# Example of bijective function from $(-1,1)$ to $\Bbb R$

I'm looking for a bijective function $$f : (-1,1) \to \mathbb{R}$$. I'm having trouble finding an example.

• $(-1,1)\to(-1,1)$, $x \mapsto x$? Nov 9, 2020 at 23:30
• What is the codomain of the envisaged function with domain $(-1,1)$? Without both pieces of data trying to think of a bijection, let alone any function, doesn't make any sense unless we automatically consider the codomain to be $(-1,1)$ - in which case the identity on this interval will do just fine. Nov 9, 2020 at 23:36
• Hi @TaylorRendon. The problem only says that the codomain belongs to the real numbers. Nov 9, 2020 at 23:43
• In your opinion could the intent of the problem be that your function be a bijection from $(-1, 1)$ to all of $\mathbb{R}.$ That is, could the problem be that every real number must be in the range of $f$? Nov 10, 2020 at 0:40
• Nov 10, 2020 at 6:34

$$f:: (-1,1) \rightarrow \mathbb{R}$$

$$f: x \mapsto \tan(\frac{\pi}{2} x)$$

• +1 for nailing it and -0.49 for making me pull out my Calculus book to confirm. I rounded up. Nov 10, 2020 at 1:05

For a different example, $$f(x) = \frac{1}{x - 1} + \frac{1}{x + 1}$$ has this property:

1. It is injective: if $$x,y\in(-1,1)$$ and $$f(x) = f(y)$$, then we have $$\frac{1}{x-1}+\frac{1}{x+1} = \frac{1}{y-1} + \frac{1}{y+1}.$$ Multiplying by $$(x-1)(x+1)(y-1)(y+1)$$ gives $$(x+1)(y-1)(y+1) + (x-1)(y-1)(y+1) = (x-1)(x+1)(y+1) + (x-1)(x+1)(y-1).$$ Now, we can simplify that down to $$x(y - 1)(y + 1) = y(x - 1)(x + 1).$$ At this point, we look to be stuck with expanding the brackets, so we'll do that, obtaining $$xy^2 - x = yx^2 - y,$$ or $$xy(y - x) = x - y.$$ If $$x = y$$, we're done, so suppose not (so that we can divide by $$x - y$$). Then we have $$xy = -1,$$ so $$|xy| = 1$$. But we know that $$|x| < 1$$ and $$|y| < 1$$, so $$|xy| = |x||y| < 1$$, a contradiction. Thus, $$f$$ is injective.

2. It is surjective: for this, we shall simply show that it is unbounded above and below, and appeal to a version of the intermediate value theorem. First, note that, for $$x = 1 - \varepsilon$$ and $$\varepsilon < 1$$, we have $$f(x) = \frac{1}{-\varepsilon} + \frac{1}{2 + \varepsilon} = \frac{2}{-\varepsilon(2 + \varepsilon)}\leq \frac{-2}{3\varepsilon} \to -\infty$$ as $$\varepsilon \to 0$$ from above. Similarly, for $$x = -1 + \varepsilon$$, we have $$f(x) \geq \frac{2}{3\varepsilon} \to +\infty$$ as $$\varepsilon \to 0$$ from above.

Thus, for any $$y \in \mathbb{R}$$, there are some $$x_0, x_1$$ such that $$f(x_0) < y < f(x_1)$$ (and, indeed, we see that $$x_0 > x_1$$). Applying the intermediate value theorem (after noting that $$f$$ is continuous on $$(-1, 1)$$ as a sum of rational functions whose denominators have no zeroes in that interval) to $$[x_1, x_0]$$ then gives us some $$x_2 \in [x_1, x_0] \subset (-1, 1)$$ such that $$f(x_2) = y$$, and so $$f$$ is surjective.

• I did not work this out, but I suppose one can also solve $f(x)=y$ to show bijectivity. It’s a quadratic equation. Nov 10, 2020 at 7:52
• @CarstenS Indeed: you'll find that for every $y$ (other than $0$), there are exactly two $x \in \mathbb{R}$ such that $f(x) = y$, exactly one of which lies in $(-1,1)$. Nov 10, 2020 at 13:15

Under the assumption that the bijection must be from
$$(-1, 1) \to \mathbb{R}$$:

$$x \geq 0:~~$$ Let

$$f(x) = - \log|x - 1|.$$

$$x < 0:~~$$ Let

$$f(x) = + \log| ~|x| - 1|.$$

HINT: If the intent of this problem is for the codomain of $$f$$ to be all of $$\mathbb{R}$$,

Define the map $$f : (-1,1) \to \mathbb{R}$$ by $$f(x) := \frac{x}{1-x^{2}}$$.