Showing the Hopf fibration has no global sections

Question

Let's consider a principle $U(1)$-bundle over $S^2$ with the transition function $g_{\infty 0} = z/|z|$ (it is known as the Hopf fibration). There is a simple topological argument showing that this bundle is not trivial by the comparison of fundamental groups. However, it should be clear directly from definitions.

So I would like to prove that this bundle has no global sections (which is equivalent to non-triviality). It means that there is no continuous function $f: \mathbb{C} \to U(1)$ such that there exists $$ \lim_{z \to \infty} \frac{z}{|z|} f(z) $$ How would one prove this?

Answer 1

Consider the function $g(z) = \frac{z}{|z|} f(z)$ on $\Bbb C \setminus \{0\}$. Your assumption is that this extends to a map on $\Bbb{CP}^1 \setminus \{0\}$. This means that $g$ has degree zero as a map to $U(1)$ (it induces multiplication-by-zero on $H_1$), as it factors through a space with $H_1 = 0$.

For $z \in S^1$, $g(z) = z \cdot f(z)$. Degree is additive under multiplication of functions to $U(1)$, and we see that $\text{deg}(f) = -1$. However, by assumption, $f$ extends across the unit disc, and so should have degree zero. This is a contradiction.

Answer 2

This fact can be also obvious claimed such way. As you noticed, Hopf bundle is nontrivial, which can be easily proved trough comparing fundamental groups. If there was a global section, it would split a one dimensional addendum from fibre at every point, which would be proof of bundles triviality( since the fibre is one dimensional itself). Contradiction