Group Homology Transfer and Low-Degree Exact Sequences

Turning a left module into a right module by $b\cdot g=g^{-1}b$ preserves exactness and takes the regular free left module to a free right module (send the basis coordinate g to $g^{-1}$). Thus the normalized bar complex $B={\bar{B}}^{\mathrm{r}\mathrm{i}\mathrm{g}\mathrm{h}\mathrm{t}}(G)\to \mathbb{Z}$ is a free right resolution: normalization preserves exactness, and its nondegenerate orbits supply a free basis.

Put $R=\mathbb{Z}G$ and $D_{pq}=B_p{\otimes }_R{P}_q$ with differential $h=d_B\otimes 1$ and $v=(-1{)}^{p}1\otimes d_P$. Then $hv+vh=0$. Tensor with $B_p$, a free right module, preserves exactness: each finite-support cycle has a primitive by lifting only its finitely many nonzero coordinates. Each projective $P_q$ is a retract of the free module on its underlying set: lift its identity through the canonical surjection. Tensor with $P_q$ is therefore a retract of tensor with a free left module and preserves exact sequences of right modules. The augmented columns of D are exact with bottom $B_p{\otimes }_{R}M$, and augmented rows are exact with left edge $\mathbb{Z}{\otimes }_R{P}_q$.

The map $B{\otimes }_{R}M\to (\bar{B}{\otimes }_{\mathbb{Z}}M{)}_G$ sending $b\otimes m$ to its diagonal orbit class is well-defined: $g^{-1}b\otimes m$ and $b\otimes gm$ are in the same orbit. Conversely $(gb)\otimes (gm)=b\otimes m$ in the balanced tensor product since $gb=b\cdot g^{-1}$. The same formula therefore defines an inverse, and both composites fix every pure tensor. Each vertex deletion commutes with these formulas, including the first and last deletions, giving a chain isomorphism.

Here is the finite comparison argument for either augmentation. For exact augmented columns, place the augmentation in vertical degree -1. In the resulting augmented total complex, a total n-cycle has finitely many components, with $0\le p\le n+1$. At its largest remaining p, the cycle equation says $v{z}_p=0$, because the component from p+1 is zero. Exactness supplies $w_p$ with $v{w}_p=z_p$. Subtract $(h+v)w_p$; the p-component vanishes and the only new component is at p-1. Repeat down to p=0, where h is zero. The cycle has become zero after finitely many boundary subtractions. Thus the augmented total complex is acyclic. The same proof with p and q interchanged works for exact augmented rows, using the largest remaining q. Signs of the primitive are absorbed into w.

The augmented total complex is the cone of the total-to-edge augmentation up to a shift and sign. Acyclicity implies that augmentation induces a homology isomorphism: a cycle on the edge lifts to a total cycle because the corresponding cone cycle bounds; if a total cycle maps to an edge boundary, pairing it with that boundary primitive makes a cone cycle, whose being a boundary says the original cycle bounds. Hence $H_{\ast }(B{\otimes }_{R}M)\cong H_{\ast }(\mathrm{Tot}D)\cong H_{\ast }(\mathbb{Z}{\otimes }_{R}P)=H_{\ast }^P(G;M)$. DC is the inherited resolution-comparison assumption; the finite elimination in step 3.1 adds no arbitrary family of choices.

The augmentations and tensor formulas commute with coefficient maps and with the bar maps induced by group homomorphisms. Maps between supplied resolutions give the same maps on homology by the supplied-resolution convention. Since inverses of isomorphisms are unique, the composite identification is natural. Degree zero gives the usual coinvariants; for $M=0$ all complexes are zero, and for $G=1$ normalized bars vanish in positive degrees.

The retraction satisfies $r(hx)=hr(x)$ for $h\in H$, since $Hx=Hhx$. For $x\in G$ and $t\in T$ write $tx=h_t{t}_x$. Right multiplication by x permutes the right cosets, so $t\mapsto t_x$ permutes T. The t-summand on $[(x{g}_0,\dots ,x{g}_n)\otimes xm]$ is $[(h_{t}r(t_x{g}_0),\dots ,h_{t}r(t_x{g}_n))\otimes h_t{t}_{x}m]$, equal in H-coinvariants to the $t_x$-summand of the original chain. This proves diagonal invariance and well-definedness.

