Generator differences form a basis of the free-group augmentation ideal

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.