Question

hi guys how to show that a function is bijective like this one

Answer 1

typically one shows it is both injective and surjective

Answer 2

Let S denotes the unit circle in the plane
\[S=\left\{ \left( x,y \right) \epsilon \mathbb{R} ^{2} : x ^{2}+y ^{2}=1\right\}\]
Define \[h:[0,2\pi) \rightarrow S \\ h( \theta) = (\cos \theta , \sin )\]

Answer 3

ooops sin (theta)

Answer 4

use \in not \epsilon to denote set membership

Answer 5

anyways, to show it is injective, observe for any \(\alpha,\beta\), if we assume \(h(\alpha)=h(\beta)\) we must prove \(\alpha=\beta\)$$h(\alpha)=h(\beta)\\\cos\alpha=\cos\beta\\\sin\alpha=\sin\beta$$Note from the first equation we have that \(\alpha=\beta,\pi-\beta\) and teh second we have \(\beta=\alpha,2\pi-\alpha\) and so it follows that \(\alpha=\beta\)

Answer 6

for surjective, just show that for all \((x,y)\in S\) we have some \(\theta\) s.t. \(h(\theta)=(x,y)\)...

Answer 7

what is \[\pi - \beta \] and \[2\pi - \alpha \]

Answer 8

@jacobian for \(\alpha,\beta\in[0,2\pi)\) we have \(\cos\alpha=\cos\beta\) only if: