I think you are right senan... If the input voltage could be changed by a relatively fast step, there will be an overshoot (likely damped one) at the output due of the opamp internal delay from input to output before the output is settled at its end value. So during the overshoot, the difference between the two input terminals is not close to zero.
Kerim, at first: Nice to hear from you (because the trouble in your country).
I believe the question from Senan is a general one - that means: He is not referring to a step response.
Of course, Barry is right in saying the opamp would "try" to equalize both voltages at the input terminals. However, this is not always understood resp. accepted by beginners.
Senan - you must try to understand the concept of negative feedback.
Perhaps it helps to start with the role of the emitter resistor RE in a common collector transistor stage (voltage follower).
The voltage developed across RE follows the input signal voltage which means: The
differential base-emitter voltage remains very small (in comparison to the signal input).
And - the higher the transconductance gm (proportional to Ic) the smaller this diff. voltage.
Something similar happens in an opamp with negative feedback, but due to the very high gain we can count with a differential voltage of 0 volts between both input terminals (in reality: some µVolts).