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Question. Does there exist an injective continuous map $f:\mathbf R^3\to \mathbf R^2$?

I am not able to settle even the following simpler version:

Does there exists a bijective continuous map $f:\mathbf R^3\to\mathbf R^2$?

I know that $f$ cannot be a homeomorphism since if it were then we get a homeomorphism $\mathbf R^3-\{\mathbf 0\}\to \mathbf R^2-\{\mathbf 0\}$. Such a homeomophism does not exist since the second homology groups of these spaces are different.

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Here is a different proof. Consider $f : \mathbb{R}^3 \to \mathbb{R}^2$ as a map $g : \mathbb{R}^3 \to \mathbb{R}^3$ by identifying $\mathbb{R}^2$ with $\mathbb{R}^2 \times \{0\}$. By the Invariance of domain, the image $U = g(\mathbb{R}^3) \subset \mathbb{R}^3$ would have to be open, but it is included in $\mathbb{R}^2 \times \{0\}$ (and of course it's nonempty). This is impossible.

More generally this shows $\mathbb{R}^n$ cannot embed in $\mathbb{R}^m$ for $n > m$, with exactly the same proof (identify $\mathbb{R}^m$ with $\mathbb{R}^m \times \{0\}^{n-m}$).

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    $\begingroup$ Amazingly this question doesn't seem to be a duplicate... $\endgroup$ – Najib Idrissi Nov 2 '15 at 10:36
  • $\begingroup$ Thanks. I just read Theorem 2.26 from Hatcher's Algebraic Topology where the author proves inexistence of a homeomorphism using local homology. So it was a natural question to ask if we can weaken the assumption of homeomorphism to injection. $\endgroup$ – caffeinemachine Nov 2 '15 at 10:56
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If $f:\mathbb R^3\to\mathbb R^2$ is injective and continuous, then the restriction of $f$ on $S^2$ is injective and continuous. But this cannot happen, from the Borsuk-Ulam theorem.

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  • $\begingroup$ That's great. Thanks. $\endgroup$ – caffeinemachine Nov 2 '15 at 10:56

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