A connected graded module coalgebra with injective unit orbit is free

Grading and the counit give, for homogeneous m∈M_d with d>0, Δ_M(m)=u⊗m+m⊗u+R, where R lies in ⊕{0<i<d}M_i⊗M{d-i}. Indeed the degree-(0,d) component is u⊗m by (ε_M⊗id)Δ_M=id, since ε_M vanishes on positive degrees; the other counit identity gives the degree-(d,0) component. Tensor bidegrees are direct by [L5]. In degree zero, Δ_M(λu)=λu⊗u, as already justified. The same argument gives the two endpoints for Δ_A(a) in positive degree.

A⁺M is the span of a·m with a of positive degree. It is graded: decomposing a,m into homogeneous components expresses every element as a finite sum of homogeneous such products. It is an A-submodule, since b(a·m)=(ba)·m and each nonzero homogeneous product ba has positive degree. It has no degree-zero component. By [L2], Q is the direct sum of Q_d=M_d/(A⁺M)_d, π is graded and surjective, its kernel is A⁺M, and Q_0=M_0. Moreover π(a·m)=ε_A(a)π(m) for every a,m: positive-degree a is killed, and degree-zero a acts as a scalar.

Choose a k-basis of each Q_d and lift each basis vector to M_d; this is allowed by [L3] and AC. In degree zero use π(u), with lift u. Extend the lifts linearly on each degree and then on the direct sum to obtain a graded section f. Conversely any graded section has f(π(u))=u, because π:M_0→Q_0 is an isomorphism. The formula for Φ is balanced and bilinear over k, so [L4] makes it a well-defined map. It is graded, and A-linearity follows from (ba)f(q)=b(af(q)).

We prove surjectivity by induction on d≥0. In degree zero Φ is the scalar isomorphism k⊗k→k·u. Suppose every M_e with e<d is in its image, and take m∈M_d. The vector m-f(π(m)) lies in (A⁺M)_d, so it is a finite sum Σ a_t m_t with homogeneous a_t of positive degree and m_t of degree d-|a_t|<d. Such an expression is obtained by projecting any finite expression in A⁺M to degree d. By induction choose y_t∈A⊗Q with Φ(y_t)=m_t. Then m=Φ(1_A⊗π(m)+Σ a_t y_t). Thus Φ is surjective in each degree, and finite degree support proves surjectivity on M.

Define T=(id_M⊗π)Δ_M:M→M⊗Q. Give M⊗Q the A-action on the first factor only. Then T is A-linear. Indeed apply id⊗π to the diagonal compatibility formula. Every summand with a₂ of positive degree vanishes by step 1.2. The surviving terms have a₂ in degree zero, so their Koszul signs are 1; the counit identity (id⊗ε_A)Δ_A(a)=a combines these terms to give T(a·m)=a·T(m). This calculation also treats a of degree zero and all inhomogeneous inputs by linearity.

For homogeneous q∈Q_d, the component of T(f(q)) with second degree d is exactly u⊗q. Every other component has second degree strictly less than d, by step 1.1. When d=0 there are no other components. Therefore TΦ(a⊗q)=ν(a)⊗q + terms of second degree less than d. This statement concerns second-factor degree, not total degree; multiplication on the first factor preserves that comparison.

Suppose z∈ker Φ. By [L5], using a homogeneous k-basis (q_j) of Q, write z uniquely as a finite sum Σ_j a_j⊗q_j with a_j∈A. If z≠0, take the largest degree d of a q_j with a_j≠0. Since TΦ(z)=0, its component of second degree d gives Σ_{|q_j|=d} ν(a_j)⊗q_j=0. The q_j in this equation are distinct basis vectors. Their coordinate functionals, tensored with id_M via [L4], give ν(a_j)=0 individually. Injectivity of ν gives a_j=0, contradicting the definition of d. Hence z=0 and Φ is injective. This finite maximum argument requires no finite-dimensionality of Q or its degree pieces.

By steps 2.1 and 4.1, Φ is a bijective graded A-linear map. Its inverse is A-linear and graded by uniqueness of preimages. Every element of A⊗Q has a unique finite expression Σ a_j⊗q_j, by [L5]; the first-factor action turns this into a free A-module with basis 1_A⊗q_j, in the sense of [L6]. Transporting that basis by Φ proves the theorem, with each generator in degree |q_j|. AC was used only for the homogeneous basis and its lifts; no further infinite selection occurs in the degree induction or finite-maximum argument.