For two H-equivariant vertex maps f,u define the prism $P_n(v_0,\dots ,v_n)={\sum }_{i=0}^n(-1{)}^i(f{v}_0,\dots ,f{v}_i,u{v}_i,\dots ,u{v}_n)$. Expansion gives $dP+Pd=u_{\ast }-f_{\ast }$: deletions away from the switch cancel the corresponding terms of Pd; the two switch faces at consecutive values of i cancel, leaving only deletion of the first f-vertex at i=0 and of the last u-vertex at i=n, with signs + and -. For n=0 this reads $d(f{v}_0,u{v}_0)=(u{v}_0)-(f{v}_0)$. Equivariance makes this descend to diagonal coinvariants. If $v_j=v_{j+1}$, each prism summand has either an adjacent equal f-pair or an adjacent equal u-pair, so it also descends to normalization.

For each deletion index $0\le j\le n$, deleting the jth vertex of $(r(t{g}_0),\dots ,r(t{g}_n))$ is exactly applying r to the tuple with $g_j$ deleted; the coefficient remains tm. Thus all faces, including j=0,n, commute with the sum and so does their alternating differential. Adjacent equal input vertices stay equal after r, so degeneracies map to degeneracies. The formula descends to normalized chains.

Let U be another transversal with retraction u. For each right coset write its representative in U as $a_{t}t$, $a_t\in H$. The corresponding U-summand becomes $[(u(t{g}_0),\dots ,u(t{g}_n))\otimes tm]$ after translating by $a_t^{-1}$ in the H-coinvariants. Thus compare r and u on the same inputs $(t{g}_0,\dots ,t{g}_n)$ and same coefficient tm. Sum the prism of step 1.2 over T. The permutation calculation of step 1.1 applies to the prism too, because both vertex maps are H-equivariant. It gives a well-defined normalized homotopy K with $dK+Kd={\mathrm{Tr}}_U-{\mathrm{Tr}}_T$.

For completeness, in the inhomogeneous tuple $(1,g_1,g_1{g}_2,\dots ,g_1\cdots g_n)$, the recursion gives $r(t{g}_1\cdots g_i)=h_1\cdots h_i$ and $r(t)=1$. Consecutive vertex ratios are therefore the printed h_i, proving the inhomogeneous formula. All sums and choices are finite; for H=G use T={1}.

Put $u(x)=x$. For each $t\in T$ define the alternating vertex prism $P_{t,n}[(g_0,\dots ,g_n)\otimes m]={\sum }_{i=0}^n(-1{)}^i[(r(t{g}_0),\dots ,r(t{g}_i),t{g}_i,\dots ,t{g}_n)\otimes tm]\mathrm{.}$ Expanding the vertex-deletion differential cancels every face away from the switch in pairs; the two endpoint faces that survive give $d{P}_t+P_{t}d=u_{t,\ast }-r_{t,\ast },$ where $u_{t,\ast }$ uses $(t{g}_0,\dots ,t{g}_n)$ and $r_{t,\ast }$ is the $t$-summand of corestriction after the transfer in [F2]. If $tx=h_t{t}_x$ with $h_t\in H$ and $t_x\in T$, then $t\mapsto t_x$ permutes $T$, $r(txg)=h_{t}r(t_{x}g)$, and both vertex strings and the coefficient $txm=h_t{t}_{x}m$ change by the same diagonal translation. Hence $P={\sum }_{t\in T}{P}_t$ descends to $G$-coinvariants. Equal adjacent input vertices make every prism summand degenerate, so it also descends to normalized chains. Thus $dP+Pd={\sum }_{t\in T}{u}_{t,\ast }-i_{\ast }{\mathrm{Tr}}_T\mathrm{.}$

Each translated summand is $[(g_0,\dots ,g_n)\otimes m]$ in diagonal G-coinvariants. The sum is therefore $[G:H]$ times that chain in every degree, including zero. Homotopic chain maps give the same map on homology because their difference on a cycle is a boundary. Thus $i_{\ast }\mathrm{Tr}=[G:H]\mathrm{id}$.

