Would a bicycle pump work underwater, with its air-input being above water? Let say I have a bicycle pump 10m underwater, with its air input following an inflexible tube to the surface. Would the bicycle pump be harder to either pull or push?
What would be the pressure inside the tube? If it is higher than the surface pressure, how could the pressure be high in one end of the tube, and low in other?
I'll place a diagram to visualize.

 A: So let's look at the forces that act on the moving piston.
We have water on one side (which we assume for simplicity that it has constant pressure over the stroke) and air on the other side.
So let's start a pump cycle with the piston in the position where there is no (or minimal) air in the cylinder. We pull out the piston. On one side we have air pressure basically the same as at the surface, on the other side of the piston we have the water pressure which depends on the depth. Because it seems simpler right now I'm using Bar instead of Pascal for pressure.
Per 10m depth we have one Bar of pressure added. 1 Bar is 1kg/cm^2. Which means that if our piston has an area of 10cm^2 at a depth of 10m we will need the equivalent force of lifting a mass of 10kg at the surface (around 100N)
Now let's see what the forces are if we want to push out that air. Again assuming that the pump, opening, piston and everything has the same pressure all around.
We now need to compress the air to the pressure of the opening which is also 1m under water. So it needs a pressure of 1bar. Again with 10cm^2 this will give a force equivalent to lifting 10kg at the surface. But at the same time we have the same force pressing from the water on the other side of the piston, so basically zero force is needed to push the air out (dismissing any friction and stuff).
So, if the pump is at the target pressure level you need force to pull air into the comression chamber but basically no force to get the air out.
A: 
Would the bicycle pump be harder to either pull or push?

Harder. Air output now is presented with additional $\Delta P_{10m} = \rho g h \approx 98~\text{kPa}$ pressure, for getting out of pump, instead of $1~\text{atm}$ while pumping in the air. In addition, while pumping in the water, - your hand will experience $\approx 816 \times$ higher drag force comparing with pumping in the air, due to a lot higher surrounding fluid density.
A: 
How could the pressure be high in one end of the tube, and low in
other?

There are two valves inside a bicycle pump (see picture below). Before you press down the inlet valve is open: the one that is connected to the air. The pressure inside the pump is equal to atmospheric pressure (I'll neglect the increase in air pressure due to depth for a second since it is only 10m). When you start pumping the inlet valve will close shut and the pressure starts increasing because you are compressing the air. Once the pressure inside the pump is higher than the pressure outside the outlet valve will open and air starts escaping.
I suspect that initially it would be equally hard to press down the handle since the pressure inside is independent of the pressure outside the pump. But while a pump functioning in normal conditions would quickly open the outlet valve, the pressure inside the underwater pump would keep increasing before it could release some air. So overall it would be harder to use the pump. But to be honest I'm not quite sure about this.

source of image: https://www.researchgate.net/publication/221249358_Lines_Blobs_Crosses_and_Arrows_Diagrammatic_Communication_with_Schematic_Figures
