$(,) = x^2$ How to determine if injective or surjective?

Question

I just learned to determine wether $f(x)=x^2$ is injective or surjective but how can I solve this when there are two variables? I've been struggling to get my head around this. I know that for a function to be injective the following condition must be met: $()≠()$ implies $≠$, or the contrapositive. Surjective: $∀ ∈ , ∃ ∈ \text{ such that } f(x) = y$.

How could I scale these definitions to handle two variables? $(,) = x^2$ mapping from $\Bbb R \times \Bbb R$ to $\Bbb R$

Edit: I want to thank you all for your time and help. It's greatly appreciated! I can clearly see through these problems now.

Answer 1

Let $f$ be the following function \begin{align} f \colon \mathbb{R} \times \mathbb{R} & \to \mathbb{R}\\ (x,y) & \mapsto x^2 \end{align} for all $(x,y) \in \mathbb{R}^2.$

Problem. Is it $f$ injective? Is it $f$ surjective?

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Sketch of the proof. Given a function $f$ we want to be able to show if $f$ is or not injective and if $f$ is or not surjective.

First, let’s recall a couple of definitions.

What does it mean to say that $f$ is injective?

Definition. Let $A$ and $B$ be any sets and $f:A \to B$ a function. We say that $f$ is injective if $$\forall x, y \in A , f(x) = f(y) \implies x = y. \tag{1}$$

And what does it mean to say that $f$ is surjective?

Definition. Let $A$ and $B$ be any sets and $f:A \to B$ be a function. We say that $f$ is surjective if $$\forall b \in B, \exists a \in A \colon f(a) = b. \tag{2}$$

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Solving. Let’s start by verifying if $f$ is or not injective.

Let $0,1 \in \mathbb{R}.$ Then $(0,0),(0,1) \in \mathbb{R}^2.$ Since $0 \neq 1,$ then $(0,0) \neq (0,1).$ By definition of $f$ It follows that $$f(0,0) = 0 ^2 = 0 = 0^2 = f(0,1).$$ Hence, $$\exists x,y,z,w \in \mathbb{R} \colon f(x,y) = f(z,w) \wedge (x,y) \neq (z,w).$$ But this is precisely the negation of $(1).$ So $f$ does not satisfies the condition $(1).$ Therefore, $f$ is not injective.

Now, let’s see if $f$ is or not surjective.

Since the domain of $f$ is $\mathbb{R}^2,$ note that for all $(x,y) \in \mathbb{R}^2$ we have that $x \in \mathbb{R}$ and, hence, $x^2 \geq 0.$ Then, $f(x,y) \geq 0$ for all $(x,y) \in \mathbb{R}^2.$ Which means that $\not\exists (x,y) \in \mathbb{R}^2 \colon f(x,y) < 0.$

This tells us that the condition $(2)$ is not satisfied by this function $f.$ Therefore, $f$ is not surjective.

(Also, note that $f$ is not injective.)

Answer 2

Hints: If $f$ is injective, $f(x,y)=f(x',y') \implies x=x',y=y'$. What are $f(-1,0)$ and $f(1,0)$?

For surjective, is there a point $(x,y)\in\mathbb{R}^2$ such that $f(x,y)=-1$?

Answer 3

Take a deep breath and do it literally.

If $f(x_1, y_1) = x_1^2 = M$ and $f(x_2, y_2) = x_2^2 = M$. so $x_1^2 = x_2^2$ does that mean $x_1 = x_2$ and $y_1 = y_2$? If yes then it is injective (so explain why). If no it is not injective (if not explain why not and/or give a counter example).

So does $x_1^2 = x_2^2$ mean $x_1 = x_2$ and $y_1 = y_2$? the answer is of course not. As the $y_k$ have nothing to do with anything they can be any thing. A counter example if $f(2, 9) = 2^2 =4$ and $f(2, \frac {\log_7 \sqrt \pi}{e^{2.7}}) = 2^2 = 4 = f(2,9)$. But $(2,9) \ne(2, \frac {\log_7 \sqrt \pi}{e^{2.7}})$ So it is not injective.

Not to mention $f(x,y) = x^2 = (-x)^2 = f(-x,tony\ the\ tiger)$ and $x\ne -x$ (if $x \ne 0$)

so not injective.

Surjective.

If $w\in \mathbb R$ then is there always as $(x,y)\in \mathbb R^2$ so that $f(x,y) = x^2 = w$.

If we try to solve for $x^2 = w$ we get that $x = \pm \sqrt w$ and $y =\text{any dang thing we want}$ will be a solution but does $(\pm \sqrt w,\text{any dang thing we want})$ always exist. Well, $\text{any dang thing we want}$ always exists, but if $w < $ then $\pm \sqrt w$ does not.

So it is not surjective. A counter example is $f(x,y) = -7$ has not solution as there are no $(x,y)$ where $x^2 = -7$ exist.