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Product of pushforward

  1. Oct 10, 2010 #1
    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)[tex]\oplus[/tex]...[tex]\oplus[/tex]T_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!
     
  2. jcsd
  3. Oct 10, 2010 #2
    I would take a natural basis in the image and "guess" what would be its pre-image - then check.
     
  4. Oct 10, 2010 #3
    But, don't we need to take component functions of arbitrary X into account?
     
  5. Oct 11, 2010 #4

    quasar987

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    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.
     
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