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For an additive contravariant left-exact functor, apply it to a projective resolution of $M$. Reversal of the arrows gives a cochain complex. Comparison maps lift using projectivity, and any two lifts are chain homotopic, so its cohomology is independent of the resolution. The functors are contravariant, vanish on projective modules in positive degrees, and send $0\to M^{\prime }\to M\to M^{\prime \prime }\to 0$ to $0\to F(M^{\prime \prime })\to F(M)\to F(M^{\prime })\to R^{1}F(M^{\prime \prime })\to \cdots$. Degreewise split resolutions from the horseshoe lemma yield this sequence.

For an additive right-exact functor on module categories, apply it to a projective resolution and take homology. Lifts of module maps exist by projectivity; any two lifts are chain homotopic, giving independence and functoriality. The zeroth derived functor is naturally $F$, positive degrees vanish on projective modules, and the horseshoe lemma produces a long exact homology sequence from each short exact sequence. A Tor functor is the left derived functor of a tensor product of modules functor.

For a right-exact additive functor out of an abelian category with enough projectives, apply the functor to a projective resolution and take homology objects. The zeroth derived functor is $F$, positive derived functors vanish on projective objects, and the degreewise split horseshoe lemma produces a long exact sequence from every short exact sequence in an abelian category. The target can be any abelian category, including an opposite category of one.

Here the usual homological algebra convention is that the right-exact functor $F$ is additive. Module categories have enough projective modules, because any module is a quotient of a free module. Choose a projective resolutionApply $F$ to the unaugmented chain complex and define the left derived functors byAdditivity ensures that $F$ preserves zero maps, sums of maps and homotopy identities, so the transformed differentials still compose to zero.

For a module homomorphism $u:M\to N$, lift it to a chain map between chosen projective resolutions. In degree zero, lift through $Q_0\twoheadrightarrow N$ using projectivity of $P_0$. In each higher degree, the already constructed map lands in the next kernel; exactness of $Q_{\bullet }$ and projectivity of $P_i$ supply the next lift. Any two lifts are chain homotopic: their degree-zero difference lands in $\ker (Q_0\to N)=\mathrm{im}{d}_1^Q$, so it lifts to $h_0:P_0\to Q_1$. Inductively lift the remaining difference after subtracting the preceding homotopy term. This givesApplying $F$ preserves this identity, so the lifts induce the same homology map. Identity lifts and composable lifts show that $L_{i}F$ is a functor. Comparison lifts between two resolutions over the identity of $M$ are inverse up to chain homotopy; hence the construction is independent of the chosen resolution up to its canonical natural isomorphism.

Right exactness identifies $L_{0}F(M)=\mathrm{coker}(F(P_1)\to F(P_0))$ naturally with $F(M)$. For a projective module $P$, the length-zero resolution gives $L_{i}F(P)=0$ for $i>0$. For a short exact sequence $0\to M^{\prime }\to M\to M^{\prime \prime }\to 0$, the horseshoe lemma builds a degreewise split short exact sequence of projective resolutions. Concretely, lift the projective generators covering $M^{\prime \prime }$ into $M$ and combine them with generators covering $M^{\prime }$; repeat on the kernels. The middle projective term in each degree is the direct sum of the two outer ones. An additive functor preserves these split sequences, and lifting cycles, taking their differentials and returning to the subcomplex produces the connecting maps inThis is the natural long exact derived-functor sequence. In particular, if $0\to K\to P\to M\to 0$ has $P$ projective, then $L_{i}F(M)\cong L_{i-1}F(K)$ for $i\ge 2$, and $L_{1}F(M)=\ker (F(K)\to F(P))$. These are dimension-shifting formulas. Positive degrees can be killed by the surjection $P\to M$. If $F$ is exact, it preserves every resolution's exactness and all its positive left derived functors vanish. Conversely, if $L_{1}F$ vanishes on all modules, the displayed long exact sequence makes $F$ exact. A natural transformation between additive right-exact functors induces transformations on the left derived functors by applying it termwise.

For extension of scalars $F(M)=B{\otimes }_{A}M$, these derived functors are the Tor functors ${\mathrm{Tor}}_i^A(B,M)$. When $B=A/(a)$ and $a$ is a non-zero-divisor, there is the length-one free resolutionAlthough the construction above resolves $M$, one may instead use this resolution of $B$. To justify that calculation, tensor a free resolution of $M$ with the two-term resolution of $B$ to form a double complex. Taking homology first along the resolution of $B$ leaves $B{\otimes }_A{P}_{\bullet }$; taking it first along the resolution of $M$ leaves $M\stackrel{a}{\to }M$. The free terms preserve exactness under tensoring. The double complex is first-quadrant with finitely many terms on each total diagonal, so both filtrations compute the same total homology. This proves the required balancing directly.

The kernel and cokernel of multiplication by $a$ now give Tor for a principal non-zero-divisor quotient:These are naturally $B$-modules. For $M=A$, the degree-one group is zero; for $M=B$, multiplication by $a$ is zero and both degrees zero and one are $B$. Thus, when $B\ne 0$, degree one explicitly detects the failure of this base-change functor to be exact.

Write $R^{i}F$ for the cohomological indexing of the right derived functor. For the contravariant construction, assume $F$ is additive and left exact, so an exact sequence $M^{\prime }\to M\to M^{\prime \prime }\to 0$ gives $0\to F(M^{\prime \prime })\to F(M)\to F(M^{\prime })$. Module categories have enough projective modules: a free module on a generating set surjects onto any module, and repeating the construction on its kernel produces a projective resolutionContravariance turns it into the cochain complexThe composite of consecutive differentials is zero because $F$ is additive and reverses composition. Applying left exactness to $P_1\to P_0\to M\to 0$ gives $F(M)=\ker (F(d_1))$, so $R^{0}F\simeq F$.

