How should I prove this map to be injective and surjective I am reading from course notes on Smooth Manifolds and I was unable to prove this which
Define $T_v : C^{\infty}(U) \to \mathbb{R} $ by $T_v(f)= v(f) = \frac{d}{dt} f(p+tv)|_{t=0}=df_p(v)=\frac{df}{dx_1}(p) v_1 +...+ \frac{df}{dx_n} (p)v_n$.
Let $M, M_1, M_2$ be manifolds and $M= M_1 \times M_2$ , $P= (P_1,P_2)$ Then Prove that $T_p M \approx T_{P_1} M_1 \oplus T_{P_2}M_2$
If $\pi_1 : M \to M_1$ , $\pi_2 : M\to M_2$, $(d\pi_1)_P: T_P M \to T_{P_1}M_1$, $(d\pi_2)_P: T_PM \to T_{P_2}M_2$, $\alpha :T_P M \to T_{P_1} M_1 \oplus T_{P_2}M_2$ ,
I thought of the map $\alpha(v)=((d\pi_1)_P(v), (d\pi_2)_P (v))$
I need help in checking map 1-1 and onto.
Let $\alpha(v)=0$ , to show v=0. $\alpha(v)=0$ => $(d\pi_1)_P(v)=0 $ and $ (d\pi_2)_P(v)=0$ but  how does definition of $df_p$ now implies that v=0. Can you please tell ?
Similarly , I am not able to think how should  I approach the surjective part.
Kindly give a rigorious proof so that I can learn the method for smooth manifolds as I think I am not much comfortable in this.
 A: Let's say $k_{1}$ and $k_2$ are the dimensions of $M_{1}$ and $M_2$ respectively, then $T_{P_{1}}(M)$ has dimension $k_1$ and $T_{P_2}(M)$  has dimension $k_2,$ while their direct sum $T_{P_{1}}(M_1) \bigoplus T_{P_2}(M_2)$ has dimension  $k_1+k_2.$ But you also know that $M=M_{1} \times M_2$ has dimension $k_1+k_2$ as a manifold, so that the tangent space $T_{P}(M)$ has dimension $k_1+k_2.$ Since the spaces $T_{P_{1}}(M) \bigoplus T_{P_2}(M)$ and $T_{P}(M)$ have the same dimension it is enough to prove that your map $\alpha$ is injective, it will follow from the rank-nullity theorem that $\alpha$ is also an isomorphism: https://en.wikipedia.org/wiki/Rank%E2%80%93nullity_theorem
To show that your map $\alpha$ is injective I will choose a system of coordinates; for a given point $P=(P_1,P_2)$ there is a chart $\varphi=(\varphi_{1},\varphi_{2})$ defined on an open set $U=U_1 \times U_2:$ where $(\varphi_1,U_{1})$ and $(\varphi_2,U_{2})$ are charts for $P_1$ and $P_2$ respectively.
Since $\alpha(v)=0$ then in particular
$$ (d\pi_{1})_{P}(v)=0$$
and so
$$ (d\pi_{1})_{P}(v)(f)=0 \quad \forall f \in C^{\infty}(M_1).$$
And by applying the definition of differential we obtain
$$ v(f \circ \pi_{1})=0 \quad \forall f \in C^{\infty}(M_1):$$
intuitively the differential 'pulls back' your function $f \in C^{\infty}(M_1)$ to the function $f \circ \pi_1 \in C^{\infty}(M_1\times M_2) $ so that you can apply your derivation $v.$
The last equality can also be written in coordinates as
\begin{equation}
 \label{cor} v^{1} \cdot (\frac{\partial}{\partial x_{1}})_{P}(f \circ \pi_{1})+ \dots +v^{k} \cdot (\frac{\partial}{\partial x_{k}})_{P}(f \circ \pi_{1})=0 \quad \forall f \in C^{\infty}(M_{1})
\end{equation}
Now if you choose as your map $f=(\varphi_{1})^{1}(u),$ the first component of $\varphi_1$, then the function $(\varphi_{1})^{1} \circ \pi_{1}$ is the first component of the chart $\varphi,$ which we can indicate simply by $x^1:$ the two functions coincide because $$\varphi(P_1,P_2)=(\varphi_{1}(P_1),\varphi_{2}(P_2)).$$
So from the equality above we deduce that
$$ v^1= v^{1}  \cdot (\frac{\partial}{\partial x_{1}})_{P}(x^1) \\
 =v^{1} \cdot (\frac{\partial}{\partial x_{1}}(x^1))_{P}+ \dots +v^{k} \cdot (\frac{\partial}{\partial x_{k}})_{P}(x^1)=0 \quad \forall f \in C^{\infty}(M_{1}):$$
the derivations $$\big(\frac{\partial}{\partial_{x_{j}}}\big)_{P}$$ with $j \neq 1$ send the function $x^1$ in zero, because if you are moving in the direction $x^1$ you are not moving along the others.
You could repeat this reasoning for all the remaining $k_1$ coordinates, then for the last $k_2$ coordinates you will use the other relation $$ (d \pi_{2})_{P}(v)=0.$$ This way we have proven that all the components of $v$ are zero, so $v=0,$ the map $\alpha$ is $1-1$ and for what we have observed is also an isomorphism.
