Crossed Homomorphisms Complements and First Cohomology

If $z,w\in Z^1(G,A)$, then $(z+w)(gh)=z(gh)+w(gh)=z(g)+w(g)+g\cdot z(h)+g\cdot w(h)=(z+w)(g)+g\cdot (z+w)(h)$, so $z+w$ is again a crossed homomorphism by [L1].

The zero map is a crossed homomorphism, and if $z\in Z^1(G,A)$ then $(-z)(gh)=-z(gh)=-z(g)+g\cdot (-z(h))$, so $-z$ is one as well. Thus pointwise addition makes $Z^1(G,A)$ a subgroup of the abelian group $A^G$. In particular it is abelian.

For $a\in A$ and $g,h\in G$, $(da)(gh)=gh\cdot a-a=g\cdot (h\cdot a-a)+(g\cdot a-a)$, so every principal crossed homomorphism is a crossed homomorphism. Thus $B^1(G,A)\subseteq Z^1(G,A)$ by [L1] and [L2].

If $da$ and $db$ are principal, then $(da+db)(g)=g\cdot (a+b)-(a+b)=d(a+b)(g)$ and $-(da)(g)=g\cdot (-a)-(-a)=d(-a)(g)$. So $B^1(G,A)$ is closed under sums and inverses, hence is a subgroup of $Z^1(G,A)$.

A function $f:G\to A$ lies in $\ker d^1$ exactly when $0=d^1(f)(g,h)=f(g)+g\cdot f(h)-f(gh)$ for all $g,h\in G$, that is, exactly when $f(gh)=f(g)+g\cdot f(h)$. So the one-cocycles in the inhomogeneous complex are exactly the crossed homomorphisms of [L1].

A function lies in $\mathrm{im}{d}^0$ exactly when it has the form $g\mapsto g\cdot a-a$ for some $a\in A$. Those are precisely the principal crossed homomorphisms in [L1].

Steps 1.1 and 2.1 identify both the numerator and denominator of the two quotient constructions. Therefore the quotients themselves are canonically the same, giving the asserted isomorphism.

Under the trivial action, the crossed-homomorphism identity becomes $z(gh)=z(g)+z(h)$, which is exactly the homomorphism law of [L2]. So crossed homomorphisms are precisely the homomorphisms $G\to A$.

Principal crossed homomorphisms are all zero, because $g\cdot a-a=a-a=0$ for every $g$ and $a$. Therefore the quotient $H^1(G,A)$ is just the group of homomorphisms from step 1.1.

Put $s={\sum }_{x\in G}z(x)\in A$. For any fixed $g\in G$, ${\sum }_{x\in G}z(gx)={\sum }_{x\in G}(z(g)+g\cdot z(x))=mz(g)+g\cdot s$, and the left-hand side is just $s$ because $x\mapsto gx$ permutes $G$. Hence $mz(g)=s-g\cdot s$.

Because multiplication by $m$ is bijective on $A$, choose $a\in A$ with $ma=-s$. Then $m(z(g)-(g\cdot a-a))=mz(g)-g\cdot (ma)+ma=(s-g\cdot s)-(-g\cdot s)+(-s)=0.$ Bijectivity of multiplication by $m$ forces $z(g)=g\cdot a-a$ for every $g$. So $z$ is principal.

Every crossed homomorphism is principal, so the quotient in [L1] is zero. Therefore $H^1(G,A)=0$.

Suppose $z$ is a crossed homomorphism. Then $(z(g),g)(z(h),h)=(z(g)(g\cdot z(h)),gh)=(z(gh),gh)$, so [L1] and [L3] show that ${\mathrm{\Gamma }}_z$ is closed under products. The identity is $(z(1),1)=(1,1)$, and inverses also stay in ${\mathrm{\Gamma }}_z$, so ${\mathrm{\Gamma }}_z$ is a subgroup.

Conversely, suppose ${\mathrm{\Gamma }}_z$ is a complement. Since it is a subgroup, the product of $(z(g),g)$ and $(z(h),h)$ again lies in ${\mathrm{\Gamma }}_z$. Comparing second coordinates gives $(z(g),g)(z(h),h)=(z(gh),gh)$, and then [L3] forces $z(gh)=z(g)(g\cdot z(h))$. So $z$ is a crossed homomorphism.

