Derived functor - Knowlpedia

Derived functors formalize the idea that a non-exact functor can be “corrected” by replacing objects with resolutions and then taking homology/cohomology.

Throughout, let $\mathcal A,\mathcal B$ be abelian categories and $F:\mathcal A\to\mathcal B$ an additive functor.

Right derived functors (from injective resolutions)

Assume $F$ is left exact and $\mathcal A$ has enough injectives (so injective resolutions exist).

For $A\in\mathcal A$ , choose an injective resolution $0\to A\to I^\bullet$ . Apply $F$ termwise to get a cochain complex $F(I^\bullet)$ in $\mathcal B$ . The right derived functors of $F$ are

R^nF(A)\;:=\;H^n\bigl(F(I^\bullet)\bigr)\qquad (n\ge 0),

where $H^n$ is cohomology.

Key facts (standard in homological algebra):

$R^0F \cong F$ .
$R^nF(A)$ is independent of the chosen injective resolution up to canonical isomorphism.
A short exact sequence in $\mathcal A$ induces a long exact sequence in the $R^nF$ .

Left derived functors (from projective resolutions)

Assume $F$ is right exact and $\mathcal A$ has enough projectives (so projective resolutions exist).

For $A\in\mathcal A$ , choose a projective resolution $P_\bullet\to A$ . Apply $F$ termwise to get a chain complex $F(P_\bullet)$ in $\mathcal B$ . The left derived functors are

L_nF(A)\;:=\;H_n\bigl(F(P_\bullet)\bigr)\qquad (n\ge 0),

where $H_n$ is homology.

Again:

$L_0F \cong F$ .
$L_nF(A)$ is well-defined up to canonical isomorphism.
Short exact sequences yield long exact sequences in the $L_nF$ .

Fundamental examples: Ext and Tor

In $\mathcal A = R\text{-Mod}$ :

The functor $\operatorname{Hom}_R(M,-)$ is left exact, and its right derived functors are $R^n\operatorname{Hom}_R(M,-)\cong \mathrm{Ext}_R^n(M,-)$ (see Ext).
The functor $-\otimes_R N$ is right exact, and its left derived functors are $L_n(-\otimes_R N)\cong \mathrm{Tor}_n^R(-,N)$ (see Tor).

Examples (explicit computations)

Example 1: Computing $\mathrm{Tor}_1^{\mathbb Z}(\mathbb Z/n,\mathbb Z/m)$ as a left derived functor

Let $F(-)= -\otimes_{\mathbb Z}\mathbb Z/m\mathbb Z$ . Take the projective resolution

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

Applying $F$ gives

0\to \mathbb Z/m \xrightarrow{\times n} \mathbb Z/m \to 0,

L_1F(\mathbb Z/n)\cong H_1\cong \ker(\times n:\mathbb Z/m\to\mathbb Z/m)\cong \mathbb Z/\gcd(m,n)\mathbb Z.

Thus $\mathrm{Tor}_1^{\mathbb Z}(\mathbb Z/n,\mathbb Z/m)\cong \mathbb Z/\gcd(m,n)\mathbb Z$ (see Tor).

Example 2: Computing $\mathrm{Ext}^1_{\mathbb Z}(\mathbb Z/n,\mathbb Z)$ as a right derived functor

Let $F(-)=\operatorname{Hom}_{\mathbb Z}(\mathbb Z/n,-)$ , which is left exact. Use the injective resolution

0\to \mathbb Z \to \mathbb Q \to \mathbb Q/\mathbb Z \to 0.

Applying $F$ gives

0\to \operatorname{Hom}(\mathbb Z/n,\mathbb Q)\to \operatorname{Hom}(\mathbb Z/n,\mathbb Q/\mathbb Z)\to 0,

with $\operatorname{Hom}(\mathbb Z/n,\mathbb Q)=0$ and $\operatorname{Hom}(\mathbb Z/n,\mathbb Q/\mathbb Z)\cong \mathbb Z/n$ . Hence

R^1F(\mathbb Z)\cong H^1 \cong \mathbb Z/n\mathbb Z,

i.e. $\mathrm{Ext}^1_{\mathbb Z}(\mathbb Z/n,\mathbb Z)\cong \mathbb Z/n\mathbb Z$ (see Ext).

Example 3: Exact functors have no higher derived functors

If $F$ is an exact functor (e.g. in module categories, localization is exact on suitable classes of modules), then applying $F$ to any resolution preserves exactness. Consequently,

R^nF=0 \ (n>0)\quad\text{and}\quad L_nF=0 \ (n>0),

whenever the relevant derived functors are defined (right/left).

Example 1: Computing Tor1Z(Z/n,Z/m)\mathrm{Tor}_1^{\mathbb Z}(\mathbb Z/n,\mathbb Z/m)Tor1Z​(Z/n,Z/m) as a left derived functor

Example 2: Computing ExtZ1(Z/n,Z)\mathrm{Ext}^1_{\mathbb Z}(\mathbb Z/n,\mathbb Z)ExtZ1​(Z/n,Z) as a right derived functor

Example 3: Exact functors have no higher derived functors

Example 1: Computing $\mathrm{Tor}_1^{\mathbb Z}(\mathbb Z/n,\mathbb Z/m)$ as a left derived functor

Example 2: Computing $\mathrm{Ext}^1_{\mathbb Z}(\mathbb Z/n,\mathbb Z)$ as a right derived functor