Question about proof on $F$-related vector fields in Lee

Question

$\textbf{Proposition:}$ Suppose $F: M \to N$ is a smooth map between manifolds with out without boundary, let $X \in \mathcal{X}(M)$ and $Y \in \mathcal{Y}(N)$. Then $X$ and $Y$ are $F$-related if and only if for every smooth valued function $f$ defined on an open subset of $N$, $$ X \left ( f \circ F \right)= \left(Yf \right) \circ F$$.

My question concerns the following statement in the proof of the above proposition,

For any $p \in M$, and any smooth real-valued $f$ defined in a nieghborhood of $F(p)$ we have, $$X \left ( f \circ F \right ) (p) = X_p \left ( f \circ F \right) = dF_p(X_p) f$$

How does one go about showing the following equality, $$X_p \left ( f \circ F \right) = dF_p(X_p) f ? $$ It looks like it could be an application of the chain rule but I am not sure...

Answer 1

The answer here is actually "by definition". If you have a smooth map $F: X \rightarrow Y$ then $$dF_p: T_p X \rightarrow T_{F(p)}Y$$ How is this defined? Take $V \in T_p X$ and send it to the tangent vector $dF_p V \in T_{F(p)}Y$, now that is just notation, but how does this act on a function? $$ dF_p (V) f := V (f\circ F)$$ See the paragraph on `differential of a smooth function' in Lee's book (p. 55 in my copy). Hope that helps!

Answer 2

That's just the definition of the differential of $F$ at $p$, $dF_p$. Take a look at page 55:

If $F\colon M\to N$ is a smooth map, for each $p\in M$, $dF_p\colon T_pM\to T_{F(p)}N$ is defined as follows: If $v\in T_pM$, and $f\in C^\infty(N)$, then $dF_p(v)\in T_{F(p)}(N)$ is the derivation that acts as $$ dF_p(v)(f)=v(f\circ F). $$

Since $X\colon M\to TM$ is a vector field, $X_p\in T_pM$, and that's your $v$ in this case.