The projection $M⋊G\to G$ restricts to a bijection ${\mathrm{\Gamma }}_z\to G$ by [L2], so ${\mathrm{\Gamma }}_z$ intersects the kernel $M$ trivially and multiplies with that kernel to all of $M⋊G$. Hence ${\mathrm{\Gamma }}_z$ is a complement to the canonical copy of $M$.

Steps 1.1-1.2 and 2.1 prove the equivalence.

In $A⋊G$, the inverse of $(-a,1)$ is $(a,1)$. Therefore $(-a,1)(z(g),g)(a,1)=(-a+z(g)+g\cdot a,g)=(z_a(g),g)$ by [L3] and the definition in [L1].

Step 1.1 shows that conjugating each element of ${\mathrm{\Gamma }}_z$ by $(-a,1)$ produces the corresponding element of ${\mathrm{\Gamma }}_{z_a}$. Hence $(-a,1){\mathrm{\Gamma }}_z(a,1)={\mathrm{\Gamma }}_{z_a}$. So kernel conjugation changes the graph exactly by a principal crossed homomorphism.

If $z:G\to A$ is a crossed homomorphism, [L2] makes ${\mathrm{\Gamma }}_z$ a complement to $A$ in $A⋊G$. By [L3], replacing $z$ by a cohomologous cocycle replaces ${\mathrm{\Gamma }}_z$ by an $A$-conjugate complement. Hence the rule $[z]\mapsto [{\mathrm{\Gamma }}_z]$ is well defined on $H^1(G,A)$.

Every complement $H\le A⋊G$ arises from some crossed homomorphism: the projection $H\to G$ is an isomorphism, so for each $g\in G$ there is a unique element of $H$ of the form $(z(g),g)$, and [L2] says the resulting map $z$ is a crossed homomorphism. Thus the map from step 1.1 is surjective on complement classes.

If ${\mathrm{\Gamma }}_z$ and ${\mathrm{\Gamma }}_w$ are $A$-conjugate, then [L3] says $w-z$ is principal, so $[z]=[w]$ in the quotient [L1]. Therefore the map of step 1.1 is injective.

Steps 1.1-2.2 give a bijection between $H^1(G,A)$ and the $A$-conjugacy classes of complements to $A$ in $A⋊G$.

For a nonabelian crossed homomorphism $z$, the graph ${\mathrm{\Gamma }}_z$ is a complement by [L1]. Conversely, every complement gives a unique graph map by the same lemma. So complements correspond exactly to nonabelian crossed homomorphisms.

Conjugating $(z(g),g)$ by $(a,1)$ in $M⋊G$ gives $(a,1)(z(g),g)(a,1{)}^{-1}=(az(g)(g\cdot a{)}^{-1},g)$, which is the graph of the cocycle $(a\ast z)(g)$ from [L2]. Hence $M$-conjugate complements correspond exactly to $M$-orbits of cocycles.

The trivial cocycle has graph $\{(1,g):g\in G\}$, the canonical complement to $M$. Therefore the bijection of step 2.1 respects the distinguished basepoint and is an isomorphism of pointed sets.

Inflation is injective. Suppose $\bar{z}:G\mathrm{/}N\to A^N$ inflates to a principal crossed homomorphism $g\mapsto g\cdot a-a$ on $G$. For $n\in N$, inflation gives $0=\bar{z}(N)=n\cdot a-a$, so $a\in A^N$. Therefore the same formula already defines a principal cocycle on $G\mathrm{/}N$, and the class of $\bar{z}$ is zero.

Conversely, let $z:G\to A$ be a crossed homomorphism whose restriction to $N$ is trivial in $H^1(N,A)$. Then there exists $a\in A$ such that $z(n)=n\cdot a-a$ for all $n\in N$. Replace $z$ by the cohomologous cocycle $z^{\mathrm{\prime }}(g)=z(g)-(g\cdot a-a)$. Now $z^{\mathrm{\prime }}{\mathrm{\mid }}_N=0$.

If $\bar{z}:G\mathrm{/}N\to A^N$ is a cocycle, then its inflation vanishes on $N$ because $nN=N$. Hence $\mathrm{Res}\circ \mathrm{Inf}=0$.

