I'm currently self-learning differential geometry. I'm going through Gadea and Muñoz "Analysis and Algebra on Differentiable Manifolds: A Workbook for Students and Teachers" to get a better understanding of the subject with more hand-on calculations. On the section about Transforming Vector Fields, problem 1.103, it is ask:
Consider the projection $p: \mathbb{R^2} \to \mathbb{R}, (x,y) \mapsto x$. Find the condition that a vector field on $\mathbb{R^2}$ must verify to be p-related to some vector field on $\mathbb{R}$.
I used the definition of F-related vector fields, $dF_{p} (X_{p})= Y_{F(p)}$, and calculate the condition to be $X^{1}(x_{0},y_{0})=Y^{1}(x_{0}) $. Yet, the solution given by the book shows a different condition for a F-related vector, which is $p_{*}(X_{(x_{0},y_{0})})= p_{*}(X_{(x_{1},y_{1})})$. Their explanation is: in order for $X \in \mathfrak{X}(\mathbb{R^{2}}) $ to be p-related to some vector field on $\mathbb{R}$, it must happen that for each couple of points $(x_{0},y_{0}),(x_{1},y_{1}) \in \mathbb{R^{2}}$ such that $p(x_{0},y_{0})=p(x_{1},y_{1})$ one has
\begin{align}
p_{*}(X_{(x_{0},y_{0})})= p_{*}(X_{(x_{1},y_{1})})
\end{align}
and they obtain the condition $X^{1}(x,y_{0})=X^{1}(x,y_{1})$ for all $x,y_{0},y_{1}$.
I don't understand how that condition is found and why it works. In subsequent F-related Vector Fields problems the book uses the same condition, I follow the calculations but not why the condition is valid. I would greatly appreciate your help, thank you.
You agree that $Y^1(x) = X^1(x,y_0)$; but $y_0$ can be arbitrary here, and so it must be the case that $X^1(x,y)$ does not depend on $y$. (The entire line $x=x_0$ projects to the point $x_0\in\Bbb R$. And you must have $p_*(X(x_0,y_0))=p_*(X(x_0,y_1))$ for all $y_0,y_1$ in order to have $Y$ well-defined.)