how to prove that $x^2 + y^2 =1$ is injective and surjective depending on the restrictions?

Question

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 $\sqrt{1-x^2} = \sqrt{1-y^2}$?

Answer 1

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.

Answer 2

An injective mapping $f$ would satisfy the property that if $f(a) = f(b)$, then $a = b$. This is obviously not the case for your mapping, defined by $$y = \sqrt{1-x^2},$$ where $\sqrt{\cdot}$ is the nonnegative square root. For if $y = 3/5$, then $x^2 = 1 - (3/5)^2 = (4/5)^2$, hence $x = \pm 4/5$. The mapping thus admits a case where $a \ne b$ but $f(a) = f(b)$, contradiction.

Surjectivity, however, is a property of the specification of the codomain: it is implied by the definition of $S$ that your mapping is $$f : [-1,1] \to [0,1],$$ hence the codomain is $[0,1]$. This does satisfy surjectivity: if $y \in [0,1]$, there is at least one $x \in [-1,1]$ such that $x^2 + y^2 = 1$. To see this, we observe that $x = \sqrt{1-y^2}$, so that $0 \le y \le 1$ implies $0 \ge -y^2 \ge -1$ and $1 \ge 1-y^2 \ge 0$. Then taking the square root (either sign works), we are guaranteed to find such an $x \in [-1,1]$.

Had the codomain been, say, $[0,\infty)$, then this would fail to be a surjective map, since $y = 2$ has no preimage in $[-1,1]$.