Projective and Injective Resolutions

Let $i_n:K_n\hookrightarrow P_n$ be the kernel inclusion, and define the new differential by $d_{n+1}:=i_{n}q:P_{n+1}\to P_n\mathrm{.}$ Because $i_n$ lands in the kernel of the previous displayed map, the composite of the new differential with that previous map is zero, so the extended row is again an augmented chain complex in the sense of [L2].

The image of $d_{n+1}$ is the image of $i_{n}q$, which is exactly $K_n$ because $q$ is epic. By [L1], this is precisely the exactness condition at $P_n$. The new term is projective by the given hypothesis and [L3].

Let ${\pi }_n:I^n\twoheadrightarrow C^n$ be the cokernel map and define the new differential by $d^n:=j{\pi }_n:I^n\to I^{n+1}\mathrm{.}$ Since ${\pi }_n$ kills the image of the previous displayed map, the composite of that previous map with $d^n$ is zero, so the extended row is again a coaugmented cochain complex in the sense of [L2].

Because $j$ is monic, the kernel of $d^n=j{\pi }_n$ is the kernel of ${\pi }_n$, namely the image of the previous displayed map. By [L1], this is exactly the required exactness at $I^n$. The new term is injective by the given hypothesis and [L3].

Start with the chosen epimorphism $P_0\twoheadrightarrow A$. Applying [L1] to the chosen epimorphism $P_1\twoheadrightarrow K_0$ makes $P_1\to P_0\to A\to 0$ exact, and repeating the same step with the chosen epimorphism onto each later kernel produces an augmented exact complex $\cdots \to P_2\to P_1\to P_0\to A\to 0$ whose terms are all projective.

By [L2], the complex assembled in step 1.1 is a projective resolution of $A$, including the case $A=0$ when the chosen initial epimorphism may be $0\twoheadrightarrow 0$.

Start with the chosen monomorphism $A\rightarrowtail I^0$. Applying [L1] to the chosen embedding $C^0\rightarrowtail I^1$ makes $0\to A\to I^0\to I^1$ exact, and repeating the same step with the chosen embedding of each later cokernel produces an exact coaugmented complex $0\to A\to I^0\to I^1\to I^2\to \cdots$ whose terms are all injective.

By [L2], the complex assembled in step 1.1 is an injective resolution of $A$, including the case $A=0$ when the chosen initial embedding may be $0\rightarrowtail 0$.

By [L1], the module $M$ has a canonical exact augmented complex of free $R$-modules ending in $M$. Because AC is assumed, [L2] makes every term of that complex projective.

By [L3], that exact augmented complex is a projective resolution of $M$, including the case $M=0$.

By [L1] and [L2], every left $R$-module $X$ admits a functorial monomorphism ${\eta }_X:X\rightarrowtail E(X)$ into an injective module. Starting from $M$, set $I^0:=E(M)$ and let $C^0$ be the cokernel of ${\eta }_M$; recursively set $I^{n+1}:=E(C^n)$ and let $C^{n+1}$ be the cokernel of $C^n\rightarrowtail I^{n+1}$.

Applying [L3] at each stage of the recursion in step 1.1 yields an exact coaugmented complex $0\to M\to I^0\to I^1\to I^2\to \cdots$ whose terms are injective.

By [L4], the complex from step 2.1 is an injective resolution of $M$, including the case $M=0$.

Put $F_0:=R^{(M)}$ and let ${\epsilon }_0:F_0\twoheadrightarrow M$ be the canonical map from [L1]. Having defined $K_n$ as the current kernel, set $F_{n+1}:=R^{(K_n)}$ and let ${\epsilon }_{n+1}:F_{n+1}\twoheadrightarrow K_n$ be its canonical free cover from [L1]. These assignments are functorial because they depend only on the underlying-set construction in [L1].

Each $F_n$ is free, hence projective under the recorded boundary [L2]. By applying [L3] successively to the canonical surjections of step 1.1, one obtains an exact augmented complex $\cdots \to F_2\to F_1\to F_0\to M\to 0.$

Therefore the iterated free-cover construction gives a functorial exact free resolution in ZF. No basis choice or arbitrary lift is used at any stage; projectivity enters only through [L2].

The augmentation ${\epsilon }_Q:Q_0\twoheadrightarrow B$ is epic, and $P_0$ is projective. Apply [L2] to the composite $u{\epsilon }_P:P_0\to B$ to obtain a lift $f_0:P_0\to Q_0$ with ${\epsilon }_Q{f}_0=u{\epsilon }_P$.

By [L1], the map $f_0$ of step 1.1 is exactly the required degree-zero part of an augmentation-preserving comparison map.

