Establish a bijective function

Question

How do I establish a bijection from $[1,0]$ to $$[1,0]\times [1,0]$$ that is continuous?

I have not been able to succeed.

Edit Sorry, it's $[0,1]$ in all cases.

Answer 1

Since $[0,1]$ is compact, such a function $f$ would yield a homeomorphism of $[0,1]$ onto $[0,1]^2$.

Now remove $1/2$ from $[0,1]$. You would get that $[0,1/2)\cup(1/2,1]$ is homeomorphic to $[0,1]^2\setminus\{f(1/2)\}$.

Note that the latter is connected, while $[0,1/2)\cup(1/2,1]$ is not.

Such a function can't exist.

Answer 2

If by $[1,0]$ you mean $[0,1]$ then there is no such function.

Indeed if there was, then the reciprocal function $g=f^{-1}$ would be continuous since $[0,1]^2$ is compact, so that it would be a homeomorphism. To see that this isn't possible, remark that if you remove an interior point of the square $[0,1]^2$, it remains connected ("in one piece"), whereas if you remove an interior point of the line segment $[0,1]$, it becomes disconnected.

(more formally, the contradiction comes from the fact the continuous image by $g$ of the connected set $[0,1]^2 - \{a\}$ should be connected).

Answer 3

A continuous bijection from a compact set to an Haussdof set is an homeomorphism.

But, there exists a continuous surjective function $[0,1] \to [0,1]^2$ : Peano's curve.