how can vfd builds more voltage while motor sharp stopping

You should at least specify whether your doubt is electrotechnical in nature, or what other.

The asynchronous motor, which has the rotor at a speed greater than that of the rotating magnetic field, is found to be working in the fourth quadrant, as a generator. The VFD inverter creates the decelerating rotating magnetic field, the rotor that is faster then discharges energy via the magnetic field to the stator windings, inducing overvoltage there by Lenz's law: (-e = d PHI / d t). Recirculating diodes on the three legs of the H-bridges return this overvoltage to the DC bus.

That said, if the question was otherwise, then sorry.

Basically, when the electric circuit feeding current to a coil is opened, the current cannot cancel or change in value in zero time; it can only quench and vary by discharging the energy stored by the coil. Consequently, the electrical potential of the coil will be forced to assume any useful value, (depending on the impedance over which the current will be able to flow).

I foundthis paper, with particular reference to Figure 8. The blue line, current, as you see, when you open the circuit does not cancel and, therefore, the coil potential, red line, reacts to varies until somewhere it manages to discharge the current. Even at the cost of producing electric arcs. (But this is not our case, because as we said the recirculating diodes send the extra potential back to the DC bus, the capacitors on it then absorb some of the energy, etc., etc., etc.