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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

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    $\begingroup$ f(0)=0, f(1)=1, thus... $\endgroup$ – Locally unskillful Apr 2 at 19:13
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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 $[f_\alpha(0),f_\alpha(1)]=[0,1]$.

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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 $ay<a+1$ 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.

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