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In Hatcher's Vector Bundles and K-theory the following description of orientability of a vector bundle $E \mapsto B$ is given:

For a vector bundle $E\mapsto B$ with $B$ path-connected, orientability is detected by the homomorphism $ \pi_1(B) \mapsto \mathbb{Z}_2$ that assigns 0 or 1 to each loop according to whether orientations of fibers are preserved or reversed as one goes around the loop. Since $\mathbb{Z}_2$ is abelian, this homomorphism factors through the abelianization $H_1 (B)$ of $\pi_1 (B)$, and homomorphisms $H_1 (B)→\mathbb{Z}_2$ are identifiable with elements of $H^ 1 (B; \mathbb{Z}_2 ).$ Thus we have an element of $H ^1 (B; \mathbb{Z}_2 )$ associated to $E$ which is zero exactly when $E$ is orientable. This is exactly the first Stiefel-Whitney class $\omega_1 (E).$

Where is the association of element of $H^1(B; \mathbb{Z}_2)$ to $E$ coming from?

I think I'll be able to see that this element is zero if $E$ is trivial if I understood what this association is.

user7090
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$\newcommand{\Ext}{\operatorname{Ext}}\newcommand{\Hom}{\operatorname{Hom}}$In general, the Universal Coefficient Theorem tells you that

$$H^n(B;R)\cong \Ext_R^1(H_{n-1}(B;R),G)\oplus \Hom_R(H_n(B;R),G)$$

for a ring $R$ and an $R$-module $G$. In the case $R=\Bbb Z$ and $G=\Bbb Z_2$, some simple homological algebra tells you that $\Ext_{\Bbb Z}^1(H_0(B;\Bbb Z),\Bbb Z_2)=0$, so you're left with $H^1(B;\Bbb Z_2)\cong \Hom(H_1(B;\Bbb Z),\Bbb Z_2)$.

Alex Mathers
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  • Awesome username! Anyway, I got the part about the UCT, but what is the association with the vector bundle $E$? – user7090 Mar 20 '17 at 17:41
  • @Janziek haha thanks! Well the reason they're related to $E$, although $E$ isn't explicitly mentioned much in the construction, is because of the way that the original homomorphism $\pi_1(B)\to\Bbb Z_2$ is defined: the definition is dependent on the way the orientations of the fibers (i.e. subsets of $E$) change as you traverse a loop. If $E$ is orientable, then the orientation is always preserved as you traverse a loop, which corresponds to the map $\pi_1(B)\to\Bbb Z_2$ being zero, which happens iff the associated map $H_1(B)\to\Bbb Z_2$ is zero, iff the first Stiefel-Whitney class is zero. – Alex Mathers Mar 20 '17 at 18:08
  • @Janziek also I think I should add, once you read (if you haven't already) the construction of all Stiefel-Whitney classes, it will be more clear why they depend on $E$. It has to do with using Leray-Hirsch to show that $H^(E;\Bbb Z_2)$ is a free $H^(B;\Bbb Z_2)$-module, and extracting unique elements of $H^(B;\Bbb Z_2)$ by writing some element of $H^(E;\Bbb Z_2)$ as a unique combination of basis elements. – Alex Mathers Mar 20 '17 at 18:17
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    Sounds good. Thanks Sheaf Keef! – user7090 Mar 20 '17 at 18:46