Here are the comparison and exactness arguments behind this construction. A map $M\to N$ lifts to a chain map between projective resolutions. In degree zero, lift through the surjection $Q_0\twoheadrightarrow N$ by projectivity of $P_0$. Inductively, the required composite from $P_i$ lands in the cycle module of the target resolution, which is the image of $Q_i$ by exactness, so projectivity supplies the next lift. Two such lifts are chain homotopic. Their difference in degree zero lands in $\ker (Q_0\to N)=\mathrm{im}{d}_1$, so it lifts to a homotopy map $P_0\to Q_1$. Subtracting the corresponding homotopy term in degree one gives another map into the next cycle module; continue inductively. Applying additive contravariant $F$ turns this chain homotopy into a cochain homotopy, so the maps on cohomology agree. Applying this to lifts of identity maps between two resolutions gives mutually inverse maps on cohomology. Thus the contravariant right derived functor is well-defined, contravariant and natural in $M$.

For $0\to M^{\prime }\to M\to M^{\prime \prime }\to 0$, construct compatible projective resolutions by the horseshoe lemma. Its construction is elementary: lift the projective cover of $M^{\prime \prime }$ into $M$ and add the projective cover of $M^{\prime }$; their sum surjects onto $M$. The kernels fit into a new short exact sequence of the corresponding kernels, so repeat at each degree. This gives a degreewise split sequenceAn additive contravariant functor takes split sequences to split sequences with their arrows reversed. The resulting short exact sequence of cochain complexes givesThe connecting map is obtained by lifting a cocycle from the quotient complex, taking its differential, and viewing the result in the subcomplex; changing a lift changes the result by a coboundary. This construction proves both exactness and naturality of the long exact sequence.

Positive derived functors vanish on projective modules, since such a module has a resolution concentrated in degree zero. For $0\to K\to P\to M\to 0$ with $P$ projective, the preceding sequence gives $R^{i+1}F(M)\simeq R^{i}F(K)$ for $i\ge 1$. This is dimension shifting. If $F$ is exact, all positive derived functors vanish because applying $F$ preserves exactness of the resolution. A natural transformation of additive left-exact contravariant functors induces natural transformations of their derived functors by applying it degreewise. These derived functors form the universal contravariant cohomological delta functor: successive projective surjections, on which the positive functors vanish, and the connecting maps determine any extension of a degree-zero natural transformation uniquely.

The displayed example in the PDF has a genuine variance mismatch. The functor $G(M)={\mathrm{Hom}}_A(A/(a),M)$ is covariant in $M$, not contravariant. For a map $u:M\to N$, its induced map sends $h$ to $u\circ h$, in the same direction. We compute that displayed example with its correct variance, and then explain the contravariant variant.

For an additive covariant left-exact functor, use an injective resolution $0\to M\to I^0\to I^1\to \cdots$ instead of a projective resolution, and set $R^{i}G(M)=H^i(G(I^{\bullet }))$. Such resolutions exist for arbitrary modules. In fact $J={\mathrm{Hom}}_{\mathbb{Z}}(A,D)$, with $(a\varphi )(b)=\varphi (ab)$, is injective: the adjunctionsends a map to its value at $1$, and has inverse $\lambda \mapsto [n\mapsto (b\mapsto \lambda (bn))]$. Extension on the right was proved in Question 5, so it gives extension on the left. Products of injective modules are injective by extending every component separately. The mapis $A$-linear and injective, because a nonzero $m$ is detected by some character at $b=1$. Applying the same embedding to successive cokernels produces the stated injective resolution. Its comparison and homotopy arguments are dual to the projective ones, and covariant derived functors have the long exact sequence in the original order of $M^{\prime },M,M^{\prime \prime }$.

Let $B=A/(a)$. Evaluation at $1+(a)$ identifiesa naturally defined $B$-module. It is left exact: a compatible element in the kernel of $G(M)\to G(M^{\prime \prime })$ is exactly an element of $M^{\prime }$ killed by $a$.

Because $a$ is a non-zero-divisor on $A$, multiplication by $a$ is surjective on every injective module $I$. Indeed, for $m\in I$, the map $aA\to I$ given by $ax\mapsto xm$ is well-defined and $A$-linear. It extends to $A\to I$, and if the extension takes $1$ to $v$, then $av=m$. Thus for an injective resolution we have a short exact sequence of cochain complexesThe cohomology of $I^{\bullet }$ is $M$ in degree zero and zero in higher degrees. Its long exact sequence therefore givesConsequently the Ext from a principal non-zero-divisor quotient computation isAll these groups are naturally $B$-modules. Equivalently they are ${\mathrm{Ext}}_A^i(B,M)$; the two-term free resolution $0\to A\stackrel{a}{\to }A\to B\to 0$ gives the same kernel and cokernel after applying ${\mathrm{Hom}}_A(-,M)$.

If one instead wants a genuinely contravariant example, take $H(N)={\mathrm{Hom}}_A(N,B)$. Then $R^{i}H(N)={\mathrm{Ext}}_A^i(N,B)$, obtained by resolving the variable $N$. There is no claim of vanishing above degree one for arbitrary $N$. At $N=B$, the same two-term resolution has zero differential after applying ${\mathrm{Hom}}_A(-,B)$, and gives ${\mathrm{Ext}}_A^0(B,B)=B$, ${\mathrm{Ext}}_A^1(B,B)=B$, and zero higher groups. This keeps the printed covariant calculation distinct from the general contravariant construction.