For the trivial group, every tuple of length at least two has adjacent equal vertices. Its normalized chain groups in positive degrees are consequently zero, so $H_n(1;\mathbb{Z})=0$ for n>0.

Apply transfer to $1\le G$. Its index is $\mathrm{\mid }G\mathrm{\mid }$ and the composite factors through the zero group of step 1.1; F1 identifies this composite with multiplication by $\mathrm{\mid }G\mathrm{\mid }$. Hence it is zero. If G=1, this says the positive homology itself is zero. Degree zero is excluded: there it is $\mathbb{Z}$.

In degree one all boundaries to degree zero are zero with trivial coefficients. Degree two has $d[x\mathrm{\mid }y]=[y]-[xy]+[x]$, with $[1]=0$. Thus H1 is the abelian group generated by symbols [x] subject precisely to $[xy]=[x]+[y]$. The assignment $[x]\mapsto x[G,G]$ induces a map to $G_{\mathrm{a}\mathrm{b}}$; conversely $x\mapsto [x]$ is a group homomorphism to an abelian group, kills every commutator, and factors through $G_{\mathrm{a}\mathrm{b}}$. The maps are inverse on generators and commute with group homomorphisms.

For R the isomorphism sends conjugation by f to $r[R,R]\mapsto fr{f}^{-1}[R,R]$. Conjugation by an element of R is trivial on this quotient, so the action factors through F/R. Taking coinvariants adds exactly the relations $fr{f}^{-1}{r}^{-1}=1$. Since $[R,R]\subseteq [F,R]$, the resulting quotient is $R\mathrm{/}[F,R]$. If R=1 it is zero, and if R=F it is $F_{\mathrm{a}\mathrm{b}}$.

For a word $w=l_1\cdots l_s$, telescoping gives $w-1={\sum }_{j=1}^s{l}_1\cdots l_{j-1}(l_j-1)$. A positive letter contributes a multiple of $x-1$, and $x^{-1}-1=-x^{-1}(x-1)$ does too. Every element ${\sum }_w{a}_{w}w$ of augmentation zero is ${\sum }_w{a}_w(w-1)$, so $\mathrm{im}\mathrm{\partial }=\ker ϵ$. Empty words contribute zero; epsilon is onto since epsilon(1)=1.

Interpret $w{e}_x$ as the oriented edge from w to wx. Its boundary is wx-w. The underlying graph is connected by reduced words and has no simple cycle: a simple cycle would give a nonempty reduced word equal to 1, impossible by F1. A finite nonzero edge chain has support in a finite forest. A nonempty finite forest with an edge has a terminal vertex (take an endpoint of a longest simple path); at that vertex its boundary coefficient is plus or minus the nonzero coefficient of its unique incident supported edge. Thus a nonzero finite edge chain cannot have zero boundary, proving injectivity.

Write $R=\mathbb{Z}F$ and denote the displayed free left resolution by $P$. Turn it into a right resolution $T$ by $t\cdot g=g^{-1}t$. This preserves the underlying exact sequence. Each left regular summand becomes a right regular summand by the coordinate map $g\mapsto g^{-1}$; thus $T_0\cong R$, $T_1\cong {⨁}_{X}R$, and the right differential sends the basis vector indexed by x to $x^{-1}-1$. Tensoring with a free module preserves exactness: the tensor product is a direct sum of copies of the original sequence, and lifting an element requires preimages only for its finitely many nonzero coordinates. For each fixed q, projectivity of $Q_q$ lifts its identity through the canonical surjection from the free left module on its underlying set, making $Q_q$ a retract of that free module. Consequently tensoring with $Q_q$ also preserves exactness, as a retract of an exact tensor functor. This uses no choice of lifts for an arbitrary basis and no simultaneous choice of splittings for all q.

