Show that $f:M_f \rightarrow A_f$ is bijective when $f(x)=\frac{x-1}{x+1}$  Problem 
Show that $f:M_f \rightarrow A_f$ is bijective when $f(x)=\frac{x-1}{x+1}$
Edit
when $M_f \in \mathbb{R}\setminus \{-1\}$ and $A_f \subset \mathbb{R}$
 Attempt to solve 
$f:M_f\rightarrow A_f$ is bijective when it is injective and surjective at the same time. 
Injection
$f:M_f \rightarrow Af$ is injective when:
$$ \forall (x,y) \in M_f : x \neq y \implies f(x)\neq f(y)  $$
when $p= x\neq y$ and $q=f(x)\neq f(y)$. If it is possible to show $\neg p \implies \neg q$ it means: $$ \neg p \implies \neg q \iff p \implies q $$
meaning
$$ x \neq y \implies f(x) \neq f(y) \iff x = y \implies f(x)= f(y) $$
I can show:
$$ \frac{x-1}{x+1}=\frac{y-1}{y+1} $$
$$ \implies  (x+1)(y+1)\frac{x-1}{x+1}=(x+1)(y+1)\frac{y-1}{y+1} $$
$$ \implies (y+1)(x-1)=(x+1)(y-1) $$
$$ \implies xy-y+x-1 = xy -x + y-1 $$
$$ \implies xy+2x = xy +2y $$
$$ \implies 2x=2y \implies x = y $$
Conclusion: $f:M_f \rightarrow A_f$ is injective since
$$ x \neq y \implies f(x) \neq f(y) \iff x = y \implies f(x)= f(y) $$
is true
Surjection
$$ f:M_f \rightarrow A_f $$ is surjective when
$$ \forall y \in A_f \exists x \in M_f : f(x)=y$$
meaning if
$$ f(x)=y $$
assuming $x \neq -1$
$$ \implies \frac{x-1}{x+1}=y $$
$$ \implies x-1=y(x+1) $$
$$ \implies x-1=yx+y $$
$$ \implies x(1-y)=1+y $$
$$ \implies x = \frac{1+y}{1-y} $$
Now we get
$$ f(x)=f(\frac{1+y}{1-y}) $$
$$\implies f(x)=\frac{(\frac{1+y}{1-y})-1}{(\frac{1+y}{1-y})+1}$$
For some reason odd reason i cannot get this into form $f(x)=y$. Should be simple elementary algebra but cannot see how this equals to $y$. In theory this expression should be equal to $y$ which would make $f:M_f\rightarrow Af$ surjective and then bijective since it was also injective. 
 A: Observe that 
$$f(x) = \frac{x - 1}{x + 1} = 1 - \frac{2}{x + 1}$$
Thus, the range of $f$ is $\mathbb{R} - \{1\}$. 
To ensure that $f: \mathbb{R} - \{-1\} \to A_f$ is surjective, we must define $A_f = \mathbb{R} - \{1\}$.
For $y \neq 1$,
\begin{align*}
f\left(\frac{1 + y}{1 - y}\right) & = \frac{\frac{1 + y}{1 - y} - 1}{\frac{1 + y}{1 - y} + 1}\\
& = \frac{\frac{1 + y}{1 - y} - 1}{\frac{1 + y}{1 - y} + 1} \cdot \frac{1 - y}{1 - y}\\
& = \frac{1 + y - (1 - y)}{1 + y + 1 - y}\\
& = \frac{2y}{2}\\
& = y
\end{align*}
Thus, the function is surjective.  Since you have shown it is injective, it is bijective.
Alternatively, define $A_f$ as above.  Define $g: \mathbb{R} - \{1\} \to \mathbb{R} - \{-1\}$ by 
$$g(x) = \frac{1 + x}{1 - x}$$
To show that $f: \mathbb{R} - \{-1\} \to \mathbb{R} - \{1\}$ is bijective, show that $g$ is $f^{-1}$ by showing that 
\begin{align*}
(g \circ f)(x) & = x~\text{for each $x$ in the domain of $f$}\\
(f \circ g)(y) & = y~\text{for each $y$ in the domain of $g$}
\end{align*}
Establishing that $(g \circ f)(x) = x$ for each $x$ in the domain of $f$ shows that $f$ is injective since if $(g \circ f)(x_1) = (g \circ f)(x_2)$, then 
$$x_1 = (g \circ f)(x_1) = (g \circ f)(x_2) = x_2$$
Establishing that $(f \circ g)(y) = y$ for each $y$ in the domain of $g$ shows that $f$ is surjective since if $g(y) = x$, then $f(x) = y$. 
A: If $A_f$ is the "set of all outputs". This means that
$$
A_f = \left\{\, \frac{x-1}{x+1} \,\middle|\, x\in\mathbb R\setminus\{-1\}\,\right\}.
$$
Hence every $y\in A_f$ is of the form
$$
y = \frac{x-1}{x+1} = f(x)
$$
for some $x\in \mathbb R\setminus\{-1\} = M_f$, so $f\colon M_f\to A_f$ is surjective.

The observation here is that every map $g\colon A\to B$ gives rise to a surjective map
\begin{align*}
        A &\longrightarrow g(A), \\
        a &\longmapsto g(a),
\end{align*}
where $g(A) = \left\{ \, g(a) \,|\, a\in A\,\right\}\subseteq B$.
When $g$ is injective, the map $A\to g(A)$ is a bijection.
