The free-presentation homology five-term sequence

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 $F_{\mathrm{a}\mathrm{b}}\to F_{\mathrm{a}\mathrm{b}}$.