Form $D_{pq}=T_p{\otimes }_R{Q}_q$ for $p,q\ge 0$, with $h=d_T\otimes 1$ and $v=(-1{)}^{p}1\otimes d_Q$. These differentials anticommute, so $d=h+v$ defines the direct-sum total complex. The two augmentations give degreewise surjective chain maps $a:\mathrm{Tot}D\to T{\otimes }_R\mathbb{Z}$ and $b:\mathrm{Tot}D\to \mathbb{Z}{\otimes }_{R}Q$, zero off q=0 and p=0 respectively. By step 3.1, all augmented columns and all augmented rows are exact. Hence $\ker a$, viewed columnwise with its degree-zero column term replaced by the augmentation kernel, has exact columns; similarly $\ker b$ has exact rows.

Both kernel total complexes are acyclic by the following finite argument. For a total n-cycle in $\ker a$, take the largest p with a nonzero component. Its vertical differential is zero, since the component at p+1 is zero. Exactness in that column supplies a vertical primitive. Subtract its total boundary: the p-component vanishes and only a component at p-1 can be introduced. Repeating ends at p=0, where h is zero. There are at most n+1 columns to remove, so the cycle is a boundary. For $\ker b$, use the largest q and horizontal primitives, decreasing q until q=0, where v is zero. Signs are absorbed into the primitives. These arguments use only finitely many existential choices for each cycle.

A degreewise surjective chain map with acyclic kernel induces a homology isomorphism: lift a target cycle; its differential is a kernel cycle, so subtract a kernel primitive to make the lift a cycle. If a source cycle maps to a boundary, lift that boundary's primitive and subtract its differential; the result is a kernel cycle and hence a boundary. This proves surjectivity and injectivity on homology, including degree zero. Applying this to a and b yields $H_{\ast }(T{\otimes }_R\mathbb{Z})\cong H_{\ast }(\mathbb{Z}{\otimes }_{R}Q)=H_{\ast }(F;\mathbb{Z})$ under the assumed DC and supplied-resolution convention.

The differential of $T{\otimes }_R\mathbb{Z}$ sends every $x^{-1}-1$ to zero. Its only nonzero terms are ${⨁}_X\mathbb{Z}$ in degree one and $\mathbb{Z}$ in degree zero, proving the asserted homology groups. If X is empty, F=1 and the degree-one term is zero.

The identity $(h+v{)}^2=0$ makes T a chain complex, and h lowers p while v preserves it, so $F_{r}T$ is a subcomplex. The quotient $F_{r}T\mathrm{/}{F}_{r-1}T$ has just differential v; its homology is the rth column homology, whose induced differential is h. This yields the stated E2 subquotients using only kernels and images. Below, an element denotes a representative in these subquotients. In an abelian category the same notation means a morphism into the indicated kernel after pulling back the epimorphism onto the image; every lift below is of this form, and equality of subobjects can be checked after such epimorphic pullbacks. Thus the calculations do not require objects to have underlying sets.

A class in $E_{20}^2$ is represented by $x\in D_{20}$ with $hx=-vy$ for some $y\in D_{11}$. Set $d_2[x]=[hy]\in E_{01}^2$. Indeed $vhy=-hvy=h^{2}x=0$. Replacing y by another such lift changes hy by h of a vertical cycle, zero in E2. Replacing x by $x+vt+hu$ with $t\in D_{21}$, $u\in D_{30}$ changes a compatible y to $y+ht$; its h-image is unchanged. These are precisely the vertical-boundary and horizontal-boundary changes allowed in $E_{20}^2$. Hence d2 is a well-defined homomorphism.

A total 2-cycle has components $(x,y,z)\in D_{20}\oplus D_{11}\oplus D_{02}$ with $hx+vy=0$ and $hy+vz=0$. Define $e_2[(x,y,z)]=[x]$. A total 3-boundary changes x by $hu+vt$ from $D_{30}$ and $D_{21}$, so e2 is well-defined. Its image is in the kernel of d2. Conversely if $d_2[x]=0$, choose y as in step 2.1. Then $hy=ha+vb$ for a vertical cycle $a\in D_{11}$ and $b\in D_{02}$. The triple $(x,y-a,-b)$ is a total cycle and maps to [x]. This proves exactness at $E_{20}^2$.

