Proving Pushforward Product Isomorphism: M1 x ... x Mk to M1 + ... + Mk

[robforsub]

Let M1,..,Mk be smooth manifolds, and let Pi_j be the projection from M1XM2X...XMk->Mj. Show that the map a:T_(p1,...,pk)(M1XM2X...Mk)->T_p1(M1)\oplus...\oplusT_pk(Mk)

a(X)=(Pi_1*X,Pi_2*X,...,Pi_k*X) is an isomorphism.

The way I am thinking to prove the statement is to show that a is a bijection, since a is already a linear map. And I have no clue how to show it, help needed!

 

[arkajad]

I would take a natural basis in the image and "guess" what would be its pre-image - then check.

 

[huyichen]

But, don't we need to take component functions of arbitrary X into account?

 

[quasar987]

Recall the manifold structure on M x N. If (U,f) is a chart of M^m around p and (V,g) is a chart of N^n around q, then (U x V, f x g) is a chart of M x N around (p,q). If ∂/∂x^i is the basis of T_pM associated with the chart (U,f) and ∂/∂y^i is the basis of T_qN associated with the chart (V,g), call (∂/∂x'^1,..., ∂/∂x'^m, ∂/∂y'^1,..., ∂/∂y'^n) the basis associated with the chart (U x V, f x g).

Just doing this should give you some room for your experiments.