In my understanding the well-known experiment of the glass full of water in equilibrium with a piece of paper, the atmospheric pressure acts on a small layer inside the glass (on the top) and under the paper (outside the glass), the hydrostatic pressure (basically the weight of the water) acts downward, so in term of forces I initially have a net force $$ \boldsymbol F = (p_A -p_A+\rho g h)A\hat z$$ where $A$ is the section of the glass, $\hat z$ is the vertical direction, and $h$ the height of the layer of water.
Now, when the paper bends under the weight of the water, and the air layer on the top increases in volume, so I can apply (with good approximation) $$pV=nRT,\ V\ \uparrow\ \Rightarrow\ p\downarrow\ \Rightarrow\ p'<p_A $$
So we have: $$\boldsymbol F = A(p'-p_A+\rho g h)\hat z$$ and since $p'-p_A<0$, it is possible to have $\boldsymbol F$ upward (clearly, depending on $h$ and $\rho$)
Now, I made the experiment with water, and the bending of the paper was upward. Can I say that the only reason is the presence of the surface tension of water? Or my reasoning is lacking somewhere else too?
Moreover, if instead of the paper I put a strongly stiff material, I can't have the same effect, no matter the weight, the geometry, etc?