transistor level design of Op-amps
The basic differential front end of a opamp works like this:



On a single chip, the resistors would be current sources, usually a current mirror.  Either way, think of how this works.  Q1 and Q2 are built to match as closely as possible.  Assume they are identical, and consider what happens as IN+ and IN- are moved up and down together.  Being identical with identical base voltages, they will share the current equally.  Note that absolute base voltage has nothing to do with this, as long as there is sufficient C-E voltage available.  Half the current appears across each resistor, and the output voltage stays the same.

If IN+ goes a little higher than IN-, then Q1 will take more of the current, Q2 will take less of the current, which means R2 will have less current thru it and less voltage across it, which means OUT goes up.  From there, OUT is amplified, integrated, and used to drive the high-current output stage.

The integrator is a important part of a opamp.  It puts a single dominant pole into the open loop transfer function, which gives the opamp predictablility and gurarnteed stability over a specific set of operating conditions.  The rest of what opamps do is accomplished with external feedback.  In any real "opamp" application (as apposed to "comparator"), this external feedback is a essential part of overall operation.

The same Q1/Q2 differential pair configutation works with FETs, or darlingtons too, but those are just implementation details that effect offset voltage, input impedance, and other tradeoffs.