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Suppose we have $f(x)=(x^2,\sin x)$ defined by $f:\mathbb{R}\rightarrow \mathbb{R} \times \mathbb{R}$. Can we show that this function is onto or one-to-one?

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5  
What have you tried? –  JavaMan Oct 25 '11 at 20:19
3  
Draw a graph... –  lhf Oct 25 '11 at 20:20
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If you stop and think about the function for a little, you should find both questions very easy. In particular, a graph will answer the first one immediately. –  Brian M. Scott Oct 25 '11 at 20:24
    
Does it seem like it should be onto? For the second one, what happens if $x=\pm\pi$? –  mathmath8128 Oct 25 '11 at 20:24

2 Answers 2

up vote 2 down vote accepted

$f$ is not onto: It's $y$ coordinate is limited by $1$, since $\sin (x) \leq \ 1 \ \forall x$.

It is not one-to-one: take $ x = -2\pi $ and $y = 2\pi $. $f(x) = (4\pi^2, \sin(-2\pi))$ $\ =$ $ (4\pi^2, 0)$. $f(y) = (4\pi^2, \sin(2\pi)) = (4\pi^2, 0) = f(x)$

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@Asaf, Thanks for fighting for a good cos. :) [Sorry, couldn't resist it.] –  Srivatsan Oct 25 '11 at 21:01

$((-2 \pi)^2, \sin (-2 \pi)) = ((2 \pi)^2, \sin (2 \pi))$

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