Notation: for $X$ a smooth manifold, let $T^n(X)$ denote the space of (global, smooth) covariant rank-$n$ tensors on $X$, i.e. sections of $(T^*X)^{\otimes n}$.
Let $X\xrightarrow{Q}Y$ be a smooth, surjective, local diffeomorphism between smooth manifolds.
Let us also assume $Q$ is "normal" in that its group of deck transformations acts transitively on each fiber.
We say $s \in T^n(X)$ is "$Q$-invariant" if $s$ is invariant under all deck transformations of $Q$. Denote by $\text{inv}_{Q}T^n(X)$ the space of $Q$-invariant covariant rank-$n$ tensors on $X$.
Then we claim for each integer $n\geq 0$ there is a unique homomorphism $Q^n_! : \text{inv}_{Q}T^n(X) \rightarrow T^n(Y)$, such that for every open $U\subset X$, if $Q|_U$ is a diffeomorphism onto $Q(U)$ with inverse $P : Q(U)\xrightarrow{\approx} U$, then for any $\omega \in \text{inv}_{Q}T^n(X)$ we have $$ (Q_! \omega)|_{Q(U)} = P^*(\omega|_U) $$ Furthermore:
For $\omega \in T^n(Y)$ we have $Q^*\omega$ is $Q$-invariant, and we should have $Q_!Q^*\omega = \omega$.
$Q^n_!$ preserves symmetric and antisymmetric tensors, thus also restricting to maps $\text{inv}_{Q}\mathrm{Sym}^n(X)\rightarrow \mathrm{Sym}^n(Y)$ and $\text{inv}_{Q}\Omega^n(X)\rightarrow \Omega^n(Y)$.
$Q^2_!$ should preserve signatures; therefore, $Q^2_!$ of a $Q$-inv Riemannian metric (or pseudo-Riemannian of some signature) is another Riemannian (or pseudo-Riemannian of same signature) metric.
For $\omega \in \text{inv}_{Q}\Omega^n(X)$ we have $d\omega \in \text{inv}_{Q}\Omega^{n+1}(X)$ and $Q^{n+1}_!(d\omega) = d(Q_!^n \omega)$.
So also, $Q_!$ gives a chain complex morphism $\text{inv}_{Q}\Omega^\bullet(X)\rightarrow \Omega^\bullet(Y)$.
Also, by the 1st bullet point, if $Q_!\omega$ is exact (on $Y$) then $\omega$ is exact (on $X$).
It should follow that $Q^n_!$ induces injective maps $H^n_{\mathrm{dR}}(X)\rightarrow H^n_{\mathrm{dR}}(Y)$, provided every de Rham class on $X$ has a $Q$-invariant representative.
My 1st question is whether the above is valid and makes sense?
My 2nd question is whether the last condition above is satisfied, and how to show this? I.e., for every closed $n$-form $\omega$ on $X$, is there an $Q$-invariant (i.e. invariant under deck transformations of $Q$) closed $n$-form $\omega'$ on $X$, which is cohomologous with $\omega$ (i.e. their difference is exact)?
Or (more weakly?) can we show the inclusion of the $Q$-invariant de Rham complex into $X$'s full de Rham complex is a quasi-isomorphism (induces isomorphisms on all cohomology groups)?
Update: Here is my thinking thus far on the 2nd question:
Let us further assume:
that the group $G$ of deck transformations is a topological group, acting continuously on $X$, and it admits a Haar measure $\mu$ with finite total volume; and
that each $g \in G$ induces the identity (not just an isomorphism) on the de Rham cohomology of $X$.
Then, given a closed $\omega \in \Omega^n_{\text{closed}}(X)$, we would like to define a form via an integral of a $G$-dependent form over $G$, $$ \omega' = \frac{1}{|G|}\int_{g\in G} (g^*\omega)\,d\mu(g) \,, $$ which we expect to be $G$-invariant by the usual arguments. Here $|G| = \mu(G)$ is the finite volume of $G$. Furthermore, since for each $g\in G$ we know $g^*$ induces the identity on $H^n_{\mathrm{dR}}(X)$ it follows that $\omega$ is cohomologous with $g^*\omega$, so also $\omega = \frac{1}{|G|}\int_{g\in G}\omega\,d\mu(g)$ is cohomologous with $\omega' = \frac{1}{|G|}\int_{g\in G} (g^*\omega)\,d\mu(g)$.
Would this be valid? Are the above assumptions necessary? The 2nd assumption bothers me, since e.g. this is not satisfied by $S^2 \xrightarrow{\text{mod $\mathbb{Z}/2\mathbb{Z}$}} \mathbb{R}\mathbb{P}^2$.