Thank you for very scientific answer. Would you be so kind to show me on this example how it is done? Do I need to define the region R?

$\bf{R}$ is the common notation for the set of real numbers. In fact, one can easily prove that $x\mapsto x^2$ is not bijective on $\bf{R}$ : you only need to find two different numbers that have the same square. You can begin with that, then I'll accompany you through the next steps !

Alright. Say we can have f(2)=4 and f(3)=9. Thus we have 2 different results 4 and 9 which are not equal and thus the function is not bijective. Is this correct? What is step 2? :)

nope, you need to find two different numbers that have the same square. For instance, you can search $a$ and $b$ such as $f(a) = f(b) = 4$. Which should be the values for $a$ and $b$ (with $a\neq b$)?

I guess that there is not such numbers that can satisfy f(a)=f(b)=4 for the given function.And since such numbers doesn't exist the condition of bijection can't be satisfied.

what about $2$ and $-2$? don't they have the same square? Or $3$ and $-3$?

Oh, yes. By some reason I was thinking about positive numbers. Thus f(2)=4 and f(-2)=4. What does it says to us then?

It basically says to us that $x\mapsto x^2$ is not bijective over $\bf{R}$. So, in order to prove that it is a homeomorphism, we need to find the largest domain $D$ included in $\bf{R}$ where $x^2$ is bijective. You naturally proposed to take $D = \bf{R_+}$ the set of positive integers, that is a smart choice. Now, the next step is to prove that $x\mapsto x^2$ is bijective for $x\geq 0$.

And in order to prove it we also need to take any positive numbers? Like f(1)=1 and f(4)=16?

Unfortunately, it's not that simple... To prove that $f$ is bijective, you have to prove that for any $x$ and $y$, you have $f(x)=f(y)$ iff $x=y$.

Let me begin it for you: