## § Vector Bundles and K theory, 1.1

- We define a sphere (in 3D) by all points with distance $1$ from the original. Call this $M$.
- The tangent plane $T_p M \equiv \{ v | v \perp p \}$.
- Define $TM \equiv \cup_{x \in M} T_M$
- We have a projection map $p$ from $TM \to M$ which sends the point $(x, v)$ to $x$.
- for a point $x \in X$, we define $U(x)$ to be the hemisphere with apex $x$. This is the portion of the sphere on one side of the hyperplane that is perpendicular to $x$.
- We want a map $p^{-1}(U(x))$ to $U(x) \times p^{-1}(x)$. The right hand is the same as $U(x) \times T_x M \times \{ x \}$, which is the same as $U(x) \times T_x M$.
- for a given $(y, v) \in TM$, that is, for a given $v \in T_y M$, we map it to $U(x) \times T_x M$ by sending $y \mapsto y \in U(x)$, and by orthogonally projecting the vector $v \in T_y M$ onto the tangent plane $T_x M$.
- Intuitively, we keep the point $y$ the same, and map the tangent plane $T_y$ to its orthogonal projection onto $T_x$.
- Since we know the basepoint $y$, it is clear that we can reconstruct the projection operator from $T_y$ to $T_x$, and that this operator is linear and surjective, and thus invertible.
- This shows us that what we have is really a fiber bundle, since we can locally straighten the $p^{-1}(U(x))$ into the trivial bundle.
- Proof?