Let A = {1, 2, 3, 4, 5, ..., 10} and f : A → A be an invertible function.

Given: f : A → A is an invertible function, where A = {1, 2, 3, 4, 5, ..., 10}

Since, $f$ is invertible   

Therefore, $f^{-1} o f(x)=x \quad \ldots(1)$

Now,

$\sum_{r=1}^{10}\left(f^{-1} o f\right)(r)=f^{-1} o f(1)+f^{-1} o f(2)+f^{-1} o f(3)+\ldots .+f^{-1} o f(10)$   $\left(\because 1+2+3+\ldots+n=\frac{n(n+1)}{2}\right)$

$=1+2+3+\ldots+10 \quad($ From $(1))$

$=\frac{10(10+1)}{2}$  

$=5(11)$

$=55$

Hence, $\sum_{r=1}^{10}\left(f^{-1} o f\right)(r)=\underline{55}$.