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

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 $U$ of $m$ 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 $TU$ with a natural structure of a differentiable manifold. Since the transition maps for $M$ are differentiable, they are for $TM$ as well, and $TM$ 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 $M$ 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 $f$ on the base, and its derivative on the fibers.