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Postmodern Algebra

Section 6.5 Derived Functors

Subsection Derived Functors: Revisited

We are now ready to define derived functors in full generality. The definitions will match the definitions over \(R\)-modules; the one notable addition from \(R\)-Mod to the general case is the new need to worry about whether the abelian category in question has enough injectives or enough projectives.

Definition 6.84. Derived functors.

Let \(\mathcal{A}\) and \(\mathcal{B}\) be abelian categories.
Let \(F: \mathcal{A} \longrightarrow \mathcal{B}\) be a covariant right exact functor. If \(\mathcal{A}\) has enough projectives, the left derived functors of \(F\) are a sequence of functors \(L_{i} F: \mathcal{A} \longrightarrow \mathcal{B}, i \geqslant 0\text{,}\) defined as follows:
  • For each object \(A\) in \(\mathcal{A}\text{,}\) fix a projective resolution \(P\) of \(A\text{,}\) and set
    \begin{equation*} L_{i} F(A):=\mathrm{H}_{i}(F(P)) \end{equation*}
  • Given an arrow \(f: A \rightarrow B\text{,}\) fix projective resolutions \(P \longrightarrow A\) and \(Q \longrightarrow B\text{,}\) and a map of complexes \(\varphi: P \rightarrow Q\) lifting \(f\text{.}\) Then
    \begin{equation*} L_{i} F(f):=\mathrm{H}_{i}(F(\varphi)) \end{equation*}
Let \(F: \mathcal{A} \longrightarrow \mathcal{B}\) be a covariant left exact functor. If \(\mathcal{A}\) has enough injectives, the right derived functors of \(F\) are a sequence of functors \(R^{i} F: \mathcal{A} \longrightarrow \mathcal{B}, i \geqslant 0\text{,}\) defined as follows:
  • For each object \(A\) in \(\mathcal{A}\text{,}\) fix an injective resolution \(E\) of \(A\text{,}\) and set
    \begin{equation*} R^{i} F(A):=\mathrm{H}^{i}(F(E)) \end{equation*}
  • Given an arrow \(f: A \rightarrow B\text{,}\) fix injective resolutions \(A \longrightarrow E\) and \(B \longrightarrow I\text{,}\) and a map of complexes \(\varphi: P \rightarrow Q\) extending \(f\text{.}\) Then
    \begin{equation*} R^{i} F(f):=\mathrm{H}^{i}(F(\varphi)) \end{equation*}
Let \(F: \mathcal{A} \longrightarrow \mathcal{B}\) be a contravariant left exact functor. If \(\mathcal{A}\) has enough projectives, the right derived functors of \(F\) are a sequence of functors \(R^{i} F: \mathcal{A} \longrightarrow \mathcal{B}, i \geqslant 0\text{,}\) defined as follows:
  • For each object \(A\) in \(\mathcal{A}\text{,}\) fix a projective resolution \(P\) of \(A\text{,}\) and set
    \begin{equation*} R^{i} F(A):=\mathrm{H}^{i}(F(P)) \end{equation*}
  • Given an arrow \(f: A \rightarrow B\text{,}\) fix projective resolutions \(P \longrightarrow A\) and \(Q \longrightarrow B\text{,}\) and a map of complexes \(\varphi: P \rightarrow Q\) extending \(f\text{.}\) Then
    \begin{equation*} R^{i} F(f):=\mathrm{H}^{i}(F(\varphi)) \end{equation*}
Finally, let \(F: \mathcal{A} \longrightarrow \mathcal{B}\) be a contravariant right exact functor. If \(\mathcal{A}\) has enough injectives, the left derived functors of \(F\) are a sequence of functors \(L_{i} F: \mathcal{A} \longrightarrow \mathcal{B}, i \geqslant 0\text{,}\) defined as follows:
  • For each object \(A\) in \(\mathcal{A}\text{,}\) fix an injective resolution \(E\) of \(A\text{,}\) and set
    \begin{equation*} L_{i} F(A):=\mathrm{H}_{i}(F(E)) \end{equation*}
  • Given an arrow \(f: A \rightarrow B\text{,}\) fix injective resolutions \(A \longrightarrow E\) and \(B \longrightarrow I\text{,}\) and a map of complexes \(\varphi: E \rightarrow I\) extending \(f\text{.}\) Then
    \begin{equation*} L_{i} F(f):=\mathrm{H}_{i}(F(\varphi)) \end{equation*}

Remark 6.85.

If \(F\) is exact, then \(\mathrm{H}_{i}(F(C))=F\left(\mathrm{H}_{i}(C)\right)\text{,}\) so \(L_{i} F=0\) for all \(i>0\text{.}\)

Remark 6.86.

