How do you prove that a line bundle (vector bundle of rank 1) on $S^2$ is isomorphic to the trivial line bundle? Can you give a reference?
Thanks.
How do you prove that a line bundle (vector bundle of rank 1) on $S^2$ is isomorphic to the trivial line bundle? Can you give a reference? Thanks. 


Since $GL(1,\mathbf R)$ and the symmetric group $S_2$ are homotopy equivalent as topological groups, there is a bijective correspondence between isomorphism classes of real line bundles and double covers. It follows that a simply connected space has no nontrivial real line bundles. 


In general, there is a bijection $$\Phi \colon [S^{k1}, \textrm{GL}_n^+(\mathbb{R})] \to \textrm{Vect}_+^n(S^k),$$ where $[X,Y ]$ denotes the set of homotopy classes of continous maps from $X$ to $Y$ and $\textrm{Vect}_+^n(Z)$ denotes the set of isomorphism classes of real (oriented) vector bundles of rank $n$ on $Z$. Using the retraction of $\textrm{GL}_n^+(\mathbb{R})$ onto $\textrm{SO}(n)$ one obtains another bijection $$\Psi \colon [S^{k1}, \textrm{SO}(n)] \to \textrm{Vect}_+^n(S^k).$$ In the case of line bundles we have $n=1$ and $\textrm{SO}(1)$ is just a point, hence the bijection $\Psi$ shows that any orientable real line bundle on $S^k$ is trivial. Furthermore, it is not difficult to prove that when $k \geq 2$ any real vector bundle on $S^k$ is orientable, so any real line bundle is trivial. Notice that this is not true for $S^1$, where there is an orientable bundle (the trivial one) and a nonorientable one (the Moebius band). Similar methods of classification can be applied to the case of complex vector bundles in order to show that $S^2$ admits nontrivial complex line bundles. In fact, one finds that there is a bijection $$\Phi_{\mathbb{C}} \colon [S^{k1}, \textrm{GL}_n(\mathbb{C})] \to \textrm{Vect}_{\mathbb{C}}^n(S^k).$$ For $n=1$, moreover, there is a bijection between $[S^{k1}, \textrm{GL}_1(\mathbb{C})]$ and $H^2(S^k, \mathbb{Z})$. In particular, if $k \neq 2$ the only complex line bundle on $S^k$ is the trivial one, whereas if $k =2$ there is a discrete family $\{L_t \}$ of nonisomorphic complex line bundles, parametrized by $t \in \mathbb{Z}$. In fact, viewing $S^2$ as the Riemann sphere with complex coordinate $z$, the transiction function of $L_t$ is $z^t$. For further details you can look at Hatcher's book Vector bundles and Ktheory, Chapter 1. 


Another solution is to consider clutching functions. The sphere $S^2$ is obtained by gluing two disks along their boundaries, it is not hard to prove that the information is encoded by a continuous map: $$f:S^1\rightarrow GL_1(\mathbb{R})$$ and that we have a map: $$[S^1,GL_1(\mathbb{R})]\rightarrow Vect^1(S^2)$$ where on the left hand side you have homotopy classes of clutching functions and on the right hand side you have isomorphism classes of real line bundles over $S^2$. As our continuous map $f$ is homotopic to a constant map the associated vector bundle is trivial. In general for $n$dimensional real vector bundles over $S^k$ you will have a map: $$[S^{k1},GL_n(\mathbb{R})]\rightarrow Vect^n(S^k).$$ This map becomes a bijection when you consider oriented vector bundles. edit: I just realize that I write my post at the same time as Francesco. 


Along the same lines. Rank $1$ real vector bundles over a compact CW complex $X$ are all pullbacks of the tautological line bundle over $\mathbb{RP}^\infty$. The space of isomorphism classes of such bundles can be identified with the space $[X,\mathbb{RP}^\infty]$ of homotopy classes of continuous maps $X\to \mathbb{RP}^\infty$. Since $\mathbb{RP}^\infty$ is the EilenbergMacLane space $K(\mathbb{Z}/2, 1)$ (the universal cover $S^\infty$ is contractible) we deduce that $[X,\mathbb{RP}^\infty]$ can be identified with the cohomology group $H^1(X,\mathbb{Z}/2)$. When $X=S^2$ this group is trivial. 

