Rational transfer identifies a finite regular cover with deck invariants

Deck transformations act freely when $Y$ is nonempty. If $g(y)=y$, then $g$ and ${\mathrm{id}}_Y$ lift the same map $p$ and agree at $y$; connectedness and uniqueness in [F2] give $g={\mathrm{id}}_Y$. Transitivity in [F1] therefore makes evaluation $g\mapsto g(y)$ a bijection from $G$ onto $p^{-1}(p(y))$. Consequently $\mathrm{\mid }G\mathrm{\mid }=d$.

A singular simplex has exactly $d$ lifts. The simplex is nonempty and convex, with paths given by segments. Intersections with sufficiently small Euclidean balls are convex relative open neighborhoods, hence path connected by segments, so it is locally path connected. Its fundamental group is trivial by [F3]. For each point above its first vertex, the criterion in [F2] gives a lift, and uniqueness says that evaluation at that vertex is a bijection between all lifts and the fiber. This includes $k=0$, where lifts are simply points. Restriction to any face is also a bijection between the lift sets: prescribe a point over any vertex of that face, lift the whole simplex based at that vertex, and use uniqueness on the face and simplex.

Use [A1] to select one lift $\tilde{\sigma }$ of each simplex. By step 1.1 and uniqueness in step 1.2, the maps $g\tilde{\sigma }$ for $g\in G$ are exactly its $d$ distinct lifts. Set $\tau (\sigma )={\sum }_{g\in G}g\tilde{\sigma }$ and extend to integer chains by finite linearity. This is the sum over the entire lift set, so changing the selected lift merely permutes the summands. Restriction to each face bijects lift sets by step 1.2; hence, with the ordinary alternating face signs, $\mathrm{\partial }\tau (\sigma )={\sum }_i(-1{)}^i\tau (\sigma {\delta }_i)=\tau \mathrm{\partial }(\sigma )$. In degree zero both boundaries vanish. Thus $\tau$ is a chain map.

Precomposition gives $T_c(\phi )=\phi \tau$ on rational cochains. The positive coboundary and $\mathrm{\partial }\tau =\tau \mathrm{\partial }$ imply $\delta T_c=T_c\delta$, so it descends to a rational-linear transfer $T:H^{\ast }(Y;\mathbb{Q})\to H^{\ast }(X;\mathbb{Q})$. On chains, $p_{\mathrm{\#}}\tau =d\mathrm{id}$, since every lift projects to the same simplex. Conversely, for a simplex $\eta$ in $Y$, all lifts of $p\eta$ are precisely $g\eta$, so $\tau p_{\mathrm{\#}}={\sum }_g{g}_{\mathrm{\#}}$. Precomposition gives the correctly typed identities $T{p}^{\ast }=d\mathrm{id}$ on $H^{\ast }(X;\mathbb{Q})$ and $p^{\ast }T={\sum }_g{g}^{\ast }$ on $H^{\ast }(Y;\mathbb{Q})$.

If $p^{\ast }x=0$, then $dx=T{p}^{\ast }x=0$, and $d\ge 1$ is invertible in $\mathbb{Q}$, so $x=0$. Every pullback is invariant since $pg=p$. If $y$ is invariant, then $p^{\ast }(d^{-1}Ty)=d^{-1}{\sum }_g{g}^{\ast }y=d^{-1}\mathrm{\mid }G\mathrm{\mid }y=y$. This proves the equality of the image and invariants in each degree. Pullback and all deck pullbacks preserve products and unit by [F5], so the equality identifies graded subalgebras; no multiplicativity of $T$ is asserted or needed.

For $d=1$ a one-sheeted covering is a bijective local homeomorphism and hence a homeomorphism, so pullback is an isomorphism and its deck group is trivial. For the empty covering there are no simplices; [F4] makes every cochain and cohomology group zero, proving the conclusion without evaluating at a point or using $\mathrm{\mid }G\mathrm{\mid }=d$. All negative-degree groups are zero; degree zero is covered by the same cochain identities. The construction uses [A1] only for the initial simultaneous selection; the full lift-sum is independent of it.