A vertical cycle $z\in D_{01}$ is a total cycle; define $j[z]=[(0,z)]$. Replacing z by $vb+ha$ with $va=0$ adds the total boundary $d(b+a)$, so j is well-defined on $E_{01}^2$. If z=hy as in step 2.1, then $z=d(x+y)$, proving $j{d}_2=0$. Conversely, if $(0,z)=d(x+y+b)$, its p=1 component gives $hx+vy=0$ and its p=0 component gives $z=hy+vb$. Therefore $[z]=d_2[x]$. This proves exactness at $E_{01}^2$.

A total 1-cycle $(a,b)\in D_{10}\oplus D_{01}$ satisfies $ha+vb=0$. Define $e_1[(a,b)]=[a]\in E_{10}^2$. Boundaries change a by $hx+vy$, so e1 is well-defined. Every $E_{10}^2$ representative a admits b with $ha=-vb$, hence e1 is onto. Clearly $e_{1}j=0$. If $[a]=0$, write $a=hx+vy$ with $x\in D_{20}$, $y\in D_{11}$. Subtract $d(x+y)$ from $(a,b)$; the result is $(0,b-hy)$, a vertical cycle, hence in the image of j. This proves exactness at $H_1(T)$ and at the final nonzero term.

All constructions commute with morphisms of bicomplexes: they use the same components, and images of chosen lifts are compatible lifts in the target, whose class is independent of the lift. Only components of total degree at most three were used. In particular $F_0{H}_1(T)=\mathrm{im}j\cong E_{01}^2\mathrm{/}\mathrm{im}{d}_2$ and $H_1(T)\mathrm{/}{F}_0{H}_1(T)\cong E_{10}^2$ by steps 3.2–4.1. These explicit subquotients establish the required low-degree filtration assertions without an infinite limiting process. Zero rows or columns are permitted throughout.

Normality of R makes left F-translation on R-orbits factor through G. A nondegenerate F-tuple has a unique form $f_0(1,a_1,\dots ,a_q)$; hence its R-orbit is specified by $\pi (f_0)$ and the relative tuple $(1,a_1,\dots ,a_q)$. Thus $C_q$ is canonically free over $\mathbb{Z}G$ on these relative tuples. Both differentials are well-defined, square to zero, and anticommute because the vertical sign changes when p decreases.

To identify $H_q(C)$ without choosing a transversal of R in F, note that the permutation module on any free R-set is tensor-exact. Each element of a tensor product has finite support in the orbit set. On the union of those finitely many orbits choose one representative per orbit; there it is a finite sum of regular free modules. Coordinate lifting proves exactness on that summand, and projection onto it shows injectivity is tested there too. Thus tensoring with the restriction of each ${\bar{B}}_q(F)$ preserves exactness even without selecting representatives of all orbits at once. The augmented complex $\bar{B}(F)\to \mathbb{Z}$ is exact also after restriction to R. Form ${\bar{B}}^{\mathrm{r}\mathrm{i}\mathrm{g}\mathrm{h}\mathrm{t}}(R){\otimes }_{\mathbb{Z}R}\bar{B}(F)$. Its exact augmented rows and columns give, by the finite elimination in F1, $H_{\ast }(C)\cong H_{\ast }(R;\mathbb{Z})$. Applying the same comparison to $\bar{B}(R)\to \bar{B}(F)$ shows that the isomorphism is induced by the literal inclusion of R-tuples.

For f in F, the maps on R-vertices $r\mapsto fr$ and $r\mapsto fr{f}^{-1}$ into F are equivariant for the same conjugated R-action. The alternating prism between them, ${\sum }_i(-1{)}^i(f{r}_0,\dots ,f{r}_i,f{r}_i{f}^{-1},\dots ,f{r}_n{f}^{-1})$, has boundary equal to their difference: off-switch faces cancel in pairs and the two surviving switch endpoints are the two maps. It preserves normalized degeneracies and descends to R-coinvariants. Hence the action on $H_q(C)$ induced by left f is conjugation by f on $H_q(R;\mathbb{Z})$. Elements of R already act trivially on C, so this is the stated G-action.

