Does Every Closed Form on U Being Exact Imply the Same for f(U)?

[rocket]

let f:U \rightarrow R^n be a differentiable function with a differentiable inverse f^{-1}: f(u) \rightarrow R^n. if every closed form on U is exact, show that the same is true for f(U).

Hint: if dw=0 and f^{\star}w = d\eta, consider (f^{-1})^{\star}\eta.
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I'm not quite sure what the hint means or how to use it. is it true that f^{\star} is basically another way of writing a differential - eg. df? I didn't really get a clear definition of it in my text.

anyway here's my thoughts so far:

consider w as a form on U. suppose w is closed. then dw = 0. since every closed form on U is exact, then there exists a \eta on U such that w = d\eta.

but how is it that f^{\star}w = d\eta (given in the hint)? like, how is this relationship derived? if w = d\eta and also f^{\star}w = d\eta, then we have w = f^{\star}w? I find that really confusing, and I'm not sure how to continue the problem. Any help is greatly appreciated. thanks in advance!

 

[AKG]

I'm rusty on this stuff, but if w is a form on f(U), then w' = (f^-1)*w is a form on U. Show that if w is closed, then so is w'. If w' is closed, then there is some h' such that w' = dh'. Try to use this to show that w is also exact. Perhaps you will find that w = d(f*h') or something like that.

 

[thepatientmental]

The hint is referring to the pullback of forms under a differentiable map. The pullback of a form w on U by a differentiable map f is defined as f^{\star}w = w \circ f . In other words, it is the form obtained by applying f to the variables in w. So, if w = d\eta on U, then f^{\star}w = f^{\star}(d\eta) = d(f^{\star}\eta) (using the properties of the pullback operation).

Now, since f is a differentiable function with a differentiable inverse f^{-1}, we can also define the pullback of forms on f(U) using (f^{-1})^{\star}. So, if we have a form \eta on f(U), then the pullback of \eta by (f^{-1})^{\star} would be (f^{-1})^{\star}\eta.

Using the definition of the pullback, we can see that (f^{-1})^{\star}\eta = \eta \circ f^{-1}.

Now, going back to the hint, if we have dw = 0 and f^{\star}w = d\eta, then we can use the properties of the pullback to obtain (f^{-1})^{\star}dw = (f^{-1})^{\star}(0) = 0. Similarly, (f^{-1})^{\star}f^{\star}w = (f^{-1})^{\star}(d\eta) = d((f^{-1})^{\star}\eta) (using the properties of the pullback).

Therefore, if we can show that (f^{-1})^{\star}\eta is a closed form on U, then we have shown that every closed form on f(U) is exact.

To show that (f^{-1})^{\star}\eta is closed, we need to show that d((f^{-1})^{\star}\eta) = 0. But we already know that d\eta = w = (f^{-1})^{\star}f^{\star}w, so we can rewrite the equation as d((f^{-1})^{\star}\eta) = (f^{-1})^{\star}f^{\star}w.

Using the properties of the pullback, we