For any injective function $f:\mathbb R \to \mathbb R^n$, is it true that $f(\mathbb R)$ is contractible?
I can show that for any Hausdorff space $X$ and injective function $f:[0,1] \to X$, $f([0,1])$ is contractible. But I am unable to figure out what happens, even just only for euclidean space as co-domain, if we take the domain to be an open interval. Please help. Thanks in advance.
If $f$ is a continuous map $X \to Y$, and either $X$ or $Y$ is contractible, then $f$ is nullhomotopic, i.e. homotopic to a constant map. So $f : \mathbb{R} \to \mathbb{R}^n$ is nullhomotopic (the injectivity hypothesis is not necessary here).
However, the image need not be contractible. For example, consider the map $g : (-1.1, 1) \to \mathbb{R}^2$ given by $g(t) = (t^2 - 1, t^3 - t)$. This is an injective map with image given by the blue curve below (taken from this Wikipedia article).
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Let $h$ be a homeomorphism $\mathbb{R}\to (-1.1, 1)$, then $f := g\circ h : \mathbb{R} \to \mathbb{R}^2$ is an injective continuous map whose image is homotopy equivalent to $S^1$, and hence not contractible.