Tensor product commutes with pullback?

[CompuChip]

Hello,

I have an exercise where we have to pullback a metric $g^{ij} \, \mathrm dx_i \, \mathrm dx_j$ under a function $f: M \rightarrow N$ (actually in this case $M = \mathbf{R}^2, N = \mathbf{R}^3$, but that's not really relevant).

I managed to do it, provided that the pullback commutes with the tensor product. That is, using that actually $\mathrm dx_i \, \mathrm dx_j = \mathrm dx_i \otimes \mathrm dx_j$, I assumed that
$$f^*(\mathrm dx_i \, \mathrm dx_j) = (f^*(\mathrm dx_i)) \otimes (f^*(\mathrm dx_j))$$
so then I could use that
$$f^*(\mathrm dx_i) = \mathrm d(f^* x_i) = \mathrm df_i = \frac{\partial f_i}{\partial x_k} \mathrm dx_k$$
and finish the exercise.

Why is this true?

Thanks a lot.

 

[Fredrik]

$$f^*(dx_i\otimes dx_j)(u,v)=dx_i\otimes dx_j (f_*u,f_*v)=dx_i(f_*u) dx_j(f_*v)$$

$$=f^*(dx_i)(u)f^*(dx_j)(v)=f^*(dx_i)\otimes f^*(dx_j)(u,v)$$

 

[CompuChip]

Thank you very much!

I missed a lot on this topic though, so I'm not really familiar with pulls/pushes. Could you clarify the first (= third) step a bit?
Basically, it doesn't matter whether we push u to TN and apply dx, or if we pull dx to T*M and apply it to u -- is this just the "d commutes with push-forwards/pullbacks" statement?

[edit]I found some info here, it seems to be pretty elementary.

Anyway, thanks again for the help. [/edit]

 

[Fredrik]

Every step in my calculation is a simple application of a definition, of either the tensor product or the pullback. If you meant the third equality sign, I'm just using the definition of a pullback there. I suppose I could have dropped some of the parentheses though and wrote the second line as

$$=f^*dx_i(u)f^*dx_j(v)=f^*dx_i\otimes f^*dx_j(u,v)$$