Let $p \colon M \to N$ be a covering space, and let $M,N$ be manifolds. Assume now that $M$ is orientable and $N$ is not orientable.
I'm asked to find a covering map $q \colon M \to N$ s.t. $p= \pi q$, where $\pi \colon \hat{N} \to N$ is the orientation cover. That's the situation:
My attempt was to define the map in the following way: $$q(m) = (p(m),p_*(o(m)))$$ where $p_* \colon H_n(M|m) \to H_n(N|p(m))$ is the induced map in homology and $o(m)$ is the given orientation in a point $ \in M$ (I'm using Hatcher's notation, and definition of orientation)
I've proved everything but surjectivity.
In fact I know that there must exist at least a $n \in N$ such that $m_1,m_2 \in p^{-1}(\{n\})$ are s.t. $p_*(o(m_1))=-p_*(o(m_2))$: In fact if it is not true, then I'm able to define an orientation on $N$ which is an absurd. Now my claim is that this must be true for each $n \in N$. In fact this would imply surjectivity (and surjectivity of the map would imply it)
If there is an orientation reversing deck transformation I'm done. I can use it to find in each fibre my two "special" points. The problem is that I do not know how to prove such existence. Obviously if one proves that the covering is normal then we are done, but this seems a result too strong to be true in this fairly general setting
Currently I'm trying to search some other methods but with no results. So the question is, is my claim correct? Does such transformation exists? Are there some other methods to prove surjectivity?
Thanks in advance