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While studying the definition of related vector fields for my course in differentiable manifolds, I noticed the following:

We gave the following propositions about the derivative of a function $f: M \to N$:

  1. We have $d_pf: T_pM \to T_{f(p)}N$

  2. Taking $X_p \in T_pM$ as a differential operator, we have that $(d_pfX_p)(g)=X_p(g \circ f)$ for $g \in C^{\infty}(N)$

How can these two definitions be compatible if $X_p$ as a differential operator is $X_p: C^{\infty}(M) \to C^{\infty}(M)$?

The only way I see for the result of $(d_pfX_p)(g)$ to be in $T_{f(p)}N$ is if $(d_pfX_p)(g)= Y_{f(p)}(g)$, were $Y\in T_{f(p)}N$. That means that there is a vector field Y f-related to X. But, as far as I understand, there is no guarantee that there is such a vector field Y. So this way of making the two propositions compatible seems wrong.

I feel I am missing something important.

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I think you have your mappings wrong.

$X_p\in T_pM$, is a map $C^{\infty}(M)\to\mathbb{R}$, so for $h:M\to\mathbb{R}$ we have $X_ph\in\mathbb{R}$

$d_pf.X_p\in T_{f(p)}N$, is a map $C^{\infty}(N)\to\mathbb{R}$, so for $g:N\to\mathbb{R}$ we have $(d_pf.X_p)g=X_p(g\circ f)\in\mathbb{R}$

So indeed $d_pf$ is a map $T_pM\to T_{f(p)}N$

You might be confusing the tangent vector $X_p\in T_pM$ with some section $Z\in\Gamma(TM)$, i.e., a map $M\to TM$ such that $\pi\circ Y=\mathrm{id}_M$ (where $\pi$ is the bundle projection) and hence $Z$ can be seen as a map $C^{\infty}(M)\to C^{\infty}(M)$ that acts like $h\mapsto Zh$ where $(Zh)_p:=Z_ph$.

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