Continuous injection from $\mathbb{R}^2$ to $\mathbb{R}^2$.

Question

Could someone give me an example of a continuous injection from $\mathbb{R}^2$ to $\mathbb{R}^2$ which does not have a continuous inverse.

It could be interesting to hear your solution. The related threads give, for example, a continuous injection from a half-open interval $\left[ a,b \right)$ to $\mathbb{R}^2$ (can use circle $\mathbb{S}^1 \subseteq \mathbb{R}^2$ as range) whose inverse is discontinuous, but no mention of a mapping like the one you ask for. — Jeppe Stig Nielsen, Jan 29 '14 at 08:22
what about some mapping, which doesn't have inversion at all? — V-X, Jan 29 '14 at 08:42

score 0 · Accepted Answer · answered Jan 29 '14 at 08:36

The theorem invariance of domain as quoted in the linked article, proves that your $f: \mathbb{R}^2 \to \mathbb{R}^2$ is a homeomorphism onto its image, i.e. if we "restrict" the codomain, $$\tilde{f}: \mathbb{R}^2 \to f(\mathbb{R}^2)$$ then the inverse of $\tilde{f}$ is continuous.

Continuous injection from $\mathbb{R}^2$ to $\mathbb{R}^2$.

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