Tangent Bun

In 'tics: the disjoint union of the tangent spaces to each point of M.
Hear: a) dTewer @bessed, wurst? what close-to-a-farmin does.

Let be a differentiable manifold. Let the tangent bundle of be(as a set) the disjoint union $\coprod_{m\in M}T_mM$ of all the tangent spaces to , i.e., the set of pairs

$\displaystyle \{(m,x)\vert m\in M, x\in T_mM\}.$

This naturally has a manifold structure, given as follows. For $M=\mathbb{R}^n$ , $T\mathbb{R}^n$ is obviously isomorphic to $\mathbb{R}^{2n}$ , and is thus obviously a manifold. By the definition of a differentiable manifold, for any $m\in M$ , there is a neighborhood of and a diffeomorphism $\varphi :\mathbb{R}^n\to U$ . Since this map is a diffeomorphism, its derivative is an isomorphism at all points. Thus $T\varphi :T\mathbb{R}^n=\mathbb{R}^{2n}\to TU$ is bijective, which endows with a natural structure of a differentiable manifold. Since the transition maps for are differentiable, they are for as well, and is a differentiable manifold. In fact, the projection $\pi:TM\to M$ forgetting the tangent vector and remembering the point, is a vector bundle. A vector field on is simply a section of this bundle. The tangent bundle is functorial in the obvious sense: If $f:M\to N$ is differentiable, we get a map $Tf:TM\to TN$ , defined by on the base, and its derivative on the fibers.