Let $F$ be a path-connected Poincaré duality space (using $\mathbb{Z}$ coefficients for singular co/homology) of dimension $n$. In particular, we have $H^n(F;\mathbb{Z}) \approx \mathbb{Z} \approx H_n(F;\mathbb{Z})$.
Let us say a given map $f : F \rightarrow F$ is orientation-preserving if the induced $f_*$ on $H_n$ is the identity. Define $\mathrm{Homeo}_+(F)$ to be the subgroup of auto-homeomorphisms of $F$ which are orientation-preserving.
Let $X \xrightarrow{\pi} Y$ be a continuous fiber bundle with fiber $F$. For $y\in Y$ we use the notation $X_y := \pi^{-1}(y)$.
We make the following definitions:
We say $\pi$ is fiber-orientable if $Y$ admits a covering by local trivializations $\pi^{-1}(U_\alpha) \xrightarrow{\approx} U_\alpha \times F$ such that the transition maps $\phi_\beta(U_\alpha) \times F \rightarrow \phi_\alpha(U_\beta) \times F$ can be viewed as maps $U_\alpha \cap U_\beta \rightarrow \mathrm{Homeo}_+(F)$.
We say a given $c \in H^n(X;\mathbb{Z})$ is a Thom class for $\pi$ if for every $y\in Y$ the pulled-back class $c|_{X_y} \in H^n(X_y;\mathbb{Z})$ is a generator.
Under the above conditions, is there a Thom isomorphism -like theorem which says the following?
The fiber bundle $\pi$ is fiber-orientable iff it admits a Thom class.
If $c$ is a Thom class, then $$ \begin{matrix} H^j(Y;\mathbb{Z}) & \rightarrow & H^{j+n}(X;\mathbb{Z}) \\ a &\mapsto& c \smile (\pi^*a) \end{matrix} $$ is an isomorphism for each integer $j\geq0$.