Mobius over the sphere is the sphere itself

Question

The Mobius band can be thought as a line bundle over $S^1$ by giving the vector spaces half a twist at some point. Now, we can do the same kind of construction by considering a line bundle over the sphere $S^2$ where we give the vector spaces half a twist at the equator. By drawing pictures, I convinced my self that the resulting space is homeomorphic to the trivial bundle $S^2\times\Bbb R$. Can we prove this intuition more rigorously?

Answer 1

I am not sure I understand your construction, but there are no non-trivial line bundles over simply connected spaces (such as the sphere).

Answer 2

Thomas Rot has explained why your construction can't work for $S^2$. You can also see more explicitly that it doesn't work for any $S^n$, even for $S^1$ (on any reasonable interpretation of what you mean by "half a twist"):

You are dividing $S^n$ into the (closed) upper hemisphere $U^n$ and the (closed) lower hemisphere $L^n$ and forming the total space $X$ of a line bundle over $S^n$ by glueing $U^n\times \mathbb{R}$ and $L^n \times \mathbb{R}$ together, identifying $(e, x) \in U^n \times\mathbb{R}$ with $(e, -x) \in L^n\times \mathbb{R}$ for points $e$ on the equator $U^n \cap L^n$ and mapping $X$ onto $S^n$ in the natural way ($[(v, x)] \mapsto v$). This gives a trivial line bundle for any $n$, via the homeomorphism of $X$ with $S^n \times \mathbb{R}$ that maps $(v, x)$ to $(v, -x)$ for $v \in L^n$.