classical mechanics - Deriving $T = F r = Ialpha$ for a rigid body

For a single point mass : $\tau=F_{t}r=ma_tr=(m r^2)\alpha = I\alpha$

For multiple point masses bound together : $\sum \tau_i = (m_ir_i^2)\alpha = I\alpha$

But how do we go from that to $I\alpha = F_tr_{app}$ (for the multiple masses case), where $r_{app}$ is the point where force is applied? That is saying that the entire sum of torques of individual masses $\sum \tau_i$ is physically equivalent to (i.e. produces the same angular acceleration of the bound rigid object as) a single torque $F_tr_{app}$ applied to the point $r_{app}$. (Or putting it another way, that applying a single force $F_t r_{app}$ will result in many $F_{t\_i}$ forces on each mass element such that $\sum F_{t\_i}r_{i}$ will be equal to $I\alpha$.)

Every place I've looked for a derivation just jumps from $\sum \tau_i =I\alpha$, to $\tau=I\alpha$, as if they are the same thing.