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