# Suppose $M$ and $N$ are smooth manifolds then $T(M \times N)$ is diffeomorphic to $TM \times TN$

I think I found a map but I cannot prove that is a diffeomorphism. That is the map :
$$F : T(M \times N) \to T(M) \times T(N)$$ defined by $$F((m,n),X) = ((m,(\pi_{1})_{\star,(m,n)}(X)),(n,(\pi_{2})_{\star,(m,n)}(X)))$$
where $$\pi_1 : M \times N \to M$$ and $$\pi_2 : M \times N \to N$$ are the projections and the "little star" represent the push-forward.

My try:

Now the idea using local charts I think is the following :

Let $$(U,\phi)$$ be a local chart for $$M$$ , $$(V,\psi)$$ be a local chart for $$N$$, and so $$(U\times V,\Phi:= \phi\times \psi)$$ is a local chart for $$M\times N$$ by definition, and $$(TU, \tilde \phi)$$ $$(TV,\tilde \psi)$$ $$(T(U\times V), \tilde\Phi)$$ will be the induced local charts in $$TM, TN$$ and $$T(M\times N)$$. Now the only problem is to check that this maps are in fact smooth. Now I will do this for $$F$$.

So in these local charts we have that $$(\tilde \phi\times \tilde \psi)\circ F\circ(\tilde \Phi)^{-1}(x_1,...x_m,y_1,...,y_n,\epsilon_1,...\epsilon_m,\eta_1,...,\eta_n)=\\= (\tilde \phi\times \tilde \psi)\circ F(\phi^{-1}(x_1,...x_m),\psi^{-1}(y_1,...,y_n),\epsilon_1,...,\epsilon_m,\eta_1,...,\eta_n)=\\=(\tilde\phi\times \tilde \psi)(\phi^{-1}(x_1,...x_m),\epsilon_1,...,\epsilon_m,\psi^{-1}(y_1,...,y_n),\eta_1,...,\eta_n)=\\ =(x_1,...,x_m,\epsilon_1,...\epsilon_m,y_1,...,y_n,\eta_1,...,\eta_n)$$

Now to clarify what is happening when we apply $$F$$ at the tangent vector that this $$v=\epsilon_1 \frac{\partial}{\partial x_1}+...+\epsilon_n \frac{\partial}{\partial x_n}+...+\eta_1 \frac{\partial}{\partial y_1}+...+\eta_n \frac{\partial}{\partial y_n}$$, we need to look at the diferentials $$d\pi_1$$ and $$d\pi_2$$. I will only look at one of them since the other is analogous. So for example at $$d\pi_1(\frac{\partial}{\partial x_i})=\sum_{j=1}^{m}\frac{(\phi\circ \pi_1 \circ (\phi\times \psi)^{-1} )^j}{\partial x_i}\frac{\partial }{\partial x_j}=\frac{\partial}{\partial x_i}$$, and $$d\pi_1(\frac{\partial}{\partial y_i})=\sum_{j=1}^{m}\frac{(\psi\circ \pi_1 \circ (\phi\times \psi)^{-1} )^j}{\partial y_i}\frac{\partial }{\partial x_j}=0$$.

My question : How could I construct $$F^{-1}$$? could anyone help?

• This math.stackexchange.com/questions/3109967/… may give you an answer. Dec 9, 2020 at 16:23
• It does defenitely differ from my question Dec 9, 2020 at 16:29
• @Reza Yes it does differ, but with some effort it answers your question. This shows that they are isomorphic as vector bundles over $M\times N$, thus as smooth manifolds. Dec 9, 2020 at 18:38

Adding to @Bargabbiati's answer. In proposition 3.14, proved in problem 3-2 of this same book, it is proven that $$\begin{equation*} \alpha (v) = (d {\pi _{1}} _{p} (v), ..., d {\pi _k}_p (v)) \end{equation*}$$ for all $$v \in T_p (M_1 \times \cdots \times M_k)$$ is an isomorphism between this tangent space and the direct sum $$\bigoplus T_{p_i} M_i.$$
That being said, such $$F$$, as you defined it, is an isomorphism (that is almost clear).