# Every smooth $n$-manifold is diffeomorphic to a properly embedded submanifold of $\mathbb{R}^{2n+1}$.

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.

• 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. Commented Jul 8, 2021 at 5:45
• @Masacroso Consider the inclusion map $(-1,1)\ni x\longmapsto (x,0,0)\in \Bbb R^3$ Commented Jul 8, 2021 at 5:49
• @Masacroso A map is proper if preimages of compact sets are compact. Commented Jul 9, 2021 at 8:52
• @Sumanta You should give an official answer to clear the question from the "unanswered" queue. Commented Jul 9, 2021 at 8:55

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.
• 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. Commented Jul 9, 2021 at 13:10