Tangent Bundle of Product Manifold Suppose $M,N$ are manifolds, and consider the product $M\times N$.
From this answer, I know that:
$T_{(m,n)}(M \times N) \cong T_m M \oplus T_n N $

Can we conclude that $T(M\times N) \cong T(M) \oplus T(N)$

 A: Let, $M \subset \Bbb R^m , N \subset \Bbb R^n$ (i.e. considered as subsets
of the Euclidean spaces of respective dimensions to start off with!)
$$T(M \times N)=\{((x,y),(v,w))\in M \times N \times \Bbb R^{n+m}: (v,w)\in T_{(x,y)}(M \times N)\}$$$$=\{(x,y,v,w)\in M \times N \times \Bbb R^{n+m}: (v,w)\in T_{(x,y)}(M \times N)\} \dots (*)$$ $$and$$ $$TM \oplus TN=\{(x,v,y,w)\in M \times \Bbb R^m \times N \times \Bbb R^n : v \in T_xM,w\in T_y N\}$$
Note that we can write $(*)$ due to the identification of $T_{(x,y)}(M \times N)=T_x M \oplus T_yN$
Let's make out intentions clear, we want to "just switch $y$ and $v$" .
Now take open sets $U,V$ in $\Bbb R^m,\Bbb R^n$ (respectively) containing $x \in X$ and $y \in Y$ respectively.Note that $U \times V$ is again an open set in $\Bbb R^{m+n}$. Then look at the "switching map" here, i.e. $$\phi: T(U \times V) \to T(U) \times T(V)$$ $$(x,y,v,w) \to (x,v,y,w)$$
Again note that, $T(U \times V)=U \times V \times \Bbb R^{n+m}$ and $T(U) \times T(V)= U \times \Bbb R^n \times V \times \Bbb R^m$ , hence the switching here makes 
 perfect sense and in fact a diffeomorphism! ( Since there is no local dilemma!) 
Hence this map $\phi$ extends the "switching map" locally and hence $\tilde{\phi}: T(M \times N) \to T(M) \times T(N)$ defined by, $(x,y,v,w) \mapsto (x,v,y,w)$ is a local diffeomorphism. It is clear that it is a bijection. Thus bijection + local diffeomorphism $\implies$ that $\tilde{\phi}$ defines a diffeomorphism from $T(M \times N) \to T(M) \times T(N)$