If \(P\) is projective, then \(0 \longrightarrow P \longrightarrow 0\) is a projective resolution of \(P\text{,}\) and thus \(L_{i} F(P)=0\) for all \(i>0\text{.}\) Similarly, if \(E\) is injective then \(R^{i} F(E)=0\text{.}\)

Proof.

  1. Let \(P\) and \(Q\) be projective resolutions of \(A\text{.}\) Theorem 5.17 gives us maps of complexes \(\varphi: P \rightarrow Q\) and \(\psi: Q \rightarrow P\) such that \(\varphi \psi\) is homotopic to \(1_{Q}\) and \(\psi \varphi\) is homotopic to \(1_{P}\text{.}\) Additive functors preserve homotopies, by Remark 7.37, so \(F(\varphi) F(\psi)\) and \(F(\psi) F(\varphi)\) are homotopic to the corresponding identity arrows. Homotopic maps induce the same map in homology, by Exercise 90. Therefore, \(F(\varphi)\) and \(F(\psi)\) induce isomorphisms in homology.
  2. Fix projective resolutions \(P\) and \(Q\) of \(M\) and \(N\text{.}\) Any two lifts \(\varphi\) and \(\psi\) of \(f: M \longrightarrow N\) to \(P \longrightarrow Q\) are homotopic, by Lemma 7.59. Additive functors preserve homotopies, by Remark 7.37, so \(F(\varphi)\) and \(F(\psi)\) are homotopic. Homotopic maps induce the same map in homology, by Exercise 90, so \(L_{i} F(\varphi)=L_{i} F(\psi)\) for each \(i\text{.}\)
  3. Given an arrow \(f\text{,}\) fix a lift \(\varphi\) of \(f\) to projective resolutions of the source and target. Since \(F\) is an additive functor, \(\mathrm{H}_{i}(F(\varphi))\) is a homomorphism for each \(i\text{,}\) and thus \(L_{i} F(f)\) is a homomorphism between the corresponding \(\Hom\)-groups, which as we’ve seen is independent of our choice of \(\varphi\text{.}\)
  4. Let \(A\) be any object and \(P\) be a projective resolution of \(A\text{.}\) Since \(P\) is right exact, and
    \begin{equation*} P_{1} \longrightarrow P_{0} \longrightarrow A \longrightarrow 0 \end{equation*}
    is exact, then so is
    \begin{equation*} F\left(P_{1}\right) \longrightarrow F\left(P_{0}\right) \longrightarrow F(A) \longrightarrow 0 \text {. } \end{equation*}
    We claim that \(\mathrm{H}_{0}(F(P))=F(A)\text{.}\) By Exercise 84, \(\operatorname{ker}\left(F\left(P_{0}\right) \longrightarrow 0\right)=1_{F\left(P_{0}\right)}\text{,}\) so the canonical arrow \(\operatorname{im} F\left(\partial_{1}\right) \longrightarrow F\left(P_{0}\right)\) is precisely the image of \(F\left(\partial_{1}\right)\text{.}\) By exactness of the last sequence we wrote above, \(\operatorname{im} F\left(\partial_{1}\right)=\operatorname{ker}\left(F\left(P_{0}\right) \longrightarrow F(A)\right)\text{.}\) On the other hand, exactness at \(F(A)\) says that \(F\left(P_{0}\right) \longrightarrow F(A)\) is an epi, by Exercise 86 . Every epi is the cokernel of its kernel, so \(F\left(P_{0}\right) \longrightarrow F(A)\) is the cokernel of \(\operatorname{im} F\left(\partial_{1}\right)\text{,}\) which we saw was exactly the canonical arrow \(B_{1}(F(P)) \longrightarrow Z_{0}(F(P))\text{.}\) Therefore, \(\mathrm{H}_{0}(F(P))=F(A)\text{,}\) the target of the cokernel of \(B_{1}(F(P)) \longrightarrow Z_{0}(F(P))\text{.}\)

Exercise 6.88.

Let \(\mathcal{A}\) be an abelian category with enough injectives, and \(F\) a covariant left exact functor.
  1. \(R^{i} F(A)\) is well-defined up to isomorphism.
  2. \(R^{i} F(f)\) is well-defined for every arrow \(f\text{.}\)
  3. \(R^{i} F(f)\) is an additive functor for every \(i\text{.}\)
  4. \(R^{0} F=F\text{.}\)

Remark 6.89.

If \(\mathcal{A}\) is an abelian category with enough injectives, then \(\mathcal{A}^{\text {op }}\) is an abelian category with enough projectives. This gives us a relationship between left derived and right derived functors: \(R^{i} F=\left(L_{i} F^{\mathrm{op}}\right)^{\mathrm{op}}\text{.}\)