# Proving $f_{\alpha}(x)$ is a bijection

Let $$f_\alpha:[0,1]\mapsto[0,1]$$ be defined as follows for $$-1\lt\alpha\lt\infty$$.

$$f_\alpha(x)=\dfrac{(\alpha+1)x}{\alpha x+1}$$

Now to prove that this is a bijection, we need to prove that it both injective and surjective. For the first part, it is quite easy. We have to prove that $$f_\alpha(x_1)=f_\alpha(x_2)\implies x_1=x_2$$, which can be done quite easily. I'm struggling in the second part however, the part wherein we're supposed to prove that it is surjective. I know that a function is surjective by definition, if all the elements of $$\mathrm{range}f$$ are a part of the ordered pairs defined by the function, but using this definition does not seem to take me anywhere. I don't know how to go about proving the surjectivity using this definition.

Any hints to proceed further are appreciated. Thanks

• f(0)=0, f(1)=1, thus... – Locally unskillful Apr 2 at 19:13

When $$x=0$$ you get $$f_\alpha(0)=0$$ and when $$x=1$$ you get $$f_\alpha(1)=1$$. Since $$f_\alpha$$ is continuous by intermediate value theorem you have that the range of $$f_\alpha$$ has to contain $$[f_\alpha(0),f_\alpha(1)]=[0,1]$$.

To answer without using calculus-level arguments, note that if $$y=\frac{(a+1)x}{ax+1}$$ then we can solve that$$x=\frac y {a+1-ay}.$$

Note that if $$y\in[0,1]$$ then $$ay\le a$$ so $$ay so $$a+1-ay>0,$$

so this is well-defined (not dividing by $$0$$),

and furthermore $$x\ge0$$ since the numerator and denominator are both non-negative,

and since $$(a+1)y\le(a+1)$$ it follows that $$y\le a+1-ay$$ and thus $$x\le1$$.

In summary, we have found a pre-image $$x \in [0,1]$$ for every $$y\in[0,1]$$, so $$f$$ is surjective.