Mike, I thought that my brief previous reply was adequate to point out the principles involved, but I'll go into greater detail to help out more. Yes, as you question, the impedance is partly the result of the crossover circuitry, not the impedance or number of the individual drivers alone. When loads(such as speaker drivers)are connected in parallel the combined impedance is such that its reciprocal is equal to the sum of the reciprocals of the impedances of the individual drivers. So, if an 8 ohm mid-woofer and and an 8 ohm tweeter were simply connected in parallel without intervening crossover circuitry, the combined impedance would in fact be 1/X=1/8+1/8, 1/X=2/8, 2X=8, X=4(ohms). The crossover circuitry changes this, however. For example, on a bass frequency the impedance of the low frequency section of the crossover would be 8 ohms, but the capacitor and other components in the high frequency section would make the impedance the amplifier "sees" there for bass frequencies far higher than 8 ohms. For a relatively simple calculation, say that the impedance of the high frequency crossover section on the bass note is 160 rather than 8 ohms. The resultant combined impedance would be 1/X=1/8+1/160, 1/X=21/160, 21X=160, X=7.6(ohms). The point therefore is that a 4 ohm speaker impedance isn't the result of two 8 ohm drivers connected in parallel.
The other point previously mentioned is that the high and low frequency sections of the crossover are always separate. They have to be or they couldn't send the separate frequency ranges to the low and high frequency drivers. This is the case regardless of whether the speaker inputs are single or dual, or if dual, regardless of whether the connecting link is or isn't removed. As was pointed out previously the only difference(when removed)is that the wiring separates at the receiver output rather than inside the enclosure.
