Is there a continuous bijection from $\mathbb{R}$ to $\mathbb{R}^2$ 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?
 A: Here is a hint: What simple space is $\mathbb{R}^2 - \{ f(\ast) \}$ homotopic to?
Edit: Just a small edit, to hopefully bump this up. I had read this as a homeomorphism, in which case it is easy. However we only have a continuous bijection from $\mathbb{R} \to \mathbb{R}^2$. 
There may be a way to argue from the fact that $\mathbb{R} - \{ \ast \}$ is disconnected and $\mathbb{R^2} - \{ f(\ast) \}$ is connected. This will work immediately to show there is no continuous bijection from $\mathbb{R}^2 \to \mathbb{R}$ as the continuous image of a connected set is connected. I am not sure about getting something out of the other direction however (perhaps the 'simplest' is Zarrax's explanation). Hopefully the experts will have something to add!
A: Suppose $f(x)$ were such a function. Note that each $A_n = f([-n,n])$ is a closed (actually compact) set, with $\cup A_n = {\mathbb R}^n$. By the Baire category theorem, there is one such $A_n$ that contains a closed ball $B$. Since $[-n,n]$ is compact, the image of any relatively closed subset of $[-n,n]$ is compact and thus closed. Hence $f^{-1}$ is continuous when restricted to $A_n$, and thus when restricted to $B$. So in particular $f^{-1}(B)$ is a connected subset of ${\mathbb R}$. Since all connected subsets of ${\mathbb R}$ are intervals, $f^{-1}(B)$ is a closed interval $I$. 
Let $x$ be any point in the interior of $B$ such that $f^{-1}(x)$ is not an endpoint of $I$. Then $B - \{x\}$ is still connected, but $f^{-1}(B - \{x\})$ is the union of two disjoint intervals, which is not connected. Since $f^{-1}$ when restricted to $B - \{x\}$ is continuous, you have a contradiction. 
A: 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.

