Smooth proper self-maps on Rn

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In summary, a smooth proper self-map on Rn is a continuous and differentiable function that maps a point in n-dimensional Euclidean space onto itself, satisfying certain conditions for well-behaved and desirable properties. These conditions include continuity, differentiability, finite limit at infinity, mapping the boundary of the space onto itself, and being proper. This type of self-map differs from a general self-map in its proven properties and has applications in mathematics, physics, and engineering. However, it is limited to continuous and differentiable functions and may not accurately model systems with discontinuities or non-differentiable behavior. Additionally, its use may be limited in higher-dimensional spaces.
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Euge
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Let ##f : \mathbb{R}^n \to \mathbb{R}^n## be a smooth proper map that is not surjective. If ##\omega## is a generator of ##H^n_c(\mathbb{R}^n)## (the ##n##th de Rham cohomology of ##\mathbb{R}^n## with compact supports), show that $$\int_{\mathbb{R}^n} f^*\omega = 0$$
 
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I haven't learned (or remember?) compactly supported cohomology well, so please let me know if there are errors here.

If we consider ##\mathbb{R}^n## as ##S^n-\{N\}## we can view ##\omega## as a top form on ##S^n## that vanishes in a neighborhood of the north pole and ##f## as a smooth map ##S^n\to S^n## that fixes the north pole. Since ##f:S^n\to S^n## misses a point it is homotopic to a constant map as ##S^n-\{\text{point}\}## is contractible. So, an application of Stokes' theorem gives
##\int_{\mathbb{R}^n} f^*\omega=\int_{S^n} f^*\omega=\int_{S^n}(\text{constant map})^*\omega=0.##

Stokes' theorem is used to say that integrating pullbacks of a closed form by homotopic maps on a closed manifold give the same answer. This is seen by considering a homotopy ##H(x,t)## and applying Stokes' theorem to the integral ##\int_{S^n\times [0,1]} d(H^*\omega).##
 
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