Let $f_1: R \rightarrow R, f_2:[0, \infty) \rightarrow R, f_3: R \rightarrow R$ and $f_4: R \rightarrow[0, \infty)$ be defined by

$f_1(x)=\left\{\begin{array}{l} |x|, \text { if } x<0 \\ e^x, \text { if } x>0 \end{array}: f_2(x)=x^2 ; f_3(x)=\left\{\begin{array}{r} \sin x, \text { if } x<0 \\ x, \text { if } x \geq 0 \end{array}\right.\right.$

and $f_4(x)=\left\{\begin{array}{l}f_2\left(f_1(x)\right), & \text { if } x<0 \\ f_2\left(f_1\left(f_1(x)\right)\right)-1, & \text { if } x \geq 0\end{array}\right.$

Then, $f_4$ is 

onto but not one-one

For $x<0$, we have

$f_1(x)=|x|=-x$  and  $f_2(x)=x^2$

∴  $f_4(x)=f_2\left(f_1(x)\right)=f_2(-x)=(-x)^2=(-x)^2=x^2$

For x > 0, we have

$f_1(x)=e^x$ and $f_2(x)=x^2$

∴  $f_4(x)=f_2\left(f_1(x)\right)-1=f_2\left(e^x\right)-1=\left(e^x\right)^2-1=e^{2 x}-1$

Thus, $f_4(x)=\left\{\begin{array}{r}x^2~~~~, \text { if } x<0 \\ e^{2 x}-1, \text { if } x \geq 0\end{array}\right.$

The graph of $f_4: R \rightarrow[0, \infty)$ is shown in Figure.