# Manifold with injective continuous map into $\mathbb R^k$ admits embedding into $\mathbb R^{k+1}$

This is Problem 4-34 from John Lee's book Introduction to Topological Manifolds:

Suppose $$M$$ is an $$n$$-manifold that admits an injective continuous map into $$\mathbb R^k$$ for some $$k$$. Show that $$M$$ admits a proper embedding into $$\mathbb R^{k+1}$$. [Hint: use an exhaustion function]

My first thought was to let $$f:M\to\mathbb R^k$$ be the injective continuous map and let $$g:M\to\mathbb R$$ be an exhaustion function, then to map $$x$$ to $$(f(x),g(x))$$, but this isn't a homeomorphism.

In the book, it is proved that every compact manifold is homeomorphic to a subset of Euclidean space. Because of this, I was thinking that, since $$g^{-1}((-\infty,c])$$ is compact for every $$c$$, maybe I could consider for each $$c$$ a homeomorphism $$h_c:g^{-1}((-\infty,c])\to\mathbb R^{k_c}$$. However, I wasn't able to show that this leads to a proper embedding into $$\mathbb R^{k+1}$$.

I guess I'm not quite sure where to start with this since I'm not very familiar with how to use exhaustion functions/why they are useful. Also, I don't know anything about immersions or smooth manifolds, so I wasn't able to understand the stuff that I found online (which talked about the Whitney immersion theorem).

Thanks!

• See the book : Introduction to Smooth Manifolds, Theorem 6.12 (Whitney Embedding Theorem)-John M. Lee – Jul 7, 2019 at 18:18
• @AllanRamos Thanks! I guess I was asking more specifically for a proof of this weaker case that avoids using this theorem, but I will definitely take a look to see if I might be able to use some of the ideas behind the proof for this problem. Jul 7, 2019 at 18:22
• I understand you. I think you will understand the demonstration, it is done step by step using the previous results. Jul 7, 2019 at 18:26
• @ hiabc . You're welcome.Good Look. Jul 7, 2019 at 18:37
• Actually, the formula $x\mapsto (f(x), g(x))$ yields a proper (since $g(x)\to\infty$ as $x\to \infty$ in $M$) injective continuous map $M\to R^{k+1}$, hence, a homeomorphism to its image. Jul 7, 2019 at 19:34

Moishe Kohan has answered your question in his comment. However, let us first note that an injective continuous map $$f : M \to \mathbb R^k$$ is not necessarily an embedding (i.e. a homeomorphism onto $$f(M)$$). As an example take $$f : M = (-1,2\pi) \to \mathbb R^2, f(x) = (\cos x,\sin x)$$ for $$x \ge 0$$ and $$f(x) = (1,-x)$$ for $$x \le 0$$. The set $$C = [\pi,2\pi)$$ is closed in $$M$$, but $$f(C)$$ is not closed in $$f(M)$$. This is a phenomenon which is possible only for non-compact $$M$$. In fact, if $$M$$ is compact, then $$f$$ is a closed map.
If $$g : M \to \mathbb R$$ is an exhaustion function, then $$F : M \to \mathbb R^{k+1}, F(x) = (f(x),g(x))$$, is an embedding. It suffices to show that $$F$$ is a closed map. So let $$C \subset M$$ be closed and let $$(y,t)$$ be contained in the closure of $$F(C)$$. Hence there exists a sequence $$(x_n)$$ in $$M$$ such that $$F(x_n) \to (y,t)$$, i.e. $$f(x_n) \to y$$ and $$g(x_n) \to t$$. W.l.o.g. we may assume that $$g(x_n) < t + 1$$ for all $$n$$. Hence $$(x_n)$$ is a sequence in $$K = g^{-1}((-\infty, t+1])$$ which is compact. Thus it has a convergent subsequence $$(x_{n_k})$$ with limit $$x \in K$$. Since $$C$$ is closed, we conclude $$x \in C$$. Thus $$f(x_{n_k}) \to f(x)$$ and therefore $$f(x) = y$$ by the uniqueness of limits. Hence $$(y,t) \in f(C)$$.