# Existence of homeomorphism between $[0,1]$ and parabola

Does a homeomorphism $$f:([0,1],d_E)\to\left(\left\{\left(x,x^2\right):x\in[0,2]\right\},d_E\right)$$ exist?

I suppose that it exists. Of course, the second set is a parabola in $\mathbb{R}^2$. But I don't know how to find a function which will be a bijection.

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First stretch $[0,1]$ to the interval $[0,2]$, then map $x$ to $(x,x^2)$

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Ok so let $f: [0,1] \rightarrow [0,2]$ and $g: [0,2] \rightarrow \left\{ (x,x^2): x \in [0,2]\right\}$ . Of course $g \circ f$ is our homeomorphism. Thanks a lot for your help. – Thomas Nov 14 '13 at 23:08

Consider $t\mapsto 2t$ and $2t\mapsto (2t,(2t)^2)$.

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You can use:

$$h(x)= -(x^{1/2}); x<0$$ and $$h(x)=x^{1/2}; x\geq 0$$

Edit: I thought it was a homeo between $\mathbb R$ and the open segment.

For a homeo as you asked (one of these days I will answer the actual question asked!.), take the square root and then multiply by $1/2$. Both are homeomorphisms: the square root is a monotone decreasing (order-preserving) bijection ( in the given domain; and so is its inverse, $x\rightarrow x^2$), and multiplication/scaling is a homeomorphism.

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The domain should be $[0,1]$. – Sigur Nov 14 '13 at 23:35
@Sigur: I just did an edit. – user99680 Nov 14 '13 at 23:40