Derived Functors

By [L1], the complex computing $L_n^{P}F(A)$ is zero in every negative degree, because a deleted projective resolution is supported only in nonnegative homological degrees. Therefore both its degree-$n$ cycle object and its degree-$n$ boundary object are zero, so $L_n^{P}F(A)=0$.

By [L2], the cochain complex computing $R_I^{n}F(A)$ is zero in every negative cohomological degree, because a deleted injective resolution begins in degree $0$. Hence its degree-$n$ cocycle and coboundary objects are zero, so $R_I^{n}F(A)=0$.

Steps 1.1 and 1.2 prove the claimed vanishing for both one-sided derived constructions.

By [L1], the objects $A$ and $B$ come with chosen projective resolutions.

Apply [L2] to $u:A\to B$ and the chosen resolutions from step 1.1. This yields an augmentation-preserving chain map ${\tilde{u}}_{\bullet }:P_{\bullet }(A)\to P_{\bullet }(B)$ lifting $u$.

By [L1], the two lifts ${\tilde{u}}_{\bullet }$ and ${\widehat{u}}_{\bullet }$ are chain-homotopic. Let $s$ be such a homotopy.

The equations in [L2] become $F({\tilde{u}}_n)-F({\widehat{u}}_n)=F(d_{n+1})F(s_n)+F(s_{n-1})F(d_n)$ after applying $F$, because [L3] lets $F$ preserve sums and zero morphisms. Hence $F(s)$ is a chain homotopy from $F({\tilde{u}}_{\bullet })$ to $F({\widehat{u}}_{\bullet })$.

By [L4], chain-homotopic maps induce the same map on homology. Using [L5] to identify those homology objects with the displayed left derived objects gives $H_n\text{\hspace{0.17em}⁣}(F({\tilde{u}}_{\bullet }))=H_n\text{\hspace{0.17em}⁣}(F({\widehat{u}}_{\bullet }))$ for every $n$.

Let ${\tilde{1}}_{\bullet }$ be any comparison lift of $1_A$ used in [L1]. By [L2], ${\tilde{1}}_{\bullet }$ is homotopic to the identity chain map on the chosen projective resolution of $A$.

Applying $F$ preserves that homotopy relation as in the construction of the left derived map, so the induced map on homology agrees with the map from the identity chain map. By [L3], that latter map is $1_{L_n^{P}F(A)}$. Therefore $L_n^{P}F(1_A)=1_{L_n^{P}F(A)}$.

Choose comparison lifts ${\tilde{u}}_{\bullet }$ of $u$, ${\tilde{v}}_{\bullet }$ of $v$, and ${\widetilde{vu}}_{\bullet }$ of $vu$ as in [L1]. By [L2], ${\widetilde{vu}}_{\bullet }$ is homotopic to ${\tilde{v}}_{\bullet }{\tilde{u}}_{\bullet }$.

After applying $F$, [L3] makes the induced homology map of $F({\widetilde{vu}}_{\bullet })$ equal to that of $F({\tilde{v}}_{\bullet }{\tilde{u}}_{\bullet })$. By [L4], the latter equals the composite of the maps induced by $F({\tilde{u}}_{\bullet })$ and $F({\tilde{v}}_{\bullet })$. Translating back through [L1] gives $L_n^{P}F(vu)=L_n^{P}F(v)\circ L_n^{P}F(u)$.

By [L1] and [L2], the assignments $A\mapsto L_n^{P}F(A)$ and $u\mapsto L_n^{P}F(u)$ already form a functor.

Let $u,v:A\to B$. Choose comparison lifts ${\tilde{u}}_{\bullet }$ and ${\tilde{v}}_{\bullet }$. By [L3], their degreewise sum ${\tilde{u}}_{\bullet }+{\tilde{v}}_{\bullet }$ is again a chain map, and it lifts $u+v$ because augmentations are additive. Thus it is a comparison lift of $u+v$.

By definition of the derived map and [L5], $L_n^{P}F(u+v)=H_n\text{\hspace{0.17em}⁣}(F({\tilde{u}}_{\bullet }+{\tilde{v}}_{\bullet }))=H_n\text{\hspace{0.17em}⁣}(F({\tilde{u}}_{\bullet }))+H_n\text{\hspace{0.17em}⁣}(F({\tilde{v}}_{\bullet }))\mathrm{.}$ If a different comparison lift of $u+v$ were chosen, [L4] and [L6] would give the same homology map. Hence $L_n^{P}F(u+v)=L_n^{P}F(u)+L_n^{P}F(v)\mathrm{.}$

Step 1.1 gives functoriality, and step 2.1 gives additivity on each hom-group. Therefore [L7] identifies $L_n^{P}F$ as an additive functor.

