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I'm trying to study line bundle over $S^2$. In this post was outlined the method based on clutching functions. But now I'm interesting in another approach.

For the sphere there is two maps : upper hemisphere and lower hemisphere with intersection as $[-\epsilon,\epsilon]\times S^1$. For the upper hemisphere and lower hemisphere its well-known that bundles over this spaces is trivial. (Any bundle over a contractible base is trivial). So to prove the fact that line bundle over $S^2$ is trivial we must create continuation of trivialization from upper hemisphere (for example) to the lower hemisphere through "border" $[-\epsilon,\epsilon]\times S^1$.

As I understand it is sufficient to continue trivialization from the "border" to the center of the "disk". (I think here it is possible to use a partition of unity, but I'm not sure).

I can't formalize this reasoning.

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A trivialization of a real line bundle over an open set $U$ is essentially a (continuous) non-vanishing function $\sigma:U \to \mathbf{R}$. If $U$ is connected (e.g., an enlarged hemisphere), such a function has constant sign, say positive.

If the upper hemisphere is trivialized, you now have a positive function $\sigma$ on the boundary of the lower hemisphere. Extending the trivialization over the lower hemisphere amounts to extending $\sigma$ to a continuous, positive function in the lower hemisphere, which may be identified with the unit disk. As you note, one method is to construct a partition of unity.

This can be done explicitly: Choose radii $0 < r_1 < r_2 < 1$ and define functions in polar coordinates: $$ f_1(r, \theta) = \begin{cases} 0 & \text{if $r \leq r_1$}, \\ \frac{r - r_1}{r_2 - r_1} & \text{if $r_1 < r < r_2$}, \\ 1 & \text{if $r_2 \leq r$}, \end{cases} $$ and $f_2(r, \theta) = 1 - f_1(r, \theta)$. Viewing $\sigma$ as a function of $\theta$ alone, the function $\sigma(\theta) f_1(r, \theta) + f_2(r, \theta)$ is a continuous extension of $\sigma$ to a positive function in the disk.

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  • $\begingroup$ Thank you! Why $\sigma$ only depends from $\theta$? $\endgroup$ – xxxxx Dec 10 '13 at 18:39
  • $\begingroup$ Initially, $\sigma$ is a function on the boundary circle; you can extend $\sigma$ continuously to the punctured disk by declaring the extension to depend only on $\theta$. $\endgroup$ – Andrew D. Hwang Dec 10 '13 at 21:11
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To make this argument formal, recall that (real) line bundles over $X$ are classified by the Cech cohomology $\check{H}^1(X, \mathbf R^*)$. Write down the Mayer-Vietoris sequence of the open cover consisting of the two hemispheres $U,V$ and their intersection, and see if you can describe $\check{H}^1(S^2, \mathbf R^*)$ from it.

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