I have the following problem:

Let M be a (non-compact) manifold and $\{\theta_i\}$ a partition of unity subordinate to a countable open cover of relatively compact sets. Then the map $f=\sum_i i\theta_i$ is a proper map

So far I only know that using $f$ is continous, I can show $f$ is proper if I can somehow show that $f^{-1}([0,N])$ is bounded, $\forall N\in\mathbb{N}$(This makes sense because M is considered as an embedded manifold in some $\mathbb{R}^p$).

Any help is appreciated, I'm quite stuck. Thanks everyone!

  • $\begingroup$ Isn't $f^{-1}([0,N])$ contained in the union of the first $N$ relatively compact sets? $\endgroup$ – Brian Moehring Oct 7 '19 at 19:29
  • $\begingroup$ Not necessarily. Because the value of some $\theta_N$ with big $N$ could positive, but very small $\endgroup$ – miraunpajaro Oct 7 '19 at 19:31
  • $\begingroup$ Yet outside the first $N$ sets in the cover, $$f = \sum_{i=N+1}^\infty i\theta_i \geq (N+1)\sum_{i=N+1}^\infty \theta_i = N+1$$ $\endgroup$ – Brian Moehring Oct 7 '19 at 19:42
  • $\begingroup$ Yeah, that seems allright to me. Thank you very much $\endgroup$ – miraunpajaro Oct 7 '19 at 19:49

Let $\{V_i\}$ denote the given countable open cover by relatively compact sets to which the partition of unity $\{\theta_i\}$ is subordinate. Then we claim $f^{-1}([0,N]) \subseteq \bigcup_{i=1}^N V_i$ for each $N \in \mathbb{N}.$

To see this, first note that if $x\in M,$ $$f(x) = \sum_{i=1}^\infty i\theta_i(x) \geq \sum_{i=N+1}^\infty i\theta_i(x) \geq \sum_{i=N+1}^\infty (N+1)\theta_i(x) = (N+1)\left(1 - \sum_{i=1}^N \theta_i(x)\right)$$

If $x \in M \setminus \bigcup_{i=1}^NV_i,$ then $\theta_i(x) = 0$ for each $i=1,\ldots,N$ and therefore $$f(x) \geq (N+1)\left(1 - \sum_{i=1}^N\theta_i(x)\right) = N+1.$$

The claim follows. Since further $$f^{-1}([0,N]) \subseteq \bigcup_{i=1}^N \overline{V_i}$$ shows that $f^{-1}([0,N])$ is contained in a compact set, it is bounded.


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