TITLE: Why is the tangent bundle defined using a disjoint union?
QUESTION [7 upvotes]: In textbooks about differential geometry, one finds often the disjoint union in the definition of the tangent bundle (e.g. in "Lee: Introduction to smooth manifolds", or "Amann, Escher: Analysis III"):
$$T\mathcal{M}:=\coprod_{p\in\mathcal{M}}T_{p}\mathcal{M}:=\bigcup_{p\in\mathcal{M}}\{p\}\times T_{p}\mathcal{M}$$
I have the following questions:
(1) Why is this necessary? Is it necessary for the definition of the smooth structure on $T\mathcal{M}$?
(2) When we define the tangent space as the set of all derivations (like in Lee's book), aren't $T_{p}\mathcal{M}$ and $T_{q}\mathcal{M}$ for $p\neq q$ allready disjoint? Why is it then necessary to use the disjoint union? Is it just a notation in this cases? How about other definitions of tangent spaces, like the geometric defined tangent space, via equivalence classes of curves?
(3) How about other bundles? Also in the definition of tensor fields and differential forms, these authors often use the disjoint union...
Thank you all!

REPLY [10 votes]: In my book Introduction to Smooth Manifolds, I define $T_pM$ as the set of all linear maps $v\colon C^\infty(M)\to\mathbb R$ that satisfy 
$$
v(fg) = f(p)vg + g(p)vf
$$ 
for all $f,g\in C^\infty(M)$. Note that the zero map is an element of all these spaces, so $T_pM$ and $T_qM$ are not disjoint for $p\ne q$.
For other definitions of tangent vectors, disjointness might follow automatically. Using the disjoint union in the definition of the bundle is just a handy way to ensure that the vector spaces associated with different points are disjoint regardless of the definition being used.