Doubt about pushforwards

Question

Given the smooth function $\psi:M\to N$, where $M$ and $N$ are two differentiable manifolds, we have introduced the pushforward map $\psi_{*,p}:T_pM\to T_{\psi(p)}N$ with $p\in M$ and the image of this map is defined as $$\psi_{*,p}(v)(f)=v(f\circ\psi)\text{ where }v\in T_pM,f\in \mathcal C^{\infty}_{loc}(M).$$ Sometimes we only write $\psi_{*,p}(v)$, where $v$ is a derivation of the tangent space $T_pM$ ($v$ can also act on the germ of the smooth functions), and we compute the image of the derivation $v$ through $\psi_{*,p}$ without considering the action of the tangent vector on a generic smooth function. I'm not really understanding how to interpret this computation.
I'll try to do an example: Let's consider two differentiable manifolds $M$ and $N$ and their (local) charts $(V\subset M,\psi)$, $(W\subset N ,\varphi)$ such that $\psi:U\to\mathbb R^m$ and $\varphi:W\to \mathbb R^n$. Suppose that $M$ has coordinate functions $x^1,\dots,x^m$ and $N$ has coordinate functions $y^1,\dots,y^n$. Then let $f:M\to N$ be a smooth function.
If I now take the tangent vector $\dfrac{\partial}{\partial x^i}\bigg|_{\psi^-1(q)}$, how can I write explicitily $f_{*}\bigg(\dfrac{\partial}{\partial x^i}\bigg|_{\psi^-1(q)} \bigg)$ in terms of the coordinate functions?

Answer 1

By definition, given a smooth function $g:M\to \mathbb{R}$, we have$$\frac{\partial}{\partial x^i}(g)\Bigg|_{q}:=\frac{\partial g\circ \psi^{-1}}{\partial x^i}\Bigg|_{\psi(q)}$$ Thus (now $g$ is a smooth function on $N$, or it is defined locally around $f(p)$, depending on your definition of tangent space, but it does not make any difference since all of our computations are local) $$\left(f_*\left(\frac{\partial}{\partial x^i}\right)\right)(g)=\frac{\partial (g\circ f\circ \psi^{-1})}{\partial x^i}=\frac{\partial (g\circ\varphi^{-1}\circ\varphi\circ f\circ \psi^{-1})}{\partial x^i}$$ Let us write $\hat f:=\varphi\circ f\circ \psi^{-1}$, i.e. the coordinate expression of $f$ and similarly $\hat g=g\circ \varphi^{-1}$. By the chain rule we get (writing ${\hat {f}}^j$ for the components of $\hat f$): $$\frac{\partial \hat g\circ\hat f}{\partial x^i}=\sum_j \frac{\partial \hat g}{\partial x^j}\frac{\partial {\hat f}^j}{\partial x^i}=\sum_j \frac{\partial}{\partial y^j}(g) \partial_i{\hat f}^j$$ Thus, in coordinates, $f_*=J\hat f$, i.e. the jacobian of $\hat f$.