Yesterday in a discussion in Komero we concluded that if f is an injective (one-to-one) map from the unit interval [0,1] to the Euclidean plane (or [tex]\mathbb{R}^n[/tex]), then the interval is homeomorphic with its image. The proof is as follows: we already know that f is continuous and one-to-one, so it remains to show that its inverse is continuous. Continuity is equivalent with the inverse images of closed sets being closed, so the question reduces to the question whether f is closed (maps closed sets to closed). But any closed subset of the unit interval is compact and the image of a compact set is compact and hence closed and we are done.

However, then I carelessly remarked that probably this extends from the interval to whole real line by representing the real line as a counhtable union of compat intervals. This was of course wrong:

**Exersice 1.** Find a contionuous injective (one-to-one) map from the real line to the plane

[tex]f\colon\mathbb{R}\to\mathbb{R}^2[/tex]

such that [tex]\mathbb{R}[/tex] is not homeomorphic to [tex]f[\mathbb{R}][/tex]

**Exersice 2.** As above but with the requirement that the fundamental group of [tex]f[\mathbb{R}][/tex] is trivial.