If $n=0$, the augmentation identity gives ${\epsilon }_Q{f}_0{d}_1^P=u{\epsilon }_P{d}_1^P=0,$ so $f_0{d}_1^P$ lands in $\ker ({\epsilon }_Q)=Z_0(Q)$. If $n>0$, the previous squares commute and $d_n^Q{f}_n{d}_{n+1}^P=f_{n-1}{d}_n^P{d}_{n+1}^P=0,$ so again $f_n{d}_{n+1}^P$ lands in $Z_n(Q)=\ker (d_n^Q)$ by [L2]. Exactness of $Q_{\bullet }$ at degree $n$ makes the canonical map $Q_{n+1}\twoheadrightarrow Z_n(Q)$ epic by [L1].

The object $P_{n+1}$ is projective, so [L3] lifts $f_n{d}_{n+1}^P:P_{n+1}\to Z_n(Q)$ across the epimorphism $Q_{n+1}\twoheadrightarrow Z_n(Q)$. Writing the lift as $f_{n+1}$ gives $d_{n+1}^Q{f}_{n+1}=f_n{d}_{n+1}^P,$ so the partial comparison map extends by one degree.

By [L1], choose $f_0:P_0\to Q_0$ with ${\epsilon }_Q{f}_0=u{\epsilon }_P$.

Starting from the degree-zero lift in step 1.1, every partial comparison map through degree $n$ extends to one through degree $n+1$ by [L2]. The successive choices depend on the previously chosen partial map, so [L4] produces maps $f_n$ in every degree.

The family $(f_n)$ is a chain map by construction, and step 1.1 gives the augmentation identity. Hence [L3] is satisfied, so $f_{\bullet }$ is a comparison map lifting $u$.

Put $c_n:=f_n-g_n-h_{n-1}{d}_n^P,$ with $h_{-1}=0$ when $n=0$. Using the chain-map identities for $f_{\bullet }$ and $g_{\bullet }$ and the already verified lower-degree homotopy equations, one gets $d_n^Q{c}_n=0$. Thus $c_n$ lands in $Z_n(Q)$ by [L2]; when $n=0$, the common augmentation condition on $f_0$ and $g_0$ says exactly that $c_0$ lands in $\ker ({\epsilon }_Q)=Z_0(Q)$.

Exactness of $Q_{\bullet }$ makes $Q_{n+1}\twoheadrightarrow Z_n(Q)$ epic, and $P_n$ is projective. By [L3], lift $c_n$ to a map $h_n:P_n\to Q_{n+1}$. Then $d_{n+1}^Q{h}_n=c_n$, which is precisely the degree-$n$ homotopy equation from [L1].

Start at degree $0$. Because $f_{\bullet }$ and $g_{\bullet }$ lift the same object map, [L1] produces $h_0:P_0\to Q_1$. Every partial homotopy through degree $n-1$ extends one degree further by [L1], and the successive choices depend on the previously chosen components. Therefore [L3] produces a family $h_n:P_n\to Q_{n+1}$ in every degree.

By construction, the family $(h_n)$ satisfies the defining chain-homotopy equation in every degree. Therefore $f_{\bullet }$ and $g_{\bullet }$ are chain-homotopic.

Apply [L1] to $1_A$ in each direction. This gives comparison maps $f_{\bullet }:P_{\bullet }\to Q_{\bullet }$ and $g_{\bullet }:Q_{\bullet }\to P_{\bullet }$, both lifting the identity on $A$.

The composites $g_{\bullet }{f}_{\bullet }$ and $f_{\bullet }{g}_{\bullet }$ also lift $1_A$, as do the identity chain maps on $P_{\bullet }$ and $Q_{\bullet }$. By [L2], $g_{\bullet }{f}_{\bullet }\simeq 1_{P_{\bullet }}\text{and}{f}_{\bullet }{g}_{\bullet }\simeq 1_{Q_{\bullet }}\mathrm{.}$ Thus the two resolutions are homotopy equivalent over $A$, including when $A=0$.

By [L1], pass to the opposite abelian category. Reversing arrows turns the given injective resolutions from [L2] into projective resolutions there, and $u:A\to B$ becomes a morphism in the opposite direction. Apply [L3] in the opposite category to obtain the required comparison map.

Translating that chain map back to the original category reverses arrows again and yields a cochain map $I^{\bullet }\to J^{\bullet }$ extending $u$.

By [L2], pass to the opposite abelian category. There the two given maps become comparison maps between projective resolutions lifting the same morphism, so [L1] makes them chain-homotopic.

Translating the resulting chain homotopy back to the original category gives the required cochain homotopy.

Apply [L1] to the identity on $A$ in both directions. This yields maps $I^{\bullet }\to J^{\bullet }$ and $J^{\bullet }\to I^{\bullet }$ extending $1_A$.

