g\] \[Z^2(G, M) = \ker d^2 \\ = \lbrace φ: G^2 → M \; \mid \; ∀g_1, g_2, g_3, \quad φ(g_2, g_3) - φ(g_1 g_2, g_3) + φ(g_1, g_2 g_3) - φ(g_1, g_2) = 0\rbrace\]Pick such a $φ$, build a group structure on $G × M$:
\[(g_1, m_1) × (g_2, m_2) \, ≝ \, (g_1 g_2, \underbrace{g_1 m_2 + g_2 m_1}_{= \, m_2 + m_1 \text{ here}} + φ(g_1, g_2))\]We have an exact sequence of groups:
\[1 ⟶ M ⟶ G ×_φ M ⟶ G ⟶ 1\]Theorem: \(H^2(G, M) ≅ \lbrace 1 ⟶ M ⟶ M ⟶ G ⟶ 1 \text{ exact sequence}\rbrace/\text{iso}\)
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
\[1 ⟶ M ⟶ H \overset{π}{⟶} G ⟶ 1\]Choose a splitting $G \xto s H$ as a set-theoretic map.
Define
\[φ(g_1, g_2) = s(g_1) s(g_2) - s(g_1 g_2) ∈ H\]Actually: $φ(g_1, g_2) ∈ M$, as $π φ = 0$ and $M = \ker π$.
Check that $φ ∈ Z^2(G, M)$.
\[\begin{xy} \xymatrix{ 1 \ar[r] & M \ar[d]^{ id }\ar[r] & H \ar[r]^{π} \ar[d]^{(π,ρ)} & G \ar[d]_{id} \ar[r] & 1\\ 1 \ar[r] & M \ar[r] & G ×_φ M \ar[r] & G \ar[r] & 1 } \end{xy}\]where
\[ρ: \begin{cases} H &⟶ M \\ h &⟼ h s π (h^{-1}) \end{cases}\]If you had chosen another section $\tilde s$, then $φ - \tilde φ$ lives in $\Im d^1$:
\[d^1: C^1(G, M) ⟶ C^2(G, M)\]Other examples of complexes
Shift
- Shift (décalage in fr):
-
$x^\bullet ∈ C(𝒞), \; p ∈ ℤ$:
\[(x^\bullet [p])^n \, ≝ \, x^{n+p}\\ d_{[p]} \, ≝ \, (-1)^p d_{x^\bullet}\]
NB: $x^\bullet[p][q] = x^\bullet[p+q], \; x^\bullet[0] = x^\bullet$
Mapping cone
$f: x^\bullet ⟶ y^\bullet ∈ C(𝒞)$.
\[\begin{xy} \xymatrix{ Mc(f)^n = x^{n+1} \ar[d]_{ -d_x } \ar[drr]^{ f } & ⊕ & y^n \ar[d]^{ d_y } \\ Mc(f)^{n+1} = x^{n+2} & ⊕ & y^{n+1} } \end{xy}\]we have $d^{n+1} d^n = 0$.
\[0 ⟶ y^\bullet ⟶ Mc(f)^\bullet ⟶ x^\bullet[1] ⟶ 0\]exact sequence in $C(𝒞)$
Complex of maps
$x^\bullet, y^\bullet ∈ C(𝒞)$.
Define $\Hom^\bullet(x^\bullet, y^\bullet) ∈ C(Mod_ℤ) ≃ C(Ab)$
\[\Hom^n(x^\bullet, y^\bullet) = \lbrace (f^p)_{p ∈ ℤ} \; \mid \; f^p: x^p ⟶ y^{p+n}\rbrace\] \[d: \begin{cases} \Hom^n(x^\bullet, y^\bullet) &⟶ \Hom^{n+1}(x^\bullet, y^\bullet) \\ (f^p)_{p∈ℤ} &⟼ (d_y^{p+n} \circ f^p + (-1)^{n+1} f^{p+1} \circ d_x^p)_{p ∈ ℤ} \end{cases}\]What is $Z^0$ of $\Hom^\bullet(x^\bullet, y^\bullet)$?
- $f^p: x^p ⟶ y^p$
- \[d_y^p f^p - f^{p+1} d_x^p = 0\]
- so \(Z^0 = \Hom_{C(𝒞)}(x^\bullet, y^\bullet)\)
Functors, homotopies
$F: 𝒞 ⟶ 𝒟$ additive, with $𝒞, 𝒟$ abelian induces
\[F: C(𝒞) ⟶ C(𝒟)\]Problem: in general, $F$ does not preserve the cohomology:
\[F(H^n(x^\bullet)) \not ≅ H^n(F(x^\bullet))\]$f: x^\bullet ⟶ y^\bullet$ morphism of complexes.
We say that $f$ is nulhomotopic/homotopic to $0$ if there are morphisms $h^n: x^n ⟶ y^{n-1}$ st
\[f^n = d_y^{n-1} h^n + h^{n+1} d_x^n\]Two morphisms $f, g: x^\bullet ⟶ y^\bullet$ are homotopic if $f-g$ is nulhomotopic.
We write $f \sim g$ (this is an equivalence relation).
A complex $x^\bullet$ is homotopic to $0$ if its identity is.
NB: if $x^\bullet ≝ 0 ⟶ x^{-1} ⟶ x^0 ⟶ x^1 ⟶ 0$, then homotopy to zero amounts to a splitting.
Exercises: $f: x^\bullet ⟶ y^\bullet$
-
$0 ⟶ y^\bullet ⟶ Mc(f)^\bullet ⟶ x^\bullet[1] ⟶ 0$ is an exact sequence
Compare the data of splitting this sequence with the data of a homotopy $f \sim 0$.
-
Recall that \(Z^0(\Hom^\bullet(x^\bullet, y^\bullet)) = \Hom_{C(𝒞)}(x^\bullet, y^\bullet)\\ Z^0(\Hom^\bullet(x^\bullet, y^\bullet)) \supseteq \Im d^{-1} = \lbrace f \text{ st } f \sim 0\rbrace ⊆ \Hom_{C(𝒞)}(x^\bullet, y^\bullet)\)
Lemma: If two morphisms $f, g: x^\bullet ⟶ y^\bullet$ are homotopic, then they induce the same morphisms.
\[H^n(x^\bullet) ⟶ H^n(y^\bullet)\]Proof: Enough to tackle the case $g = 0$.
Corollary: $F: 𝒞 ⟶ 𝒟$ additive, $𝒞, 𝒟$ abelian.
$x^\bullet ∈ C(𝒞)$ if $x^\bullet$ is homotopic to $0$, then:
- $x^\bullet$ is acyclic: $H^n(x^\bullet) = 0 \quad ∀ n$
- $F(x^\bullet) \sim 0$, hence $H^n(F(x^\bullet)) = 0 \quad ∀ n$
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