A silicon diode will drop about 0.7V, so the 12V from your adapter is not 12V, but 11.3V, depending on current draw.
Not enough to charge your battery, as stated previously in another topic.
A diode will only conduct if the anode is more positive (again, aprox 0.7V) then the cathode.
So if you take care your adapter puts out 13.8+0.7V=14.5V, the battery will charge. As it becomes charged it will draw less and less current, but I guess it won't fall to zero.
Any voltage higher then the present float voltage of the battery will cause a charging current to flow. The higher the voltage, the more current will flow, depending on state of charge. For a 12V lead-acid battery, the absolute maximum charging voltage is 14.4V or so.