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Let $f:\mathbb{R} \to \mathbb{R}\times\mathbb{R}_{0}^{+} \space$ be defined as $f(x)=(x+1,x^2)$.

To prove that this function is surjective I started by the definition.

$$\forall \space (a,b) \space \exists \space c \space (f(c)=(a,b))$$

Then,

$a=c+1$

$b=c^2$

Solving the first for $c$ because there exists one $c$ for each $a$, $ \space c=a-1$. Now I guess that I should substitute $c$ in the second equation. Stays, $b=(a-1)^2$. But I don't know what conclusions should I make.

Can you help me? Thanks

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Do you have to prove or to decide? Is the point $(1,1)$ on the image? –  Sigur Jan 24 '13 at 21:50
    
Sigur, in the presence of my calculus,the difficult is to decide. –  João Jan 24 '13 at 21:57

2 Answers 2

up vote 4 down vote accepted

To show that $f$ is surjective, you need to show that for all $(y,z) \in \mathbb{R} \times\mathbb{R}_{\ge 0}$, there exists an $x \in \mathbb{R}$ such that $f(x) = (y,z)$. Hence, you need to solve $y=x+1$ and $z=x^2$ simultaneously. Clearly $y=x+1$ if and only if $x=y-1$ and hence $z=(y-1)^2$.

The function $f$ is not surjective. Its image is the parabola $z=(y-1)^2$:

$$\{ (y,(y-1)^2) \in \mathbb{R} \times\mathbb{R}_{\ge 0} : y \in \mathbb{R} \}.$$

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This is not true. For example $(1,1)\in\mathbb{R}\times\mathbb{R}_0^+$, and if $x\in\mathbb{R}$ should satisfy $x^2=1$ and $1+x=1$, then $x=1$ and $x=0$ which is absurd.

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What a coincidence? Both of us thought the same point to counter example... lol –  Sigur Jan 24 '13 at 21:52
    
@Sigur: it was probably the easiest example besides $(0,0)$ :D –  Stefan Hansen Jan 24 '13 at 21:55

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