Their composites and the identity cochain maps all extend $1_A$, so [L2] makes the composites homotopic to the identities. Hence the two injective resolutions are homotopy equivalent under $A$, including when $A=0$.

The projective statement is exactly [L1], and the injective statement is exactly [L2].

Thus either kind of resolution is unique only up to homotopy equivalence. The preceding comparison theorems show that one may choose many actual lifts inside that homotopy class, so the equivalence is not unique on the nose.

Let $F_{\bullet }$ lift $u$ and $G_{\bullet }$ lift $v$. By [L1], choose a comparison map $H_{\bullet }$ lifting the composite $vu$. The composite $G_{\bullet }{F}_{\bullet }$ also lifts $vu$.

Since $H_{\bullet }$ and $G_{\bullet }{F}_{\bullet }$ lift the same morphism $vu$, [L2] makes them chain-homotopic. This is exactly the claimed compatibility with composition up to homotopy.

By [L2], both $F_{\bullet }$ and $1_{P_{\bullet }}$ are comparison maps lifting $1_A$.

Apply [L1] with both factors equal to the identity morphism on $A$, taking the chosen lift of the composite to be $F_{\bullet }$ and the two factor lifts to be the literal identity chain maps. Then $F_{\bullet }$ is homotopic to $1_{P_{\bullet }}\circ 1_{P_{\bullet }}=1_{P_{\bullet }}$.

Since $p:A\twoheadrightarrow A^{\mathrm{\prime }\mathrm{\prime }}$ is epic and $P_0^{\mathrm{\prime }\mathrm{\prime }}$ is projective, [L2] lifts the augmentation ${\epsilon }^{\mathrm{\prime }\mathrm{\prime }}:P_0^{\mathrm{\prime }\mathrm{\prime }}\to A^{\mathrm{\prime }\mathrm{\prime }}$ to a map $s:P_0^{\mathrm{\prime }\mathrm{\prime }}\to A$ with $ps={\epsilon }^{\mathrm{\prime }\mathrm{\prime }}$.

Define ${\lambda }_0(x,y):=i{\epsilon }^{\mathrm{\prime }}(x)+s(y)\mathrm{.}$ To hit a given $a\in A$, first choose $y\in P_0^{\mathrm{\prime }\mathrm{\prime }}$ with ${\epsilon }^{\mathrm{\prime }\mathrm{\prime }}(y)=p(a)$. Then $a-s(y)$ lies in $\ker (p)=\mathrm{im}(i)$, so $a-s(y)=i(a^{\mathrm{\prime }})$ for some $a^{\mathrm{\prime }}\in A^{\mathrm{\prime }}$, and some $x\in P_0^{\mathrm{\prime }}$ satisfies ${\epsilon }^{\mathrm{\prime }}(x)=a^{\mathrm{\prime }}$. Hence ${\lambda }_0(x,y)=a$, so ${\lambda }_0$ is epic. Its source is projective by [L3].

The map ${\lambda }_0$ from [L1] fits into a commutative diagram $0\to P_0^{\mathrm{\prime }}\to P_0^{\mathrm{\prime }}\oplus P_0^{\mathrm{\prime }\mathrm{\prime }}\to P_0^{\mathrm{\prime }\mathrm{\prime }}\to 0$ over $0\to A^{\mathrm{\prime }}\to A\to A^{\mathrm{\prime }\mathrm{\prime }}\to 0,$ where both rows are exact and the top row is split exact. Applying the snake lemma [L3] gives an exact sequence $0\to \ker ({\epsilon }^{\mathrm{\prime }})\to \ker ({\lambda }_0)\to \ker ({\epsilon }^{\mathrm{\prime }\mathrm{\prime }})\to 0.$

By [L2], these three kernels are exactly ${\mathrm{\Omega }}^1(A^{\mathrm{\prime }})$, $K$, and ${\mathrm{\Omega }}^1(A^{\mathrm{\prime }\mathrm{\prime }})$. Hence the displayed kernel sequence is short exact.

Apply [L1] to the short exact sequence $0\to K_n^{\mathrm{\prime }}\to K_n\to K_n^{\mathrm{\prime }\mathrm{\prime }}\to 0$ and to the degree-zero tails of the already chosen projective resolutions of $K_n^{\mathrm{\prime }}$ and $K_n^{\mathrm{\prime }\mathrm{\prime }}$. This produces a projective object $P_{n+1}=P_{n+1}^{\mathrm{\prime }}\oplus P_{n+1}^{\mathrm{\prime }\mathrm{\prime }}$ together with an epimorphism $P_{n+1}\twoheadrightarrow K_n$.

Applying [L2] to that new surjection gives the next short exact sequence of kernels. Therefore the horseshoe construction advances by one degree.

