Injective, surjective or bijective?: $f: \mathbb{R}^{2} \rightarrow \mathbb{R}^{2}, (x,y) \mapsto (x+y,x+y^{2})$

Question

Is the following mapping injective, surjectiv or bijective? Say why and if necessary, calculate the inverses.

$f: \mathbb{R}^{2} \rightarrow \mathbb{R}^{2}, (x,y) \mapsto (x+y,x+y^{2})$

I have looked for some tasks I don't understand in old exams that have been written years ago and this is one I couldn't solve at all. Can you tell me an easy way of doing it?

Anyway I give it a try, somehow..:

Say we have $x=2$ and $y=2$ then insert it into $f:$

$$(2,2)=(2+2,2+2^{2})$$

$$(2,2)=(4,6)$$

Now let's take $x=1$ and $y=-7$:

$$(1,-7) =(1-7,1+(-7)^{2})$$

$$(1,-7)=(-6,50)$$

Something is injective if for every $(x+y,x+y^{2})$ we get at most one $(x,y)$ value (?)

Something is surjective if for every $(x+y,x+y^{2})$ we get at least one $(x,y)$ value (?)

So I would say this is only surjective.

Answer 1

To prove that something is injective there are three approaches:

• Direct approach: Assume that $f(a,b)=f(c,d)$, prove that it must be the case that $a=c$ and $b=d$.

• Contrapositive approach: Assume that $(a,b)\not=(c,d)$ and prove that $f(a,b)\not=f(c,d)$.

• Contradiction approach: Assume that $(a,b)\not=(c,d)$ and $f(a,b)=f(c,d)$ and derive a contradiction.

For your problem, I suggest going with the direct approach. Therefore, assume that $$ f(a,b)=f(c,d). $$ By substituting our definition, this means that $$ (a+b,a+b^2)=(c+d,c+d^2) $$ Therefore, \begin{align*} a+b&=c+d\\ a+b^2&=c+d^2. \end{align*} Solving the first equation for $c$ gives $c=a+b-d$. Substituting this into the second equation, we get $$ a+b^2=a+b-d+d^2. $$ This simplifies to $$ b^2-b=d^2-d. $$ We now ask if it is possible to find two values for $b$ and $d$ that satisfy this equation. We see that if $b=0$ and $d=1$, the left and right sides of the equality are equal. Now, if we choose $a=0$, it must be that $c=-1$.

Observe that $f(0,0)=(0+0,0+0^2)=(0,0)$ and $f(-1,1)=(-1+1,-1+1^2)=(0,0)$, so we have two different inputs that give the same output, so the function is not injective. Even though we tried to prove that this was injective, our work lead us to an example where f was not injective (two different inputs gave the same output)

To prove that something is surjective

• Take an element $(a,b)$ that is the right form for the output and try to construct the inputs that lead to that output.

In this case, you want to find $(x,y)$ so that $f(x,y)=(a,b)$. In other words, $x+y=a$ and $x+y^2=b$. Subtracting the first equation from the second, we get that $y^2-y=b-a$ must be solved. This quadratic, however, cannot be solved (in the reals) for all $b$ and $a$'s. In fact, the minimum value occurs at $y=1/2$ with a value of $-1/4$. So if $b-a$ is less than $-1/4$, we cannot find $(x,y)$ so that $f(x,y)=(a,b)$.

For example, consider $(1,0)$, then we would need $f(x,y)=(1,0)$ or that $x+y=1$ and $x+y^2=0$. Solving the first equation for $x$ gives $x=1-y$ and substituting this into the second equation gives $1-y+y^2=0$ or that $y^2-y=-1$, but this cannot be solved. Hence the function is not surjective.

Answer 2

Hint for surjection.

take$(a,b) \in \mathbb R^2$ and try to find

$(x,y)$ such that

$$(x+y,x+y^2)=(a,b)$$

this yields to

$$y^2-y+a-b=0$$

so, to find at least one $y$, the discriminant must be $\geq0$.

$\Delta=1+4(b-a)\geq 0$.

thus, for example with $b=0,a=1$

$y$ doesn't exist. $f$ is not surjective.

Answer 3

The Jacobian Matrix of $f$ is given by $$J_f = \left( \begin{array}{cc} 1 & 1 \\ 1 & 2y \end{array}\right)$$ The Jacobian is singular, i.e. $\det(J_f)=0$, when $2y-1=0$.

The critical points of $f$ are the points $(x,y)$ for which $2y-1=0$, i.e. $y=\frac{1}{2}$.

The image of the critical points $y=\frac{1}{2}$ is given by $f(x,\frac{1}{2})=(x+\frac{1}{2},x+\frac{1}{4})$. If $f(x,\frac{1}{2})=(u,v)$ then $u-v=\frac{1}{4}$ is the equation of the critical value set.

We have a fold singularity, so points with $u-v>\frac{1}{4}$ have no pre-image, points with $u-v<\frac{1}{4}$ have two pre-images, and points with $u-v=\frac{1}{4}$ have a single repeated pre-image.