Let $U \subset \mathbb R^n$ be open and $f: U \to \mathbb R^m$ be smooth.
Consider the vector fields ${\partial f \over \partial x_i}$.
I think these are an example of a vector field along $f$ but I can't seem to prove it.
Concretely: Along a map means that the projection $\pi_{TM}: TM \to M$ of the vector field equals the map at a given point. But I tried an example and it seems to be clearly false:
The projection takes $(p,v)$ to $v$, that is, $\pi(p, {\partial f \over \partial x_i}) = {\partial f \over \partial x_i}$ but clearly, this is not equal to $f$.
What am I missing? And how can I prove this (assuming it is actually true)?
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Maybe if possible you could given an example of a concrete $f$ and the map $\pi_{TM}$, I think that would really help me.
Let $f:M\to N$ be a smooth map between two smooth manifolds. A vector field along $f$ is a smooth function $X:M\to TN$ such that for every $m\in M$ we have $\pi(X(m))=f(m)$, where $\pi:TN\to N$ is the projection. Intuitively, $X$ eats a point on $M$ and spits out a tangent vector at its image under $f$. This is clearly the case in the example you give, by definition.