For fixed q, the horizontal complex ${\bar{B}}^{\mathrm{r}\mathrm{i}\mathrm{g}\mathrm{h}\mathrm{t}}(G)\otimes C_q$ has zero positive homology since C_q is free; its zero homology is $(C_q{)}_G=({\bar{B}}_q(F){)}_F$. The augmentation to this column induces a total homology isomorphism by the finite row elimination of F1. For fixed p, the first factor is free, so vertical homology is ${\bar{B}}_p^{\mathrm{r}\mathrm{i}\mathrm{g}\mathrm{h}\mathrm{t}}(G)\otimes H_q(C)$ (the sign does not change kernels or images). Taking horizontal homology and using step 3.1 gives exactly the asserted E2 terms. Thus the total computes $H_{\ast }(F;\mathbb{Z})$, and F3 applies.

The map from $E_{01}^2$ to total H1 sends the bar cycle $(1,r)$, r in R, to $y=(1)\otimes [(1,r){]}_R\in D_{01}$. The horizontal augmentation sends this to $[(1,r){]}_F$, which represents r in $F_{\mathrm{a}\mathrm{b}}$. F2 identifies its domain with R/[F,R], so this edge is the homomorphism induced by the subgroup inclusion $R\hookrightarrow F$; it is not asserted to be injective.

For arbitrary f in F put $g=\pi (f)$ and $y=(1)\otimes [(1,f){]}_R\in D_{01}$, $x=(1,g^{-1})\otimes [(1){]}_R\in D_{10}$. Balancing uses $(g^{-1})=(1)\cdot g$, so $hx=(1)\otimes ([f{]}_R-[1{]}_R)=vy$. Thus y-x is a total cycle whose horizontal augmentation represents [f]. The vertical augmentation sends it to $-[(1,g^{-1})]$, representing $-[g^{-1}]=[g]$ in $G_{\mathrm{a}\mathrm{b}}$. Since the [f] generate H1, the other edge is exactly the quotient map. If g=1 then x is degenerate and zero, consistent with step 5.1.

Group maps of presentations act vertexwise on bars and on R-orbits, commuting with all augmentations and component maps. The constructed homology and E2 identifications, including the edge maps, are therefore natural. Empty X, trivial R, and trivial G cause no failure in the formulas; when G=1 all horizontal positive bars vanish.

The total homology equals H*(F) by F1. The free resolution of F2 has no terms above degree one, so after tensoring it has zero homology there; in degree one it gives the free abelian group on the free generators, which is $F_{\mathrm{a}\mathrm{b}}$. Hence the asserted total vanishing holds even for an infinite free generating set.

The zero homology of R with trivial integers is $\mathbb{Z}$: every vertex is identified, and each degree-one boundary is a difference of vertices. Conjugation fixes this generator. Consequently $E_{p0}^2=H_p(G;\mathbb{Z})$. Also $E_{01}^2=H_0(G;H_1(R;\mathbb{Z}))=R\mathrm{/}[F,R]$ by F3. The edge calculations of F1 send r to its class in $F_{\mathrm{a}\mathrm{b}}$ and f to its image in $G_{\mathrm{a}\mathrm{b}}$, with positive signs.

Apply F2 to the first-quadrant free-presentation bicomplex. Substitute $H_2(T)=0$, $E_{20}^2=H_2(G;\mathbb{Z})$, $E_{01}^2=R\mathrm{/}[F,R]$, $H_1(T)=F_{\mathrm{a}\mathrm{b}}$ and $E_{10}^2=G_{\mathrm{a}\mathrm{b}}$ from F1. This identifies the groups in the displayed sequence and supplies exactness at the last three nonzero terms; the next two steps define the first arrow by the printed representative formula and prove the two adjacent exactness claims directly.

