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The terminology involved in this post comes from the book on smooth manifolds written by John M. Lee. By the Whitney embedding theorem, every smooth $n$-manifold $M$ admits a proper smooth embedding into $\mathbb{R}^{2n+1}$. And Professor Lee then concludes that $M$ is diffeomorphic to a properly embedded submanifold of $\mathbb{R}^{2n+1}$. I have no question about why $M$ is diffeomorphic to an embedded submanifold of $\mathbb{R}^{2n+1}$, which is just a direct application of Whitney's theorem. But somehow I can't understand the reason why the submanifold is properly embedded. In order for this to come true, one must be able to show that the inclusion map $\iota:F(M)\hookrightarrow\mathbb{R}^{2n+1}$ is a proper map, where $F$ is the diffeomorphism. Does anyone have an idea? Thank you.

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    $\begingroup$ Consider $F\colon M\hookrightarrow \Bbb R^{2n+1}$ the smooth embedding with $F$ proper, and the inclusion $\iota\colon F(M)\hookrightarrow \Bbb R^{2n+1}$. Let $\varphi\colon M\xrightarrow{\simeq} F(M)$ be the diffeomorphism obtained from $F$ restricting the codomain. So, $F=\iota\circ\varphi\implies F\circ\varphi^{-1}=\iota$, i.e., $\iota$ is a composition of two proper maps, hence itself proper. $\endgroup$
    – Sumanta
    Commented Jul 8, 2021 at 5:45
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    $\begingroup$ @Masacroso Consider the inclusion map $(-1,1)\ni x\longmapsto (x,0,0)\in \Bbb R^3$ $\endgroup$
    – Sumanta
    Commented Jul 8, 2021 at 5:49
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    $\begingroup$ @Masacroso A map is proper if preimages of compact sets are compact. $\endgroup$
    – Paul Frost
    Commented Jul 9, 2021 at 8:52
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    $\begingroup$ @Sumanta You should give an official answer to clear the question from the "unanswered" queue. $\endgroup$
    – Paul Frost
    Commented Jul 9, 2021 at 8:55

1 Answer 1

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Consider a smooth embedding $F:M\hookrightarrow \Bbb R^{2n+1}$ with $F$ proper, and the inclusion $\iota:F(M)\hookrightarrow \Bbb R^{2n+1}$. Let $φ:M\xrightarrow{≃}F(M)$ be the diffeomorphism obtained from $F$ restricting the codomain. So, $F=\iota\circ φ$ i.e., $F\circ φ^{-1}=\iota$, i.e., $\iota$ is a composition of two proper maps, hence itself proper.

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  • $\begingroup$ Is it easy to show that a composition of two proper maps is still proper? $\endgroup$
    – Boar
    Commented Jul 9, 2021 at 12:43
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    $\begingroup$ Let $f\colon X\to Y$ and $g\colon Y\to Z$ be two proper maps. Then, $(g\circ f)^{-1}(\text{compact}_1)=f^{-1}\big(g^{-1}(\text{compact}_1)\big)=f^{-1}(\text{compact}_2)=\text{compact}_3$. Here, $\text{compact}_1,\text{compact}_2,\text{compact}_3$ are compact sets. $\endgroup$
    – Sumanta
    Commented Jul 9, 2021 at 13:10
  • $\begingroup$ The first equality is exactly what I need. Actually, that holds for any set. Right? $\endgroup$
    – Boar
    Commented Jul 9, 2021 at 13:46
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    $\begingroup$ Yup, See here math.stackexchange.com/questions/1606868/… $\endgroup$
    – Sumanta
    Commented Jul 9, 2021 at 14:03

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