If $n\in N$ and $g\in G$, then $z^{\mathrm{\prime }}(ng)=z^{\mathrm{\prime }}(n)+n\cdot z^{\mathrm{\prime }}(g)=n\cdot z^{\mathrm{\prime }}(g)$, while also $z^{\mathrm{\prime }}(ng)=z^{\mathrm{\prime }}(g(g^{-1}ng))=z^{\mathrm{\prime }}(g)+g\cdot z^{\mathrm{\prime }}(g^{-1}ng)=z^{\mathrm{\prime }}(g)$ because $g^{-1}ng\in N$ and $z^{\mathrm{\prime }}{\mathrm{\mid }}_N=0$. Therefore $n\cdot z^{\mathrm{\prime }}(g)=z^{\mathrm{\prime }}(g)$ for all $n\in N$, so $z^{\mathrm{\prime }}(g)\in A^N$. Also $z^{\mathrm{\prime }}(gn)=z^{\mathrm{\prime }}(g)+g\cdot z^{\mathrm{\prime }}(n)=z^{\mathrm{\prime }}(g)$, so $z^{\mathrm{\prime }}$ is constant on cosets of $N$.

Define $\bar{z}:G\mathrm{/}N\to A^N$ by $\bar{z}(gN)=z^{\mathrm{\prime }}(g)$. Step 2.2 shows this is well defined, and the crossed-homomorphism identity for $z^{\mathrm{\prime }}$ implies that $\bar{z}$ is a cocycle on $G\mathrm{/}N$. By construction, inflating $\bar{z}$ gives $z^{\mathrm{\prime }}$, so the original class of $z$ lies in the image of inflation. Hence $\ker \mathrm{Res}=\mathrm{im}\mathrm{Inf}$.

Steps 1.1, 2.1, and 3.1 prove exactness of the displayed sequence.

Define $z:C_2\to \mathbb{Z}$ by $z(1)=0$ and $z(t)=1$. Then $z(t^2)=z(1)=0=1+t\cdot 1=z(t)+t\cdot z(t)$, so [L1] shows that $z$ is a crossed homomorphism.

But $z$ is not an ordinary homomorphism, because a homomorphism $C_2\to \mathbb{Z}$ must send $t$ to an element of order dividing $2$, hence to $0$, whereas $z(t)=1$. Therefore the claim is false.

Under the trivial action, a nonabelian cocycle is exactly a homomorphism to $S_3$. Define cocycles $z,w:C_2\times C_2\to S_3$ by $z(s)=(12),z(t)=1,w(s)=1,w(t)=(23)\mathrm{.}$ Their pointwise product $u=zw$ satisfies $u(s)=(12)$ and $u(t)=(23)$. Since $st=ts$, a homomorphism would require $u(s)u(t)=u(t)u(s)$, but $(12)(23)\mathrm{\ne }(23)(12)$. Thus $u$ is not a cocycle.

Thus cocycles are not even closed under the proposed pointwise operation. Fact [L1] accordingly defines nonabelian $H^1$ only as a pointed orbit set, and [L2] identifies its natural classification target as a pointed set of complement classes. Therefore pointwise multiplication does not induce the asserted group structure.

Fact [L1] speaks only about complements to the canonical copy of $A$, meaning subgroups whose projection to $G$ is an isomorphism.

A subgroup such as the kernel copy $A$ itself or the trivial subgroup does not project isomorphically onto $G$ unless $G=1$. Therefore such subgroups are outside the classification of [L1]. The claim that all subgroups are classified is false.

For every integer $m$, the map $z_m:C_2\to \mathbb{Z}$ given by $z_m(1)=0$ and $z_m(t)=m$ is a crossed homomorphism. Conjugation by the kernel element $(1,1)$ changes the graph of $z_1$ to the graph of $z_3$, since the corresponding principal cocycle has value $t\cdot (-1)-(-1)=2$ at $t$. Hence $z_1$ and $z_3$ represent the same class under the kernel-conjugacy relation of [L1].

The quotient copy has only the elements $(0,1)$ and $(0,t)$. Conjugating the graph of $z_m$ by $(0,t)$ gives the graph of $z_{-m}$, so quotient-copy conjugacy sends ${\mathrm{\Gamma }}_{z_1}$ only to itself or to ${\mathrm{\Gamma }}_{z_{-1}}$, never to ${\mathrm{\Gamma }}_{z_3}$. Thus quotient-copy conjugacy misses a pair that [L1] identifies, and it is not the equivalence relation defining $H^1$.

Fact [L1] shows that the cochain model comes with a specific degree-one differential whose sign and action conventions matter.

Fact [L2] is therefore not vacuous: one must check that the cocycle equation and principal cocycles match the crossed-homomorphism formulas. So the claim that no comparison proof is needed is false.