Let $f:\mathrm{ℝ}\to \mathrm{ℝ}$ be defined as $f\left(x\right)=x-1$  and $g:R\to \left\{1,-1\right\}\to \mathrm{ℝ}$ be defined as $g\left(x\right)=\frac{x^2}{x^2-1}$. Then the function $fog$ is:

$fog\left(x\right)=f\left(g\left(x\right)\right)$  $=f\left(\frac{x^2}{x^2+1}\right)$

$=\frac{x^2}{x^2+1}-1$  $=\frac{x^2-x^2-1}{x^2+1}$ $=\frac{-1}{x^2+1}$

We know that, $0\le x^2<\infty , \forall x\in R$

$\Rightarrow 1\le x^2+1<\infty , \forall x\in R$ $\Rightarrow 1\ge \frac{1}{x^2+1}>0, \forall x\in R$  $\Rightarrow -1\le \frac{-1}{x^2+1}<0, \forall x\in R$

So, range of $fog\left(x\right)$ is $\left[-1, 0\right)\subset R$.

 Hence, the function $fog\left(x\right)$ is into function and $fog\left(-x\right)=f\left(g\left(-x\right)\right)=\frac{-1}{{\displaystyle {\left(-x\right)}^2}+1}=\frac{-1}{x^2+1}=f\left(g\left(x\right)\right)$

$\therefore fog\left(x\right)$ is an even function. So, it is a many one function.

Hence, $fog\left(x\right)$ is neither one-one nor onto function.