In the bicomplex of [F3], represent a class in $E_{20}^2$ by $x\in D_{20}$ with $hx=-vy$ for some $y\in D_{11}$, and define $d_2[x]=[hy]\in E_{01}^2$. This is a vertical cycle because $vhy=-hvy=h^{2}x=0$. Replacing $y$ by another lift changes $hy$ by the horizontal boundary of a vertical cycle. Replacing $x$ by $x+vt+hu$, with $t\in D_{21}$ and $u\in D_{30}$, permits the compatible replacement $y+ht$ and leaves $hy$ unchanged. Thus the displayed formula is well-defined.

This $d_2$ is injective here. If $[hy]=0$ in $E_{01}^2$, write $hy=ha+vb$ for a vertical cycle $a\in D_{11}$ and $b\in D_{02}$. Then $(x,y-a,-b)$ is a total 2-cycle mapping to $[x]$. Since $H_2(T)=0$ by [F1], its class and hence $[x]$ vanish. Moreover the edge map $j:E_{01}^2\to H_1(T)$ sends a vertical cycle $z$ to the total cycle $(0,z)$. The equation $j{d}_2[x]=0$ follows from $(0,hy)=d(x+y)$. Conversely, if $j[z]=0$, write $(0,z)=d(x+y+b)$ with $x\in D_{20}$, $y\in D_{11}$, and $b\in D_{02}$. Its components give $hx=-vy$ and $z=hy+vb$, hence $[z]=d_2[x]$. This proves exactness at the first two nonzero terms for the stated representative formula; [F2] supplies exactness at the remaining terms.

The edge computations in F1 identify the next arrows with r mapped into $F_{\mathrm{a}\mathrm{b}}$ and f mapped to its quotient class. All constructions in steps 1.2–2.1 commute with the vertexwise maps induced by maps of presentations, so the sequence and the displayed transgression are natural. No terminal surjectivity beyond the printed $G_{\mathrm{a}\mathrm{b}}\to 0$ is needed. For R=1 the sequence reduces to zero H2 of F and the identity of its abelianization; for G=1 it reduces to the identity $F_{\mathrm{a}\mathrm{b}}\to F_{\mathrm{a}\mathrm{b}}$.

In degrees zero and one the differential reads $(\delta a)(g)=ga-a$ and $(\delta d)(g,h)=gd(h)-d(gh)+d(g)$. Thus $\delta d=0$ is exactly the crossed-homomorphism identity. Setting g=h=1 gives d(1)=0, so every crossed homomorphism is normalized. Also $(gh)a-a=(ga-a)+g(ha-a)$, so every principal map is crossed. Both sets are additive groups and the principal maps form a subgroup.

The cycles and boundaries of normalized degree one are therefore exactly the two groups in the Definition, so their quotient is bar H1 and hence, under the stated convention, the cohomology of F2. For trivial action the crossed identity is the homomorphism identity and every principal map is zero, giving $H^1(G,A)=\mathrm{Hom}(G,A)$. For G=1 or A=0 it is zero.

If a is N-fixed, then $n(ga)=g(g^{-1}ng)a=ga$, since N is normal. Thus ga is N-fixed. Replacing g by gn with n in N does not change ga, proving the Q-action on $A^N$. For a crossed map c into $A^N$, $c(\pi (gh))=c(\pi g)+gc(\pi h)$; hence its inflation is crossed. Principal c from a in $A^N$ inflates to $g\mapsto ga-a$. Restriction plainly preserves the crossed identity and sends the principal map from a to the principal map from the same a.

For $d\in Z^1(N,A)$, write $u=g^{-1}ng$, $w=g^{-1}mg$. Then $(g\cdot d)(nm)=g(d(u)+ud(w))=(g\cdot d)(n)+n(g\cdot d)(m)$, so the action preserves crossed maps. It sends $\delta a$ to $\delta (ga)$ and satisfies $(gh)\cdot d=g\cdot (h\cdot d)$ by substitution. For $t\in N$, using $d(t^{-1})=-t^{-1}d(t)$ gives $(t\cdot d)(n)=d(n)+nd(t)-d(t)$. Hence N acts trivially on H1, and the action there factors through Q.

For a global crossed map D on G the identical expansion gives $gD(g^{-1}ng)=D(n)+nD(g)-D(g)$. Consequently conjugation changes its restriction by the principal map $\delta (D(g))$, so the restriction class is Q-invariant. All formulas are additive in the crossed map, and therefore descend to the stated homomorphisms. The zero module and either trivial end group obey the same identities.

