I want to check my arguments for the following proof. Let $M$ and $N$ be smooth manifolds where $\dim M=n$, $\dim N = m$, and $\pi_1:M\times N\to M$, $\pi_2:M\times N\to N$ be corresponding projective maps. I want to show that the map $$\alpha:T_{(p,q)}(M\times N)\to T_p M\oplus T_q N$$ given by $\alpha(v)=(d(\pi_1)_p(v),d(\pi_2)_p(v))$ is an isomorphism where $(p,q)\in M\times N$.
It is enough to argue that $\alpha$ is injective since the $\dim(T_{(p,q)}(M\times N))=\dim(T_p M\oplus T_q N)$. In other words, we want to show that if $\alpha(v)=0$, then $v=0$. We can see that if $\alpha(v)=0$, then $$d(\pi_i)_p(v)=0\text{ for }i=1,2.$$ But, for example, if we are going to write $d(\pi_1)_p$ in local coordinates for the map $\pi_1:M\times N\to M$, then we will obtain that $$d(\overline{\pi}_1)_p=[I\text{ }0]$$ where $\overline{\pi}_1:U_1\times U_2\to U_1$ with $\overline{\pi}_1(u_1,u_2)=u_1$, $u_1\in U_1$ and $u_2\in U_2$. So, if we are going to write $v$ in terms of the basis elements, then we can show that first $n$ components of $v$ are zero by applying $\overline{\pi}_1(v)=0$. By doing the same for $\overline{\pi}_2$, we will obtain that $v=0$.
I know the second approach showing this by constructing the inverse map, but I wonder if this approach is also correct. Thank you!