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I come across a statement like

Let $L$ be a complex line bundle on a manifold. $c_1(L)=0$ mod $2$ if and only if there exists a line bundle $K$ such that $L\cong K^{\otimes 2}$.

How can one prove this statement? Does one need some further assumption?

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    You are right. I usually work in algebraic or holomorphic category and am being confused. Thanks.2012-12-27

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Mariano's comment has essentially answered the question, but I'll go ahead and flesh it out.

On any manifold $X$, there is an isomorphism of groups $\mathrm{Pic}(X) \xrightarrow{~\cong~} H^2(X; \mathbb{Z}),$ $L \mapsto c_1(L).$ Now if $c_1(L) \equiv 0 \pmod 2,$ then there is some element $a \in H^2(X; \mathbb{Z})$ such that $c_1(L) = 2a.$ The above isomorphism tells us that there exists a complex line bundle $K \in \mathrm{Pic}(X)$ such that $c_1(K) = a$ and $K \otimes K \mapsto 2a = c_1(L).$ Then $K \otimes K \cong L,$ so that $K$ is a square root of $L$.

For a proof of the above isomorphism, see for example Proposition 3.10 in Allen Hatcher's unfinished book Vector Bundles and $K$-Theory.

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    @MichaelAlbanese Yes, thanks for noticing the typo.2012-12-27