On the existence of a continuous bijection $f\colon [0,1]\to [0,1]\times [0,1]$ Let $f$ be a continuous function on $[0,1]$ such that $f([0,1])=[0,1]\times[0,1].$ Then show that $f$ is not one-one.
Hints will be appreciated.
 A: Assume $f:K\rightarrow X$ where $K$ is a compact topological space, $X$ is a topological Hausdorff space and $f$ is continuous then for any closed set $F\subseteq K$ we have that $F$ is compact and hence $f(F)$ is compact in $X$ and hence closed in $X$. 
Now if $f$ is one-to-one and onto this shows that $f^{-1}$ is continuous. And hence an homeomorphism. 
Hint : apply this to your $f$ with $K=[0,1]$ and $X=[0,1]\times [0,1]$. You will get that $f$ is a homeomorphism between $K$ and $X$, finally you just have to show that it is not possible ($K-\{0.5\}$ and $X-\{f(0.5)\}$ should be homeomorphic as well).
A: Hint: For each $x\in [0,1]^2$, 
$$
[0,1]^2 - \{x\} 
$$ is connected, but
$[0,1] - \{1/2\}$ is not.
A: Assume to the contrary that $f\colon [0,1]\rightarrow [0,1]\times [0,1]$ is a continuous bijection. Then its inverse is also continuous, i.e. it is a homeomorphism. Suppose that $f(\frac{1}{2})=(x,y)$.
Then $f$ induces a homeomorphism $g\colon [0,1]\setminus \{\frac{1}{2}\}\to ([0,1]\times[0,1])\setminus \{(x,y)\}$, but the domain is not connected, while the image is connected, which is a contradiction.
A: Hint: The space $[0,1]^2$ contains non-trivial closed jordan-curves. However, $[0,1]$ does not.
