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I am curious about two questions below

Let $M$, $N$ be two topological manifold.

  • If $\dim M>\dim N$, is there exist an injective continuous map $f: M\rightarrow N$?

  • If $\dim M<\dim N$, is there exist an surjective continuous map $f: M\rightarrow N$?

  • To the first question, I think we can construct a map $F: M\rightarrow N\times\mathbb R^{m-n}$, which satisfies $F(p)=(p,0)$. If $f$ is injective, then $F$ is injective too. Then from invariance of domain, we can get $F$ is an embedding. Also, $f$ is an embedding. So is there exist embedding from $M$ to $N$?

  • Second question can be transfered $N$ as $\mathbb R^n$ by using local coordination. Then I have know idea.

Any advice is helpful. Thank you.

gaoxinge
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1 Answers1

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For the first question, the answer is no, there is no continuous injective map from a larger manifold to a smaller one.

Suppose there existed such an $f$. Then restricting $f$ to a chart $U\subseteq M$ we get an injective map $f|_U:U\rightarrow N$. By further restricting $U$, we may assume $f(U)$ is a subset of a chart $V$ on $N$.

Thus, by composing with chart maps (which are homeomorphisms, and therefore injective), we may assume $f$ maps an open subset of $\mathbb{R}^m$ to a subset of $\mathbb{R}^n$ (with $m = \dim M$ and $N = \dim n$). Abusing notation, I'll still write $U$ for this open subset of $\mathbb{R}^m$.

Now, consider the map $g:U\subseteq \mathbb{R}^m$ to $\mathbb{R}^m$ given by $g(x) = (f(x), \vec{0})$ with $\vec{0}\in\mathbb{R}^{m-n}$. This will also be injective with image a subset of $\mathbb{R}^n\times \{\vec{0}\}$. By Invariance of Domain, $g(U)$ is an open subset of $\mathbb{R}^m$. But every open subset which contains a point of the form $(f(x), 0)$ also contains a point of the form $(f(x), v)$ for some tiny $v\in \mathbb{R}^{m-n}$. This contradiction implies no such $f$ exists.

For the second question, yes, there are examples. Perhaps the most famous are the Peano curves. With some technical details sorted out, I believe one can prove that every (connected) topological manifold is the continuous image of $\mathbb{R}$.

Jason DeVito - on hiatus
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    For the second question, is everything is smooth the answer changes: There are no smooth surjective maps from a smaller dimensional smooth manifold to a larger dimensional smooth manifold. – Jason DeVito - on hiatus Nov 03 '14 at 18:32
  • Nice, and thanks. En, there is a little mistake: if a topological manifold is compact, there does not exist a surjective continuous map between it and $\mathbb R$. – gaoxinge Nov 04 '14 at 08:35
  • I think there is still a surjective continuous map with domain $\mathbb{R}$. For example, there is a surjective continuous map from $\mathbb{R}$ to $\mathbb{R}^2$, obtained by iterating the usual Peano curve. There is an easy surjective map $\mathbb{R}^2\rightarrow S^2$ given by wrapping a plane around $S^2$ infinitely many times. The composition gives a surjective map from $\mathbb{R}$ to $S^2$. (More generally, one can iterate this to find a surjective continuous map $\mathbb{R}\rightarrow \mathbb{R}^n$. Then, after choosing a complete Riemannian metric, the exponential map gives – Jason DeVito - on hiatus Nov 04 '14 at 15:36
  • a surjective map from $\mathbb{R}^n$ to whatever manifold you want. The composition is a surjective continuous map from $\mathbb{R}$ to whatever manifold you want. (You're right that a compact manifold can't surject onto a non-compact manifold.) – Jason DeVito - on hiatus Nov 04 '14 at 15:38