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From $0\rightarrow \mathbb{Z}\rightarrow \mathbb{Z}\rightarrow \mathbb{Z}_{2}\rightarrow 0$, we have the long exact sequence $H^{1}(X,\mathbb{Z})\rightarrow H^{1}(X,\mathbb{Z}_{2})\rightarrow H^{2}(X,\mathbb{Z})$. Meanwhile $H^{1}(X,\mathbb{Z}_{2})\cong H^{1}(X,\mathbb{Z})\otimes\mathbb{Z}_{2} \oplus Tor(H^{2}(X,\mathbb{Z}),\mathbb{Z}_{2})$.

My question is: if $L$ is a real line bundle on $X$, does $w_{1}(L)$ have component in $H^{1}(X,\mathbb{Z})\otimes\mathbb{Z}_{2}$ or $Tor(H^{2}(X,\mathbb{Z}),\mathbb{Z}_{2})$ under the above isomorphism.

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    $\begingroup$ The splitting in the UCT isn't canonical. $\endgroup$
    – Paul
    Commented Jun 7, 2013 at 4:03

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Every element in $H^1(X,\mathbb Z/2)$ is the first Stiefel-Whitney class of exactly one line bundle. This is just because $H^1(X,\mathbb Z/2)$ parameterizes double covers of $X$, and for every double cover there is a unique line bundle that trivializes on it. So $w_1(L)$ can have components in either, neither, or both.

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  • $\begingroup$ Thanks Will. As we know, $w_{1}(L)$ goes to the first Chern class of the complexfication of $L$ under the above Bockstein homomorphism. My original motivation is when $c_{1}(L_{\mathbb{C}})=0$ implies $w_{1}(L)=0$. $\endgroup$
    – Allen
    Commented Jun 7, 2013 at 3:14

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