A standard silicon diode, like those used in most isolators will have a .6 to .7 volt drop in forward voltage, depending on current. This is a fact (look it up anywhere you like, or test it yourself). Some (very few) isolators are made with Shottky diodes, which have lower drops, between .2 and .4 v.

While this drop will make battery charging more difficult, it will still happen. Actually, depending on where the voltage sense for the Alt is, it will try and maintain the regulator setpoint, and will just kick its output up by .7 v.

And define 'charge'. If you define it as a flow of current into the battery, then even if the regulator doesn't pick up the slack, the battery will still charge. Say the alternator is 13.5v. A .7v drop will put 12.8 at the battery. If the battery is down to anything less than 12.8, current will flow into it, and it will charge. Sure, it won't have the same maximum voltage potential as it would if given the full 13.5V, and the current flow will be lower, but it will still charge.

Italiancorolla...sorry I didn't get to your PM sooner...I'm in Vegas, at the Sema show, and haven't been on the internet much. Looks like you got your question answered, though.

While a relay will work just fine, if I were you, I would use the 1/2 of the isolator. Hook the front (main starting) battery to terminal one, and then hook both the new 4 ga (or larger) wire from your alternator and rear battery to terminal two. This will prevent your system from ever draining your front battery, while also giving your system full voltage. I like this better than a relay, because it is a solid state device, and is always 'on'...no contacts to burn up, no extra wires to hook up, no switches to leave in the wrong position, etc.
Toby