Is there a one-one onto continuous function $f:[0,1]\rightarrow[0,1]^2$? Is there a one-one onto continuous function $f:[0,1]\rightarrow[0,1]^2$? 
I was trying to prove that there is no such function, but failed.
Any suggestions?
 A: If such a function existed, it would have a continuous inverse. Because the preimage of a closed set under the inverse would just be the forward image of a closed set under the original function. Now every closed subset of $[0,1]$ is compact, so it will be mapped by a continuous function to a compact, and hence closed, subset of $[0,1]^2$.
So it suffices to show that no continuous function $f:[0,1]^2\to[0,1]$ is injective. For this, you can find two different paths in $[0,1]^2$ that only agree at the endpoints and use the intermediate value theorem to show that $f$ is not injective.
A: Here's an alternate solution using properties of connected sets: 
Suppose for contradiction such a function $\varphi$ existed, then it would have a continuous inverse (cf. baby Rudin 4.17) $$\varphi^{-1} : [0,1]^2 \rightarrow [0,1]$$ Now, $[0,1]^2 \setminus\{\varphi(1/2)\}$ is connected (since $\varphi$ is a bijection we are just removing one point from the square) and by properties of continuous functions we know that $\varphi^{-1}$ maps connected sets to connected sets (cf. baby Rudin 4.22) . But $$\varphi^{-1}([0,1]^2 \setminus\{\varphi(1/2)\}) = [0, 1/2)\cup(1/2, 1]$$ which is not connected, so we have a contradiction and hence no such map can exist. 
A: Use the generalized Implicit Function Theorem. This is the sort of problem it should be useful for.
"Various forms of the implicit function theorem exist for the case when the function f is not differentiable..."
http://en.wikipedia.org/wiki/Implicit_function_theorem