Start at degree zero with the original short exact sequence $0\to A^{\mathrm{\prime }}\to A\to A^{\mathrm{\prime }\mathrm{\prime }}\to 0.$ Every current short exact kernel sequence extends one more degree by [L1], and the next choice depends on the previously constructed degree. Therefore [L4] produces the whole augmented complex for $A$, whose degree-$n$ term is $P_n^{\mathrm{\prime }}\oplus P_n^{\mathrm{\prime }\mathrm{\prime }}$ and whose kernel sequences remain short exact in every degree.

Each term $P_n^{\mathrm{\prime }}\oplus P_n^{\mathrm{\prime }\mathrm{\prime }}$ is projective by [L3], and the short exact kernel sequences from step 1.1 are exactly the data needed for exactness of the middle augmented complex. Therefore [L2] identifies the resulting complex as a projective resolution of $A$.

By [L3], pass to the opposite abelian category. The given injective resolutions from [L2] become projective resolutions there, so [L1] supplies the dual horseshoe resolution in the opposite category.

Translating back to the original category reverses arrows again and turns the opposite-category coproducts into finite products, which in an abelian category are the same biproducts. Hence one obtains the asserted injective horseshoe resolution.

By hypothesis and [L3], the two chosen middle maps are comparison maps between the same pair of projective horseshoe resolutions, they lift the same middle-object morphism, and together with the fixed side maps they define morphisms of short exact sequences of complexes.

Any two such middle comparison maps lifting the same object morphism are chain-homotopic by [L2]. Therefore the horseshoe construction is compatible with morphisms only up to homotopy, not canonically on the nose.

By [L2], identify $A$ with $A^{\mathrm{\prime }}\oplus A^{\mathrm{\prime }\mathrm{\prime }}$. Applying [L1] to the chosen projective resolutions and the chosen splitting maps gives a middle resolution whose degree-$n$ term is $P_n^{\mathrm{\prime }}\oplus P_n^{\mathrm{\prime }\mathrm{\prime }}$ and whose augmentation is the direct-sum augmentation.

Under that identification, the differentials are exactly the direct-sum differentials of the two side resolutions. Hence the split short exact sequence admits the direct-sum resolution.

Form the pullback $X=P{\times }_A{P}^{\mathrm{\prime }}$ of the two epimorphisms onto $A$ as in [L1]. By [L2], the projections $X\twoheadrightarrow P$ and $X\twoheadrightarrow P^{\mathrm{\prime }}$ are epimorphisms. Their kernels are $K^{\mathrm{\prime }}$ and $K$ respectively, so there are short exact sequences $0\to K^{\mathrm{\prime }}\to X\to P\to 0,0\to K\to X\to P^{\mathrm{\prime }}\to 0.$

Since $P$ and $P^{\mathrm{\prime }}$ are projective, [L3] splits both short exact sequences. Therefore $X\cong K^{\mathrm{\prime }}\oplus P\cong K\oplus P^{\mathrm{\prime }},$ which yields the claimed isomorphism $K\oplus P^{\mathrm{\prime }}\cong K^{\mathrm{\prime }}\oplus P\mathrm{.}$

The first syzygies of the two resolutions are the kernels of two projective presentations of $A$ by [L1]. Therefore [L2] gives a stable isomorphism between them.

Proceed by induction. Suppose ${\mathrm{\Omega }}_P^{n-1}(A)\oplus E\cong {\mathrm{\Omega }}_Q^{n-1}(A)\oplus F$ for projective objects $E,F$. After identifying these sums with a common object, the two exact rows $0\to {\mathrm{\Omega }}_P^n(A)\to P_{n-1}\oplus E\to {\mathrm{\Omega }}_P^{n-1}(A)\oplus E\to 0$ and $0\to {\mathrm{\Omega }}_Q^n(A)\to Q_{n-1}\oplus F\to {\mathrm{\Omega }}_Q^{n-1}(A)\oplus F\to 0$ are projective presentations of that common object. Applying [L2] gives ${\mathrm{\Omega }}_P^n(A)\oplus Q_{n-1}\oplus F\cong {\mathrm{\Omega }}_Q^n(A)\oplus P_{n-1}\oplus E,$ so the $n$th syzygies are stably isomorphic. Together with step 1.1 this proves the claim in every degree.

By [L2], pass to the opposite abelian category. The two injective copresentations become projective presentations there, so [L1] applies.

Translating the resulting stable isomorphism back to the original category gives $C\oplus I^{\mathrm{\prime }}\cong C^{\mathrm{\prime }}\oplus I,$ which is the dual Schanuel statement.

Consider the augmented complex $0\to P\stackrel{1_P}{\to }P\to 0,$ with $P$ placed in degree zero. It is exact because the augmentation is the identity, and its only nonzero term is projective by [L3].