Uniqueness of Pushforward of a smooth map Below is what I asked originally. I edited my question thanks to 0-th user. I will add a new question after the original question.
(Original question:)
Let $M, N$ be smooth manifold of dimension $m,n$ respectively. Suppose $F: M \rightarrow N$ is smooth. For $p \in M$, let $(U_1,\varphi_1)$ and $(U_2, \varphi_2)$ be charts about $p$, and for $F(p) \in N$, let $(V_1, \psi_1)$ and $(V_2, \psi_2)$ be charts about $F(p)$. For convenience, suppose $V_1 \supset F(U_1)$ and $V_2 \supset F(U_2)$.
Now, define $\tilde F_1:= \psi_1 \circ F \circ \varphi_1^{-1}: \varphi_1(U_1) \rightarrow \psi(V_1)$, and define $\tilde F_2$ likewise. Then, am I guaranteed that $\tilde F_1 = \tilde F_2$? If so, I would appreciate a little explanation.
(Back to new question:)
If representation of $\tilde F_1$ and $\tilde F_2$ is not unique, then is there a reason to believe that pushfoward $F_*$ is unique? I learned from class that matrix representation of $F_*$ is the same as matrix representation of $(\tilde F_1)_*$ and $(\tilde F_2)_*$, and if $\tilde F_1$ does not match $\tilde F_2$, then there's no reason to believe in the uniqueness of $F_*$.
 A: Remember that $F_*$ is defined as map on tangent spaces, so you need to use the charts to put coordinates on the tangent spaces as well. You have to just unwind all of this to see that $F_*$ is in fact well-defined.
I'll assume that $\phi_i(p) = \psi_i(f(p))=0$ for convenience. Recall that $F_{*p}\colon T_pM\to T_{f(p)}N$ is defined to be $(\psi_1^{-1})_{*0}\circ (\tilde F_1)_{*0}\circ(\phi_1)_{*p}$. You're wondering why this is equal to $(\psi_2^{-1})_{*0}\circ (\tilde F_2)_{*0}\circ(\phi_2)_{*p}$.
At this point, it will help to draw some diagrams ("commutative squares") following the mappings I give you. Let $\phi_{12} = \phi_1\circ\phi_2^{-1}\colon \phi_2(U_1\cap U_2)\to \phi_1(U_1\cap U_2)$ and similarly for $\psi_{12}$. Then $\tilde F_1 = \psi_{12}\circ\tilde F_2\circ\phi_{12}^{-1}$. (Make sure you work that out carefully for yourself.) Therefore (take a deep breath!)
\begin{align*} 
(\psi_1^{-1})_{*0}\circ (\tilde F_1)_{*0}\circ(\phi_1)_{*p} &= (\psi_1^{-1})_{*0}\circ (\psi_{12})_{*0}\circ(\tilde F_2)_{*0}\circ(\phi_{12}^{-1})_{*0}\circ(\phi_1)_{*p} \\
&= (\psi_2^{-1})_{*0}\circ (\tilde F_2)_{*0}\circ (\phi_2)_{*p},
\end{align*}
as required.