By [L1], the objects $A$ and $B$ come with chosen injective resolutions.

Apply [L2] to $u:A\to B$ and the resolutions from step 1.1. The resulting coaugmentation-preserving cochain map ${\tilde{u}}^{\bullet }:I^{\bullet }(A)\to I^{\bullet }(B)$ is the required comparison extension.

By [L1], the two comparison extensions are cochain-homotopic. Using [L2], read that cochain homotopy as a chain homotopy after reindexing the complexes.

Applying $F$ to the homotopy equations from [L3] preserves their sum-and- zero form by [L4]. Hence the two reindexed chain maps $F({\tilde{u}}^{\bullet })$ and $F({\widehat{u}}^{\bullet })$ remain chain-homotopic.

By [L5], these two maps induce the same homology map on the reindexed complexes. Translating back through [L2] and [L6], that is exactly equality of the induced maps on cohomology $R_I^{n}F(A)\to R_I^{n}F(B)$.

Identity and composition are proved exactly as on the projective side: choose comparison extensions for the relevant morphisms, compare the extension of a composite or identity with the obvious chain-level candidate, and use [L5] after reindexing by [L3]. Therefore the assignments in [L1] form a functor.

Let $u,v:A\to B$. Choose comparison extensions ${\tilde{u}}^{\bullet }$ and ${\tilde{v}}^{\bullet }$. By [L4], their degreewise sum is a cochain map and extends $u+v$, so it is a comparison extension of $u+v$.

Reindexing by [L3], applying $F$ degreewise by [L4], and using homotopy invariance from [L5], the induced cohomology map of ${\tilde{u}}^{\bullet }+{\tilde{v}}^{\bullet }$ equals the sum of the induced cohomology maps of ${\tilde{u}}^{\bullet }$ and ${\tilde{v}}^{\bullet }$. Hence $R_I^{n}F(u+v)=R_I^{n}F(u)+R_I^{n}F(v)\mathrm{.}$

Steps 1.1 and 2.1 give the functoriality and hom-group additivity required by [L6]. Therefore $R_I^{n}F$ is an additive functor.

For each object $A$, the components ${\alpha }_{P_k(A)}$ form a chain map ${\alpha }_{P(A{)}_{\mathrm{d}\mathrm{e}\mathrm{l}}}:F(P(A{)}_{\mathrm{d}\mathrm{e}\mathrm{l}})\to G(P(A{)}_{\mathrm{d}\mathrm{e}\mathrm{l}}),$ because [L1] makes them commute with each differential of the chosen deleted resolution.

By [L3], step 1.1 induces a morphism ${\mathbf{L}}_n^P(\alpha {)}_A:L_n^{P}F(A)\to L_n^{P}G(A)$ for each $A$.

Let $u:A\to B$, and choose a comparison lift ${\tilde{u}}_{\bullet }:P_{\bullet }(A)\to P_{\bullet }(B)$. Naturality in [L1] gives $G({\tilde{u}}_k){\alpha }_{P_k(A)}={\alpha }_{P_k(B)}F({\tilde{u}}_k)$ for every degree $k$, so the square of chain maps commutes. Passing to homology and using [L4] gives $L_n^{P}G(u)\circ {\mathbf{L}}_n^P(\alpha {)}_A={\mathbf{L}}_n^P(\alpha {)}_B\circ L_n^{P}F(u)\mathrm{.}$ Thus the components from step 2.1 are natural.

For each object $A$, the components ${\alpha }_{I^k(A)}$ commute with the cochain differentials by [L1], so they form a cochain map $F(I(A{)}_{\mathrm{d}\mathrm{e}\mathrm{l}})\to G(I(A{)}_{\mathrm{d}\mathrm{e}\mathrm{l}})$. Reindexing by [L2] turns this into a chain map.

By [L3], step 1.1 induces a map on homology of the reindexed complexes, hence on cohomology: ${\mathbf{R}}_I^n(\alpha {)}_A:R_I^{n}F(A)\to R_I^{n}G(A)$.

Let $u:A\to B$, and choose a comparison extension ${\tilde{u}}^{\bullet }:I^{\bullet }(A)\to I^{\bullet }(B)$. Naturality in [L1] gives degreewise commutative squares with the maps ${\alpha }_{I^k(A)}$ and ${\alpha }_{I^k(B)}$. Passing to cohomology and translating through [L4] gives $R_I^{n}G(u)\circ {\mathbf{R}}_I^n(\alpha {)}_A={\mathbf{R}}_I^n(\alpha {)}_B\circ R_I^{n}F(u)\mathrm{.}$ Thus the components from step 2.1 are natural.

