Derived functors (see derived functors
) turn short exact sequences into long exact sequences in homology/cohomology. Conceptually, this is a systematic abstraction of the snake lemma
and its boundary map; see also connecting homomorphisms
.
Theorem (left derived functors) Let A , B \mathcal A,\mathcal B A , B abelian categories
and let
F : A → B
F:\mathcal A \to \mathcal B
F : A → B be an additive functor that is right exact. Assume A \mathcal A A L n F L_nF L n F
Then for every short exact sequence
0 → A ′ → u A → v A ′ ′ → 0
0 \to A' \xrightarrow{u} A \xrightarrow{v} A'' \to 0
0 → A ′ u A v A ′′ → 0 in A \mathcal A A
δ n : L n F ( A ′ ′ ) → L n − 1 F ( A ′ )
\delta_n: L_nF(A'') \to L_{n-1}F(A')
δ n : L n F ( A ′′ ) → L n − 1 F ( A ′ ) such that the following sequence is exact:
⋯ → L 2 F ( A ) → L 2 F ( A ′ ′ ) → δ 2 L 1 F ( A ′ ) → L 1 F ( A ) → L 1 F ( A ′ ′ ) → δ 1 L 0 F ( A ′ ) → L 0 F ( A ) → L 0 F ( A ′ ′ ) → 0.
\cdots \to L_2F(A) \to L_2F(A'') \xrightarrow{\delta_2} L_1F(A') \to L_1F(A) \to L_1F(A'') \xrightarrow{\delta_1} L_0F(A') \to L_0F(A) \to L_0F(A'') \to 0.
⋯ → L 2 F ( A ) → L 2 F ( A ′′ ) δ 2 L 1 F ( A ′ ) → L 1 F ( A ) → L 1 F ( A ′′ ) δ 1 L 0 F ( A ′ ) → L 0 F ( A ) → L 0 F ( A ′′ ) → 0. Moreover, L 0 F ≅ F L_0F \cong F L 0 F ≅ F
Theorem (right derived functors) Let G : A → B G:\mathcal A\to\mathcal B G : A → B A \mathcal A A R n G R^nG R n G
Then every short exact sequence 0 → A ′ → A → A ′ ′ → 0 0\to A'\to A\to A''\to 0 0 → A ′ → A → A ′′ → 0
0 → G ( A ′ ) → G ( A ) → G ( A ′ ′ ) → δ 0 R 1 G ( A ′ ) → R 1 G ( A ) → R 1 G ( A ′ ′ ) → δ 1 R 2 G ( A ′ ) → ⋯ .
0 \to G(A') \to G(A) \to G(A'') \xrightarrow{\delta^0} R^1G(A') \to R^1G(A) \to R^1G(A'') \xrightarrow{\delta^1} R^2G(A') \to \cdots.
0 → G ( A ′ ) → G ( A ) → G ( A ′′ ) δ 0 R 1 G ( A ′ ) → R 1 G ( A ) → R 1 G ( A ′′ ) δ 1 R 2 G ( A ′ ) → ⋯ . Important special cases Examples Example 1: Computing T o r 1 Z ( Z / n , A ) \mathrm{Tor}^{\mathbb Z}_1(\mathbb Z/n,A) Tor 1 Z ( Z / n , A ) Start with the short exact sequence of Z \mathbb Z Z
0 → Z → × n Z → Z / n → 0.
0 \to \mathbb Z \xrightarrow{\times n} \mathbb Z \to \mathbb Z/n \to 0.
0 → Z × n Z → Z / n → 0. Apply the right exact functor ( − ) ⊗ Z A (-)\otimes_{\mathbb Z} A ( − ) ⊗ Z A T o r i Z ( − , A ) \mathrm{Tor}^{\mathbb Z}_i(-,A) Tor i Z ( − , A )
T o r 1 Z ( Z , A ) → T o r 1 Z ( Z / n , A ) → Z ⊗ A → × n Z ⊗ A → ( Z / n ) ⊗ A → 0.
\mathrm{Tor}^{\mathbb Z}_1(\mathbb Z,A)\to \mathrm{Tor}^{\mathbb Z}_1(\mathbb Z/n,A)\to
\mathbb Z\otimes A \xrightarrow{\times n} \mathbb Z\otimes A \to (\mathbb Z/n)\otimes A \to 0.
Tor 1 Z ( Z , A ) → Tor 1 Z ( Z / n , A ) → Z ⊗ A × n Z ⊗ A → ( Z / n ) ⊗ A → 0. Since T o r 1 Z ( Z , A ) = 0 \mathrm{Tor}^{\mathbb Z}_1(\mathbb Z,A)=0 Tor 1 Z ( Z , A ) = 0 Z ⊗ A ≅ A \mathbb Z\otimes A\cong A Z ⊗ A ≅ A
0 → T o r 1 Z ( Z / n , A ) → A → × n A → A / n A → 0 ,
0 \to \mathrm{Tor}^{\mathbb Z}_1(\mathbb Z/n,A)\to A \xrightarrow{\times n} A \to A/nA \to 0,
0 → Tor 1 Z ( Z / n , A ) → A × n A → A / n A → 0 , so
T o r 1 Z ( Z / n , A ) ≅ ker ( A → × n A ) = A [ n ] .
\mathrm{Tor}^{\mathbb Z}_1(\mathbb Z/n,A)\cong \ker(A \xrightarrow{\times n} A)=A[n].
Tor 1 Z ( Z / n , A ) ≅ ker ( A × n A ) = A [ n ] . Example 2: Computing E x t Z 1 ( Z / n , A ) \mathrm{Ext}^1_{\mathbb Z}(\mathbb Z/n,A) Ext Z 1 ( Z / n , A ) Apply the left exact functor H o m Z ( − , A ) \mathrm{Hom}_{\mathbb Z}(-,A) Hom Z ( − , A )
0 → Z → × n Z → Z / n → 0.
0 \to \mathbb Z \xrightarrow{\times n} \mathbb Z \to \mathbb Z/n \to 0.
0 → Z × n Z → Z / n → 0. The induced long exact sequence in right derived functors yields the segment
0 → H o m ( Z / n , A ) → H o m ( Z , A ) → × n H o m ( Z , A ) → E x t 1 ( Z / n , A ) → 0.
0 \to \mathrm{Hom}(\mathbb Z/n,A)\to \mathrm{Hom}(\mathbb Z,A)\xrightarrow{\times n}\mathrm{Hom}(\mathbb Z,A)\to \mathrm{Ext}^1(\mathbb Z/n,A)\to 0.
0 → Hom ( Z / n , A ) → Hom ( Z , A ) × n Hom ( Z , A ) → Ext 1 ( Z / n , A ) → 0. Using H o m ( Z , A ) ≅ A \mathrm{Hom}(\mathbb Z,A)\cong A Hom ( Z , A ) ≅ A
E x t Z 1 ( Z / n , A ) ≅ c o k e r ( A → × n A ) ≅ A / n A .
\mathrm{Ext}^1_{\mathbb Z}(\mathbb Z/n,A)\cong \mathrm{coker}(A \xrightarrow{\times n} A)\cong A/nA.
Ext Z 1 ( Z / n , A ) ≅ coker ( A × n A ) ≅ A / n A .