I read your post again and I am at a loss. I have never said anything about the effect of moving the passive crossover from one end of the wire to the other. Please read my post again if you don't believe. Regardless, like your point on the Fourier transform topic, there is no disagreement! As an aside, it allows us to work back and forth between the time domain and frequency domain, works great.
All I was trying to say was: The speaker has links that shorts out two sets of terminals, one for the LF the other for the HF crossovers/drivers. Without bi-wiring, the single pair of speaker cable must carry all the signals from the amp to the speaker. At the joined (common) point, the XOs filter the signals to the LF and HF drivers. In this case, the single pair of wires has no choice but to carry signals of full range of frequencies, e.g. 20 to 20,000 Hz.
Using bi-wire, and with the links removed, the XO networks are now separated, it is now like having two separate speakers, but one will accept only LF the other only HF because of the different XO/filter networks. It is the characteristics of the impedance of each individual XO/filter network that forces a desired frequency band of signal to flow through the pair of cable that connects to the respective XO. The EMF that you mentioned at the amp output terminals are the same, but the impedance of the two signal paths are not different and are frequency dependent due to the two different crossovers in each path.
Why wouldn't you agree that each (remember they are now separated, isolated period)crossover that is connected to its own pair of cable would choose its own signals of the preferred frequencies. To sum up, bi-wiring means neither pair of wires will carry the full range of frequency. Whether this makes the sound different or not I don't know and I am staying out of it. I hope this help clarify one specific point.