TITLE: Is there a continuous bijection from $\mathbb{R}$ to $\mathbb{R}^2$
QUESTION [34 upvotes]: I need a hint. The problem is: is there a continuous bijection from $\mathbb{R}$ to $\mathbb{R}^2$
I'm pretty sure that there aren't any, but so far I couldn't find the proof.
My best idea so far is to consider $f' = f|_{\mathbb{R}-\{*\}}: \mathbb{R} - \{*\} \to \mathbb{R}^2 - \{f(*)\}$, and then examine the de Rham cohomologies: $$H^1_{dR}(\mathbb{R}^2 - \{f(*)\}) = \mathbb{R} \  \xrightarrow{H^1_{dR}(f')} \ 0 = H^1_{dR}(\mathbb{R} - \{*\}),$$ but so far I failed to derive a contradiction here. Am I on the right path? Is it possible to complete the proof in this way e.g. by proving that $H^1_{dR}(f')$ must be a mono? Or is there another approach that I missed?

REPLY [2 votes]: From the Wikipedia Space-filling curves page:

A non-self-intersecting continuous curve cannot fill the unit square
  because that will make the curve a homeomorphism from the unit
  interval onto the unit square (any continuous bijection from a compact
  space onto a Hausdorff space is a homeomorphism). But a unit square
  has no cut-point, and so cannot be homeomorphic to the unit interval,
  in which all points except the endpoints are cut-points.