Let $A = \{(x,y) : x^2+y^2 = 1\}$ and let $f : A \to \Bbb R$ be a continuous function. Then prove or disprove the following: 1. $f$ is one to one. .

Question

Let $A = \{(x,y) : x^2+y^2 = 1\}$ and let $f : A \to \Bbb R$ be a continuous function. Then prove or disprove the following:

$f$ is one to one.
$f$ is onto.

If I take $f(x,y) = e^{x+y}$ which is continuous function but $\nexists x,y$ s.t. $f(x,y) =0 \in \Bbb R$ so $f$ is not onto.

We know that if $f$ is one to one then $f(x_1,y_1) = f(x_2,y_2) \implies x = y$, where $x = (x_1,x_2), y = (y_1,y_2)$. Also if $f$ is onto then for each $y \in \Bbb R, \exists x \in A$ such that $f(x) = y$ But I have no idea how to use conditions of one to one and onto function. Help me.

$f(x, y) = 0$ is not one-to-one and not onto. – Martin R Jun 28 '21 at 12:46 — Martin R, Jun 28 '21 at 12:46
This question is very similar. Perhaps it will help you. – TonyK Jun 28 '21 at 12:48 — TonyK, Jun 28 '21 at 12:48

score 2 · Answer 1 · answered Jun 28 '21 at 12:47

$f$ cannot be surjective since $A$ is compact and its image by a continuous function is compact.

The image of $f$ is a closed interval $[a,b]$ since $A$ is connected.

Let $x: f(x)=a, f(y)=b$. $A$ is the union of two connected arcs $C_1$, $C_2$ whose ends are $x,y$. The image of the restriction of $f$ to $C_1$ is the image of the restriction of $f$ to $C_2$, $f$ cannot be one to one.

Let $A = \{(x,y) : x^2+y^2 = 1\}$ and let $f : A \to \Bbb R$ be a continuous function. Then prove or disprove the following: 1. $f$ is one to one. .

1 Answers1