Long exact sequence for derived functors - Imported Knowlpedia Content

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 $\mathcal A,\mathcal B$ be abelian categories and let

F:\mathcal A \to \mathcal B

be an additive functor that is right exact. Assume $\mathcal A$ has enough projectives, so the left derived functors $L_nF$ exist.

Then for every short exact sequence

0 \to A' \xrightarrow{u} A \xrightarrow{v} A'' \to 0

in $\mathcal A$ , there are natural connecting morphisms

\delta_n: L_nF(A'') \to L_{n-1}F(A')

such that the following sequence is exact:

\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.

Moreover, $L_0F \cong F$ .

Theorem (right derived functors)

Let $G:\mathcal A\to\mathcal B$ be additive and left exact, and assume $\mathcal A$ has enough injectives so that the right derived functors $R^nG$ exist.

Then every short exact sequence $0\to A'\to A\to A''\to 0$ yields a natural long exact sequence

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.

Important special cases

Taking $F(-)=(-)\otimes_R N$ (right exact; see tensor is right exact) gives the long exact sequence in Tor.
Taking $G(-)=\mathrm{Hom}_R(-,N)$ (left exact; see Hom is left exact) gives the long exact sequence in Ext.

Examples

Example 1: Computing $\mathrm{Tor}^{\mathbb Z}_1(\mathbb Z/n,A)$ from the long exact sequence

Start with the short exact sequence of $\mathbb Z$ -modules

0 \to \mathbb Z \xrightarrow{\times n} \mathbb Z \to \mathbb Z/n \to 0.

Apply the right exact functor $(-)\otimes_{\mathbb Z} A$ . The left derived functors are $\mathrm{Tor}^{\mathbb Z}_i(-,A)$ , and the long exact sequence includes the segment

\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.

Since $\mathrm{Tor}^{\mathbb Z}_1(\mathbb Z,A)=0$ and $\mathbb Z\otimes A\cong A$ , this simplifies to

0 \to \mathrm{Tor}^{\mathbb Z}_1(\mathbb Z/n,A)\to A \xrightarrow{\times n} A \to A/nA \to 0,

\mathrm{Tor}^{\mathbb Z}_1(\mathbb Z/n,A)\cong \ker(A \xrightarrow{\times n} A)=A[n].

Example 2: Computing $\mathrm{Ext}^1_{\mathbb Z}(\mathbb Z/n,A)$ from the long exact sequence

Apply the left exact functor $\mathrm{Hom}_{\mathbb Z}(-,A)$ to the same short exact sequence:

0 \to \mathbb Z \xrightarrow{\times n} \mathbb Z \to \mathbb Z/n \to 0.

The induced long exact sequence in right derived functors yields the segment

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.

Using $\mathrm{Hom}(\mathbb Z,A)\cong A$ , we get

\mathrm{Ext}^1_{\mathbb Z}(\mathbb Z/n,A)\cong \mathrm{coker}(A \xrightarrow{\times n} A)\cong A/nA.

Example 1: Computing Tor1Z(Z/n,A)\mathrm{Tor}^{\mathbb Z}_1(\mathbb Z/n,A)Tor1Z​(Z/n,A) from the long exact sequence

Example 2: Computing ExtZ1(Z/n,A)\mathrm{Ext}^1_{\mathbb Z}(\mathbb Z/n,A)ExtZ1​(Z/n,A) from the long exact sequence

Example 1: Computing $\mathrm{Tor}^{\mathbb Z}_1(\mathbb Z/n,A)$ from the long exact sequence

Example 2: Computing $\mathrm{Ext}^1_{\mathbb Z}(\mathbb Z/n,A)$ from the long exact sequence