Stable Thom cohomology is a square-module coalgebra

For x∈M, take its sufficiently high finite-rank components x_n. The Whitney coalgebra is induced by the finite-rank direct-sum Thom pullback μ*, so naturality gives Δ_M(Sq^k x)=μ*(Sq^k x)=Sq^k(μ* x). Write μ*x under the Künneth isomorphism as a finite sum Σx_a⊗y_b. Cartan gives $S{q}^k(x_a\times y_b)={\sum }_{i+j=k}S{q}^i(x_a)\times S{q}^j(y_b)$.

Therefore Δ_M(Sq^k x)=Σ_{i+j=k}(Sq^i⊗Sq^j)Δ_M(x) =Sq^k·Δ_M(x), where the last action is the diagonal A-action defined from Δ_A(Sq^k). This is the required compatibility for the generators. For a product ab∈A, Δ_A(ab)=Δ_A(a)Δ_A(b); the module law on M and the already-verified generator compatibility give Δ_M((ab)x)=(ab)·Δ_M(x). Extend by linearity to all a∈A. The diagonal action is unital because ${\mathrm{\Delta }}_A(1)=1\otimes 1$; its composition law follows by expanding ${\mathrm{\Delta }}_A(ab)={\mathrm{\Delta }}_A(a){\mathrm{\Delta }}_A(b)$ and using the action law in each factor. For homogeneous $a,m$, ${\epsilon }_M(am)={\epsilon }_A(a){\epsilon }_M(m)$: both sides vanish if either degree is positive, and in total degree zero this is the scalar-action identity. Linearity gives the same conclusion for all inputs. Thus all hypotheses involving the action and coproduct are satisfied.