Ever 17 - The Out of Infinity -

(i am assuming that you assumed that takeshi closed the door and tsugumi has to bilge the water out of the compartment takeshi was previously in, in order to float)

Consider the submarine as a whole. Two chambers, two humans in one, both are filled with air. This assembly has slightly negative buoyancy. The air contained within the exit chamber is highly positively buouant. When Takeshi leaves, the chamber goes from a positively buoyant status to a neutrally buoyant status, and the whole assembly loses buoyancy.

I thought of a less esoteric example than my styrofoam discussion earlier. Think of an inflated balloon that contains a small weight (takeshi) and has a large weight(the rest of the mass) affixed to the exterior. Let's assume that the two masses are close to infinitely dense, thus they have no buoyancy of their own, and let's also assume the balloon and the air have essentially no mass and that the air is non-compactable. Now, if the internal mass is removed without any air escaping the assembly will stop sinking and will begin to float.

However, it is impossible to jettison the small internal weight without also expelling some air. Let's say removing the internal mass requires you to expel 1/10 of the air in the balloon (probably close to what opening the exit chamber door would be like). The balloon has now lost 10% of it's lift force. That's fine if the internal mass was greater than 10% of the total system mass, it will now float. However, I'm assuming that the takeshi is less than 10% of the total submarine mass, and thus the lost air volume needs to be replaces somehow or else the system will simply sink faster. Pumping out the water taken in is the same as inflating the balloon back to its original size.