Suppose we have
$S=\{(x,y) \in [-1,1]\times[0,1]: x^2 + y^2 = 1\}$
I know this is a function since the domain(s)= $[-1,1]$ and I know this should be surjective and injective since the restriction makes the image out to be a top half circle, but I am having trouble proving that this function is a injective and onto function.
For injective I am usually working with just $X$ variables for example if I want to prove that
$x+3$ is injective for all real numbers then I just let $x+3=y+3$ and calculate $x=y$ since the definition of injective is if $f(x)=f(y)$ then $x=y$ but i get stumped with this equation of the circle.
Should I just set $x^2 + y^2 =1$ to $y=\sqrt{1-x^2}$ and set
$\sqrt{1-x^2} = \sqrt{1-y^2}$?
Best Answer
You need to be much more careful with your function/relation.
The set of points satisfying $x^2 + y^2 = 1$ is a relation; in order to say something is injective or surjective, you must be very precise specifying the domain, the range, and in particular, you need a function.
If it's the top half of a circle, the equation $f(x) = \sqrt{1 - x^2} : [-1, 1] \to [0, 1]$ will work, but the reason you have a function is not because the domain is $x \in [-1, 1]$.
At any rate, you're going to have a hard time showing that distinct $x$'s get sent to distinct $y$'s; that $f(x_1) = f(x_2)$ implies $x_1 = x_2$. Think geometrically.