If an inflated c is principal, say $c(\pi g)=ga-a$, restriction to N gives $na=a$ for every n, because c(1)=0. Thus $a\in A^N$ and c itself is principal on Q. Inflation is injective. An inflated cocycle restricts to zero on N, so its class belongs to the kernel of restriction.

Conversely let D restrict to a principal map $n\mapsto na-a$. Replace D by $D_0=D-\delta a$, so $D_0{\mathrm{\mid }}_N=0$. For n in N, $D_0(gn)=D_0(g)$, and $D_0(ng)=n{D}_0(g)$. But $ng=g(g^{-1}ng)$, so these equations also give $n{D}_0(g)=D_0(g)$. Thus $D_0$ is constant on quotient fibers and takes values in $A^N$. Define c(q) as this unique common value; no representatives need be chosen. For any g,h above q,r the crossed identity yields $c(qr)=c(q)+qc(r)$. Hence $D_0=\inf c$ and [D] lies in its image.

For an extension E, apply AC to the nonempty fibers of $E\to G$ and set $s(1)=1$. Unique kernel coordinates define f by $s(g)s(h)=i(f(g,h))s(gh)$. Then $f(1,g)=f(g,1)=0$. Associativity of $s(g)s(h)s(l)$ gives $f(g,h)+f(gh,l)=gf(h,l)+f(g,hl)$, exactly $\delta f=0$. The action term follows from $s(g)i(a)s(g{)}^{-1}=i(ga)$, the prescribed action.

Conversely, for a normalized cocycle f define $E_f=A\times G$ with $(a,g)(b,h)=(a+gb+f(g,h),gh)$. The first coordinates of the two triple products differ by $f(g,h)+f(gh,l)-gf(h,l)-f(g,hl)=0$, so multiplication is associative. The identity is (0,1). The inverse is $(a,g{)}^{-1}=(-g^{-1}a-g^{-1}f(g,g^{-1}),g^{-1})$; the right product is the identity, and the left product is too since the cocycle identity gives $f(g^{-1},g)=g^{-1}f(g,g^{-1})$. Inclusion $a\mapsto (a,1)$ and projection to G are exact and conjugation induces ga.

A new normalized section $s_b(g)=i(b(g))s(g)$ changes f to $f_b=f+\delta b$, where $(\delta b)(g,h)=b(g)+gb(h)-b(gh)$. The map $E_f\to E_{f_b}$, $(a,g)\mapsto (a-b(g),g)$ is an isomorphism: substitution in the two multiplication laws gives first coordinate $a+gc+f(g,h)-b(gh)$ on both sides. It fixes A and G and has inverse adding b(g). If two normalized cocycles differ by a coboundary, the cochain b is normalized as well, since its coboundary at (1,g) equals b(1).

The map $E_f\to E$, $(a,g)\mapsto i(a)s(g)$, is a homomorphism by the factor-set equation; unique kernel coordinates in each fiber make it a bijection fixing A and G. Conversely any equivalence carries a chosen section to a section and preserves its factor set. Thus the two constructions induce inverse bijections on the quotient by coboundaries and on extension classes. By F2 this quotient is the indicated H2.

If the class is zero, choose a normalized b with $f+\delta b=0$; the section $s_b$ is then a homomorphism. A homomorphic section conversely has f=0 and hence zero class. For A=0 there is the unique extension G, and for G=1 the unique extension A; both have zero class. Only step 1.1 uses arbitrary choice; supplied sections suffice for an individual construction.

The pullback is $P=\{(e,h):p(e)=\alpha (h)\}\le E\times H$. Its projection to H is onto, its kernel is $\{(i(a),1)\}$, and the conjugation action is the restricted action. Choose a normalized section s of E using AC; $(s(\alpha (h)),h)$ is a section of P. Its factor set is $(h,l)\mapsto f(\alpha (h),\alpha (l))$. By F1 this represents the bar pullback, including when alpha is not injective or surjective.