Regarding diffeomorphism on manifolds

[Sarosh]

I am trying to show that:
If M, N are smooth manifolds without boundary.
Then T(M × N) is diffeomorphic to TM × TN.

an element of TM is of the form (x,v) where x is in M, and same for TN: (y,u) where y is in N.
An elemnt of TM × TN is: ((x,v),(y,u)).
And of T(M × N) is: ((x,y),(v,u)).
Now we define a map
f: TM × TN —>T(M × N) by
f((x,v),(y,u))=((x,y),(v,u)).

We want to show that f is a diffeomorphism, i.e. it is smooth, a bijection and its inverse is smooth.
It is obvious that f is injective and surjective. I am stuck of how to show formally smoothness on a chart.

 

[blamocur]

If you consider manifold charts containing $x$ and $y$ then $f$ can be represented by $\tilde f : \mathbb R^{2n+2m} \rightarrow \mathbb R^{2n+2m}$, where $m$ and $n$ are the dimensions of $M$ and $N$.

 

[Sarosh]

What is f^~ here?
Can you please provide additional information?
I also saw that the diffeomorphism
T_(p,q)(M×N) = T_p M×T_q N implies the diffeo I need to show, but I had no idea how it is related!

 

[blamocur]

What is f^~ here?
Can you please provide additional information?
I also saw that the diffeomorphism
T_(p,q)(M×N) = T_p M×T_q N implies the diffeo I need to show, but I had no idea how it is related!

$\tilde f$ is the "affine version", or induced map, of $f$. A map from one $k$-dimensional manifold to another is considered smooth iff the corresponding charts induce smooth maps in $\mathbb R^k$. In your case one chart maps $T_{p,q}(M\times N)$ to $\mathbb R^{2(m+n)}$ and another maps $T_p(M) \times T_q(N)$ to $\mathbb R^{2n} \times \mathbb R^{2m}$, so you have to show that the induced map (called $\tilde f$ in my previous post) from (neighborhood of) $\mathbb R^{2(m+n)}$ to $\mathbb R^{2n} \times \mathbb R^{2m}$ is smooth.

 

[Sarosh]

Can you please explain how to show this?

Thanks

 

[blamocur]

Can you please explain how to show this?

Thanks

Sorry, but you have to work out the details on your own. If you have more specific questions I might be able to answer them.