By [L1] and [L2], there exist comparison maps $c_{\bullet }:P_{\bullet }(A)\to Q_{\bullet }(A)$ and $d_{\bullet }:Q_{\bullet }(A)\to P_{\bullet }(A)$ whose composites are homotopic to the identity chain maps on the two resolutions.

Apply $F$ degreewise and pass to homology. By [L3], the induced maps $H_n(F(c_{\bullet }))$ and $H_n(F(d_{\bullet }))$ are inverse because their composites equal the homology maps of chain maps homotopic to the identities. Using [L4], this yields the claimed isomorphism ${\theta }_{P,Q}(A):L_n^{P}F(A)\stackrel{\sim }{\to }{L}_n^{Q}F(A)$.

Choose objectwise comparison maps $c_A:P_{\bullet }(A)\to Q_{\bullet }(A)$ and $c_B:P_{\bullet }(B)\to Q_{\bullet }(B)$ that define the isomorphisms in [L2], and choose comparison lifts ${\tilde{u}}^P:P_{\bullet }(A)\to P_{\bullet }(B)$ and ${\tilde{u}}^Q:Q_{\bullet }(A)\to Q_{\bullet }(B)$ from [L1].

Both composites $c_B{\tilde{u}}^P$ and ${\tilde{u}}^Q{c}_A$ are comparison maps from $P_{\bullet }(A)$ to $Q_{\bullet }(B)$ lifting the same morphism $u$, so [L3] makes them chain-homotopic. Passing to homology gives ${\theta }_{P,Q}(B)\circ L_n^{P}F(u)=L_n^{Q}F(u)\circ {\theta }_{P,Q}(A)\mathrm{.}$ Therefore the family ${\theta }_{P,Q}(A)$ is natural.

By [L2], each object $A$ carries an isomorphism ${\theta }_{P,Q}(A):L_n^{P}F(A)\stackrel{\sim }{\to }{L}_n^{Q}F(A)$.

By [L3], the family from step 1.1 is natural. Together with [L1], this is exactly a natural isomorphism of additive functors $L_n^{P}F\Rightarrow L_n^{Q}F$.

For the pair $(P,P)$, one valid comparison map is the identity chain map on $P_{\bullet }(A)$. Any comparison map used to define ${\theta }_{P,P}(A)$ also lifts $1_A$, so [L2] makes it homotopic to the identity chain map. By [L3], the induced homology map is therefore the identity on $L_n^{P}F(A)$.

For the triple $(P,Q,R)$, the chain map defining ${\theta }_{Q,R}(A)\circ {\theta }_{P,Q}(A)$ is the composite of two comparison maps $P_{\bullet }(A)\to Q_{\bullet }(A)\to R_{\bullet }(A)$ lifting $1_A$. The chain map defining ${\theta }_{P,R}(A)$ is another comparison map lifting $1_A$. By [L2] they are homotopic, so [L3] gives equality of the induced homology maps: ${\theta }_{Q,R}(A)\circ {\theta }_{P,Q}(A)={\theta }_{P,R}(A)\mathrm{.}$

Since $A$ was arbitrary, steps 1.1 and 1.2 prove the identity and cocycle laws for the natural isomorphisms.

Fix an object $A$. By [L2], there are comparison maps $c_A^{\bullet }:I^{\bullet }(A)\to J^{\bullet }(A)$ and $d_A^{\bullet }:J^{\bullet }(A)\to I^{\bullet }(A)$ whose composites are cochain- homotopic to the identities. Using [L4], these induce inverse isomorphisms ${\theta }_{I,J}(A):R_I^{n}F(A)\stackrel{\sim }{\to }{R}_J^{n}F(A)\mathrm{.}$

For a morphism $u:A\to B$, choose comparison extensions ${\tilde{u}}^I:I^{\bullet }(A)\to I^{\bullet }(B)$ and ${\tilde{u}}^J:J^{\bullet }(A)\to J^{\bullet }(B)$ from [L5]. Both composites $c_B^{\bullet }{\tilde{u}}^I$ and ${\tilde{u}}^J{c}_A^{\bullet }$ extend $u$, so [L3] makes them cochain-homotopic. By [L4], their induced cohomology maps agree, which is exactly the naturality square ${\theta }_{I,J}(B)\circ R_I^{n}F(u)=R_J^{n}F(u)\circ {\theta }_{I,J}(A)\mathrm{.}$

Steps 1.1 and 2.1 produce a natural isomorphism $R_I^{n}F\Rightarrow R_J^{n}F$, and [L1] records that both sides are additive functors.

