Let $f:\left[0,1\right]\to R$ be an injective continuous function that satisifes the condition $-1<f\left(0\right)<f\left(1\right)<1$ Then, the number of functions $g:\left[-1,1\right]\to \left[0,1\right]$ such that $\left(gof\right)\left(x\right)=x$ for all $x\in \left[0,1\right]$ is

We have,

$-1<f\left(0\right)<f\left(1\right)<1$

$g=\left[-1,1\right]\to \left[0,1\right]$

$\mathit{gof}\left(x\right)=x,\forall x\in \left[0,1\right]$

Only condition that $g\left(x\right)$ should satisfy for $gof\left(x\right)=x,\forall x\in \left[0,1\right]$ is that $g\left(x\right)$ should attain all values in $\left[0,1\right]$ when range of $f\left(x\right)$ a subset of $\left(-1,1\right)$ is used as image for $g\left(x\right)$. Thus, there can be infinite such functions $g\left(x\right)$ with domain $\left[-1,1\right]$ and range $\left[0,1\right]$