Let h\* be a multiplicative generalized cohomology theory and $E \rightarrow X$ a real vector bundle.


1. Is it true that, if $E$ is orientable with respect to h\*, then it is also orientable with respect to the singular cohomology with coefficients in $h^{0}(pt)$, for $(pt)$ a space with one point?
2. Is it true that, if $E$ is orientable with respect to the singular cohomology with coefficeints in $\mathbb{Z}\_{p}$, for $p > 2$, then it is also orientable with respect to the singular cohomology with coefficients in $\mathbb{Z}$?


**Added later**: I thought to a possible simple answer using weak orientability, actually proving what stated in the last comment. I wrote this as a comment to answer 3.



Orientation and generalized cohomology theories
Assuming $X$ is connected and the dimension $k$ of $\xi$ is positive, here is a proof of the yes answer to (1).


Consider the relative Atiyah-Hirzebruch spectral sequence


$$H^p(D\xi,S\xi;h^q(\mathrm{pt}))\Rightarrow h^\ast(D\xi,S\xi)$$


where $D\xi$ and $S\xi$ are the disk and sphere bundles in some Euclidean metric. The pair $(D\xi,S\xi)$ is $(k-1)$-connected, so $E\_2^{p,q}$ is zero for $0<p<k$. Note that $$E\_2^{0,k}=H^0(D\xi,S\xi;h^k(\mathrm{pt}))\cong \tilde{H}^0(T\xi;h^k(\mathrm{pt})),$$
 the reduced cohomology of the Thom space in dimension zero, which should be trivial by our dimension hypothesis (the Thom space is connected). Hence the edge homomorphism


$$H^k(D\xi,S\xi;h^0(\mathrm{pt}))\to h^k(D\xi,S\xi)$$


is an isomorphism, and you can check that a Thom class corresponds to a Thom class.


**Edit** Now that I've carefully read Charles' answer, I see that I was over-simplifying things by assuming the spectrum was connective (so that $h^q(\mathrm{pt})=0$ for $q$ negative). If not, either we take the connective cover and appeal to the claim in Charles' answer, or we've got stuff in the lower quadrant of our spectral sequence. Even so, there is still an "edge homomorphism"


$$h^k(D\xi,S\xi)\twoheadrightarrow E^{k,0}\_\infty \hookrightarrow E^{k,0}\_2 = H^k(D\xi,S\xi;h^0(\mathrm{pt}))$$


which should take a Thom class to a Thom class.


**Added later:** Here is a proof of a yes answer to (2) (again assuming connected, positive dimension). The homology group $H\_k(T\xi;\mathbb{Z})$ is $\mathbb{Z}$ if $\xi$ is orientable (in the usual sense, with $\mathbb{Z}$ coefficients) and $\mathbb{Z}\_2$ otherwise. This follows from the twisted Thom isomorphism $H\_0(X;\mathbb{Z}^\xi)\cong H\_k(T\xi;\mathbb{Z})$, where $\mathbb{Z}^\xi$ denotes the local coefficient system on $X$ determined by $\xi$, which holds regardless of orientability. (I'm getting lazy with tildes here.)


Then the universal coefficient theorem gives $H^k(T\xi;\mathbb{Z}\_p)\cong\mathrm{Hom}(H\_k(T\xi),\mathbb{Z}\_p)\cong 0$ if $\xi$ is non-orientable (and assuming $p$ is **odd**). So non-orientable in the usual sense implies there can't be a Thom class with $\mathbb{Z}\_p$ coefficients.


