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Proper functions are defined over topological spaces.

For functions $f:\mathbb{R}^d\rightarrow \mathbb{R}^d$ the conditions "simplify" to $\lim_{x\rightarrow \infty}|f(x)|=\infty$.

Are similar "simplications" know for functions over the n-sphere $S^{d}=\{x\in\mathbb{R}^{d+1} \mid \|x\|=1\}$, i.e., $f:S^{d}\rightarrow S^{d}$?

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Alexander Mathiasen
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1 Answers1

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A function $f$ from $\mathbb{S}^d$ to itself is proper iff it is continuous.

Note that $\mathbb{S}^d$ is compact and so every closed subset of it is compact and viceversa

  • Continuous$\rightarrow$ proper. Every continuous function $f:\mathbb{S}^d\to \mathbb{S}^d$ is proper, since $f^{-1}$ sends closed (=compact) sets to closed (=compact) sets.

  • Proper $\rightarrow$ continuous. Every proper function $f:\mathbb{S}^d\to \mathbb{S}^d$ is such that $f^{-1}$ sends compact (=closed) sets to compact (=closed) sets. This is one of the possible definition of continuity

The result is true in a more general form: given an Hausdorff compact space $X$, $f:X\to X$ is proper iff it is continuous,