Tangent map $F_{*}: T_{p}(M) \to T_{F(p)}(N)$ are linear tranformations. How to show that tangent map $F_{*}: T_{p}(M) \to T_{F(p)}(N)$ are linear tranformations?
I know that it suffices to show that
$$F_{*}(ax_{u}+bx_{v})=aF_{*}(x_{u})+bF_{*}(x_{v})$$
where $x$ is a patch in $M$.
By definition,
$$F_{*}(ax_{u}+bx_{v})=\frac{d}{dt}F(\alpha(t))_{t=0}$$
where $\alpha'(0)=ax_{u}+bx_{v}$.
I also know that if $y=F(\alpha(t))$, then $y'=F_{*}(\alpha'(t))$.
We can express $\alpha(t) =x(a_{1}(t),a_{2}(t))$, then $$\alpha'(t) =a_{1}'(t)[x_{u}(a_{1}(t),a_{2}(t))]+a_{2}'[x_{v}(a_{1}(t),a_{2}(t))]$$
For $t=0$, $\alpha'(0) =a_{1}'(0)[x_{u}(a_{1}(0),a_{2}(0))]+a_{2}'[x_{v}(a_{1}(0),a_{2}(0))]=ax_{u}+bx_{v}$.
Thus, I get $a_{1}(0)=u$, $a_{2}(0)=v$, $a_{1}'(0)=a$, $a_{2}'(0)=b$.
Now, we can say that
$$F_{*}(ax_{u}+bx_{y})=\frac{d}{dt}F(x(a_{1}(t),x(a_{2}(t)))_{t=0}$$
$$=a_{1}'(0)\frac{\partial}{\partial{u}}F(x(a_{1}(t),x(a_{2}(t)))+a_{2}'(0)\frac{\partial}{\partial{v}}F(x(a_{1}(t),x(a_{2}(t))$$
If $\frac{\partial}{\partial{u}}F(x)=F(x_{u})$
, done. but is it right?
Thanks in advance.
 A: To see that $F_*$ is linear, let us describe its image in the natural basis of $T_{F(p)}N$ coming from the choice of a coordinate patch. Let $\tilde{x}$ be a coordinate patch around $F(p)$, and let $\phi,\psi$ be the (smooth) functions on the domain of $x$ such that
$$F(x(u,v)) = \tilde{x}(\phi(u,v),\psi(u,v)).$$
Also, let $\tilde{\alpha} = F \circ \alpha$ and write $\tilde{\alpha}(t) = \tilde{x}(\tilde{a}_1(t),\tilde{a}_2(t))$ where $\tilde{a}_1(t) = \phi(a_1(t),a_2(t))$, and $\tilde{a}_2(t) = \psi(a_1(t),a_2(t))$. Then
$$F_*(\alpha'(0)) = \tilde{\alpha}'(0) = \tilde{a}_1'(0) \tilde{x}_u + \tilde{a}_2'(0) \tilde{x}_v \\ =(a\phi_u + b\phi_v) \tilde{x}_u + (a\psi_u+b\psi_v)\tilde{x}_v.$$
As an added bonus, this gives you a matrix representation of $F_*$ with respect to the bases given by $x$ and $\tilde{x}$.
A: You could for example go into the definition of $ F_* $ in local coordinates and use the physical description of tangent vectors. Then everything cooks down to matrix multiplication.
Or you use the description of how $ F_* $ acts on equvivalence classes of curves $[\gamma ]$ , i.e. $F_*[\gamma]=[\gamma \circ F]$. You just have to take care how you get the vector space structure in that case.