For any ordered pair $(S,T)$, [L1] gives comparison extensions $c_A:S^{\bullet }(A)\to T^{\bullet }(A)$ and $d_A:T^{\bullet }(A)\to S^{\bullet }(A)$ of $1_A$. Their composites extend $1_A$, so [L2] and [L3] show that the induced cohomology maps are inverse. Any other choice of $c_A$ extends the same identity and hence induces the same map. For a morphism $u:A\to B$, choose within-data comparison extensions using [L4]. The two composites from $S^{\bullet }(A)$ to $T^{\bullet }(B)$ both extend $u$, so [L2] and [L3] give the naturality square. Thus the displayed construction specifies a well-defined natural isomorphism ${\theta }_{S,T}$.

For the pair $(I,I)$, the identity cochain map on $I^{\bullet }(A)$ is a comparison extension of $1_A$. Any comparison extension used in step 1.1 to define ${\theta }_{I,I}(A)$ extends the same identity morphism, so [L2] makes it cochain-homotopic to the identity. By [L3], the induced map on cohomology is therefore the identity on $R_I^{n}F(A)$.

For the triple $(I,J,K)$, the cochain map defining ${\theta }_{J,K}(A)\circ {\theta }_{I,J}(A)$ is the composite of two comparison extensions of $1_A$, while the map defining ${\theta }_{I,K}(A)$ is another comparison extension of $1_A$. By [L2] these are cochain-homotopic, so [L3] gives ${\theta }_{J,K}(A)\circ {\theta }_{I,J}(A)={\theta }_{I,K}(A)\mathrm{.}$

Since $A$ was arbitrary, steps 2.1 and 2.2 prove the identity and cocycle laws for the natural isomorphisms constructed in step 1.1.

By [L1], the morphism $P_1(A)\to P_0(A)\to A\to 0$ is exact. Applying $F$ and using [L3] gives an exact sequence $F(P_1(A))\to F(P_0(A))\to F(A)\to 0.$ Hence $F(A)$ is the cokernel of $F(P_1(A))\to F(P_0(A))$.

By [L2], that same cokernel is exactly $H_0(F(P(A{)}_{\mathrm{d}\mathrm{e}\mathrm{l}}))=L_0^{P}F(A)$. Therefore there is a canonical isomorphism $L_0^{P}F(A)\stackrel{\sim }{\to }F(A)$.

The construction in steps 1.1 and 2.1 is functorial in $A$, and [L4] already supplies the functoriality of $L_0^{P}F$. Thus the isomorphism is natural in $A$.

By [L1], the morphism $0\to A\to I^0(A)\to I^1(A)$ is exact. Applying $F$ and using [L3] gives an exact sequence $0\to F(A)\to F(I^0(A))\to F(I^1(A))\mathrm{.}$ Therefore $F(A)=\ker (F(I^0(A))\to F(I^1(A)))$.

By [L2], the zeroth cohomology of $F(I(A{)}_{\mathrm{d}\mathrm{e}\mathrm{l}})$ is that kernel, because the zero-th coboundary object is $0$. Hence $R_I^{0}F(A)=H^0(F(I(A{)}_{\mathrm{d}\mathrm{e}\mathrm{l}}))\cong F(A)$.

Step 2.1 is natural in $A$, and [L4] records the functoriality of $R_I^{0}F$. Therefore the displayed isomorphism is natural.

By [L1], the object $Q$ has a projective resolution concentrated in degree $0$. Its deleted complex therefore has only one nonzero term, namely $Q$ in degree $0$.

Let $P^{\mathrm{\prime }}$ be the supplied projective resolution datum on the same domain as $P$ that agrees with $P$ away from $Q$ and assigns the length-zero resolution from step 1.1 to $Q$. By [L2], the derived object computed from $P$ is isomorphic to the one computed from $P^{\mathrm{\prime }}$. By [L3], the deleted resolution in $P^{\mathrm{\prime }}(Q)$ is the one-term complex from step 1.1, whose homology is zero in every positive degree. Hence $L_n^{P}F(Q)=0$ for $n>0$.

The coaugmented complex $0\to J\stackrel{1_J}{\to }J\to 0\to 0\to \cdots$ is exact and all its terms are injective by [L2], so [L1] makes it an injective resolution of $J$. After applying $F$, its deleted cochain complex has only one nonzero term, namely $F(J)$ in degree $0$.