Let say I want to show two differential forms $\omega_1$ and $\omega_2$ on a smooth manifold $M$ are equal. Of course it suffices to show $\omega_1=\omega_2$ locally, i.e. the equality holds over every open subset $U_\alpha$ of a smooth atlas $\{(U_\alpha, \varphi_\alpha)\}$ of $M$.
Since each $U_\alpha$ is a smooth manifold given by an atlas consisting of a single chart, $U_\alpha$ is orientable (however, $M$ is NOT assumed to be orientable, or the charts of the atlas $\{(U_\alpha, \varphi_\alpha)\}$ are NOT assumed to be consistently oriented). Now suppose I have shown that $\omega_1=\omega_2$ on each $U_\alpha$, where the proof involves Poincare duality: $H^k(U_\alpha)\cong H_c^{n-k}(U_\alpha)^*$, where $n=\dim(M)$ and $0\leq k\leq n$. I am not sure if I can conclude that $\omega_1=\omega_2$ on $M$.
I have seen some arguments of proving two differential forms are equal, which is roughly the same as above: prove the equality locally, and locally one assumes the orientability, and it is used in an essential way (and with some other topological condition rather than orientability).
The reason I am not sure whether my argument in above is correct or not is, well, I feel like it's cheating, in the sense that Poincare duality is kind of topological. Of course, Poincare duality holds for non-orientable manifolds (whose statement involves the orientation bundle of $M$), but I cannot use this version of Poincare duality in my argument for some reason.
