Why is the Collector current of a BJT transistor independent of the Collector Base voltage?
You seem to think that because the depletion area is large for the base collector junction that there must be a large field there. That is the opposite of how it works. The voltage difference between the emitter, base, and collector is dropped exclusively in the depletion region. Thus the larger the depletion region, the more spread out the field is and so the less strong the field is. In fact, back injection is one of the reasons for hetero junction transistors. In the non depletion regions, diffusion is the dominant mechanism. Thus by creating a large electric field that essentially acts like a reverse flow barrier, the BJT forces diffusion to occur. The reason the current flows from C->E (NPN) or E->C (PNP) is because the BJT uses different doping levels to create depletion widths for different purposes. The reverse biased C-B depletion region, being large, is dropped almost exclusively on the collector's side. That means there's very little length for diffusion. That small length means that the Collector becomes an excellent source of minority carriers for the base compared to the emitter, whose large electric field is almost insurmountable. When those minority carriers cross, they drive diffusion in the base. Once the base provides that minority carrier to the emitter, it can't get it back, so it must get another one from the collector.

In essence the collector current is dependent only on the amount that flows out of the emitter, not on the field, which is designed to drop almost entirely over the collector.

Please leave a comment on this if you don't understand, because I needed about 10 pictures to get this down in college.

Edit: The BJT picture to rule them all!