# Is the tangent bundle the DISJOINT union of tangent spaces?

Let $M$ be a smooth manifold and consider the Lee's definition of the tangent space $T_pM$ (so $T_pM$ is the vector space of derivations at $p$). The canonical definition of tangent bundle (as set) of $M$ is: $$TM=\bigcup_{p\in M}\{ p\}\times T_pM$$ so it is the disjoint union of all tangent spaces; but L.W.Tu in his "Introduction to Manifolds" says that the tangent spaces are already disjoint and for this reason he defines $$TM=\bigcup_{p\in M} T_pM$$

Why we can't find a common derivation between $T_pM$ and $T_qM$ if $q\neq p$? I think that Tu's statement is not true.

• How did Lee and Tu respectively define $T_pM$? – Neal Nov 4 '12 at 12:32
• This should also be differential-topology, not differential-geometry. – M.B. Nov 4 '12 at 12:33
• I just noticed this question. I'll add that with the definition of $T_pM$ that I give in my book (derivations of $C^\infty(M)$ at $p$), the zero derivation is a derivation at $p$ for every $p$, so defining the tangent bundle as a simple union would not work. I understand the advantages of defining $T_pM$ as the set of derivations of the space of germs (and I often think of it that way myself), but for pedagogical reasons I made the decision to use the conceptually simpler definition involving $C^\infty(M)$, which I thought would be a little easier for novices to wrap their heads around. – Jack Lee Jun 9 '17 at 18:42
• – An old man in the sea. Nov 3 '18 at 13:48

A derivation at $p\in M$ is in particular a linear map $\partial: C^\infty_p \to \mathbb R$ defined on the set of germs of smooth functions at $p$, and similarly for $q$.
So the sets of derivations at $p$ and $q$ are disjoint simply because they consist of maps with different domains (namely $C^\infty_p$ and $C^\infty_q$). And maps with different domains cannot be equal, as follows from the set-theoretical definition of "map".
• Tu's definition of derivation is different from that in Lee's book. The first, uses germs instead the latter says that a derivation is a linear function with domain $C ^{\infty} (M)$ – Dubious Nov 4 '12 at 15:58
• As an algebraic/analytic geometer I much prefer the definition with germs. The other definition works in differential geometry because of the specific result that the canonical map $\mathcal C^\infty(M) \to \mathcal C^\infty_p$ is surjective. This is completely false in the algebraic/analytic category. – Georges Elencwajg Nov 4 '12 at 17:20
• But how are we sure that we always have $C^\infty_p \neq C^\infty_q$ for $q\neq p$? – An old man in the sea. Nov 3 '18 at 12:50
• in a hausdorff space p and q are separated by open neighborhoods which i suspect make up the domains of the elements of $C_p^\infty$ and $C_q^\infty$. – peter Feb 13 at 10:29