Why derivations obey chain rule.

Question

Let $X, Y, Z$ be smooth manifolds and suppose we have smooth maps $$ F:X\to Y, $$ $$ G:Y\to Z. $$ By derivation at $x\in X$ I mean a linear map $$ \mathfrak{d}:C^\infty(X) \to \mathbb{R}, $$ such that for any $f_1,f_2\in C^\infty(X)$ we have $$ \mathfrak{d}(f_1f_2)=\mathfrak{d}(f_1) f_2(x)+ f_1(x)\mathfrak{d}(f_2). $$ Define the derivative of F as usual: $$ DF|_x: Der_x X\to Der_y Y, $$ $$ \mathfrak{d}\mapsto DF|_x (\mathfrak{d}), $$ where $$ DF|_x(\mathfrak{d}): C^\infty(Y) \to \mathbb{R}, $$ $$ g\mapsto \mathfrak{d}(g\circ F). $$ I want to prove that the chain rule holds, that is, for $x\in X$ and $y=F(x)\in Y$ we have $$ D(G\circ F)|_x=DG|_y \circ DF|_x. $$ Suppose we have $\mathfrak{d}\in Der_xX$, then LHS gives $$ D(G\circ F)|_x (\mathfrak{d}): C^\infty (Z) \to \mathbb{R}, $$ $$ h\mapsto \mathfrak{d}(h\circ G\circ F) $$ Then on RHS we first have $$ DF|_x(\mathfrak{d}): C^\infty(Y) \to \mathbb{R}, $$ $$ g\mapsto \mathfrak{d}(g\circ F). $$ Denote $\mathfrak{d}':= DF|_x (\mathfrak{d})$. Therefore, $$ DG|_y(\mathfrak{d}'):C^\infty(Z) \to \mathbb{R}, $$ now I don't understand what is the image of $f\in C^\infty(Z)$ under this map. It should be $\mathfrak{d}(f\circ G\circ F)$, but I don't understand how.

Answer 1

It may be best to name the second object, to help abstract your understanding so you can follows the abstract (too much in my opitinion) definition. Let $\mathfrak{d}'(g) = \mathfrak{d}(g \circ F)$. Then, by definition of the differential, $DG(\mathfrak{d}') = \mathfrak{d}''$, where $\mathfrak{d}''(h) = \mathfrak{d'}(h \circ G) = \mathfrak{d}((h \circ G) \circ F)$.

As an alternate method of proof, just take a triple of coordinate systems, and apply the chain rule in euclidean space, which then lifts back to the manifold. The chain rule in euclidean space is the real reason why manifolds obey the chain rule, and the reason the chain rule holds in euclidean space is that the composition of two linear functions is linear.