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I need to know whether There exists any continuous onto map from $(0,1)\to (0,1]$

could any one give me any hint?

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Hint: $(0,1) = (0,\frac 12] \cup [\frac 12, 1)$. Can you map each part onto $(0,1]$?

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    $\begingroup$ hint: $\times 2$ $\endgroup$ – Dan Rust Jul 28 '13 at 12:36
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    $\begingroup$ Taxi: you have no idea how to send $(0,\frac12]$ continuously to $(0,1]$? Really no idea? $\endgroup$ – Did Jul 28 '13 at 12:37
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    $\begingroup$ @TaxiDriver Yes, you do. $\endgroup$ – Git Gud Jul 28 '13 at 12:42
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    $\begingroup$ @TaxiDriver ... and now: can you map $[\frac 12, 1)$ onto $(0,1]$? $\endgroup$ – martini Jul 28 '13 at 12:42
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    $\begingroup$ @Taxi Driver no, you're joking! $\endgroup$ – W_D Jul 28 '13 at 12:43
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Find a polynomial that:

  1. Crosses the x-axis at $x=0$ and $x=1$.
  2. Has an absolute maximum of $f(x)=1$.

$$f(x)=-4(x^2-x),x\in(0,1)$$

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From The Hint of Martini the Map $f(x)=2x; x\in (0,{1\over 2}]$ and $f(x)=1;x\in [{1\over 2},1)$ will work

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$f(x)=|\sin (\pi x)|$ will work

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