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First, it's very easy to show this "the discrete way": Let $r$ be real and assume $r \ne 0.$ If $r > 0$, then $\displaystyle{\frac{r}{1 + r} \in (0, 1) \text{ and } f\left(\frac{r}{1 + r}\right) = r}$. The other case is just as easy.

On the other hand, the proof in quotes below is a bit confusing as I don't know calculus yet. I do know how to work with the rigorous definition of limit, but that's no help here. Does the red part in the proof below imply the blue part or do the colored parts together imply the conclusion? Can somebody, please, refer me to (or elaborate) theorems and definitions used in the proof in quotes? Thanks.

The function $f$ is defined on $(-1, 1)$. Also, $\color{red}{f(0) = 0, f(x) > 0 \text{ when } 0 < x < 1, \ f(x) < 0 \text{ when } -1 < x < 0}$. This function also has the property that $\color{red}{\displaystyle{\lim_{x \to 1^-}f(x) = +\infty, \ \lim_{x \to -1^+} = -\infty}}$. If you recall enough information about continuous functions from calculus, you might see that $\color{blue}{\text{ this function is continuous on the interval $(−1,1)$}}$. From this information, it follows that $f((−1,1)) = \mathbb R$ and that $f$ is onto.

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  • $\begingroup$ Are you familiar with the Intermediate value theorem? $\endgroup$ – bjorn93 Mar 15 at 21:39
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    $\begingroup$ The theorem in question is the intermediate value theorem. Given the red part and the fact (which needs proving) that $f$ is continuous, you can show that for any $y \in \Bbb{R}$, there are $a, b \in (-1, 1)$ such that $y \in [f(a), f(b)]$ and then the intermediate value theorem gives you an $x$ such that $f(x) = y$. $\endgroup$ – Rob Arthan Mar 15 at 21:41
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The key theorem here is the intermediate value theorem. The point is that since $f$ is continuous, and $f(0)=0$, $\lim_{x\to 1^-}f(x)=+\infty$, by the intermediate value theorem, for any $c\in [0,+\infty)$, there is an $a\in [0,1)$ so that $f(a)=c$. Same goes on $(-1,0)$.

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The red part does not imply the blue part. To get the blue part, notice (one piece of this is taken from the red part): $$ f(x) = \begin{cases} \frac{x}{1+x} ,& x > 0 \\ 0 ,& x = 0 \\ \frac{x}{1-x} ,& x < 0 \\ \end{cases} \text{.} $$ That $f$ is continuous on $(0, \infty)$ and $(-\infty, 0)$ is a consequence of common continuity properties:

  • constant functions are continuous ($1$),
  • the identity function is continuous ($x$),
  • sums and differences of continuous functions are continuous ($1+x$, $1-x$), and
  • quotients, where the denominator is nonzero, are continuous. ($\frac{x}{1+x}$, $\frac{x}{1-x}$)

To finish showing continuity, we must verify that $$ \lim_{x \rightarrow 0^-} f(x) = f(0) = \lim_{x \rightarrow 0^+} f(x) \text{,} $$ that is, the function agrees with its limits (some variant of which is typically taken as defining continuity) at the boundaries of the pieces in the piecewise definition.

The colored parts show that $f$ is continuous and $f$ takes arbitrarily small ($\lim_{x \rightarrow -1^+} f(x) = -\infty$) and arbitrarily large ($\lim_{x \rightarrow 1^-} f(x) = \infty$) values, so by the intermediate value theorem it takes every value. That is, $f$ is a surjection onto $\Bbb{R}$.

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