proper base change theorem

Question

Let $f: X \to S$ be proper with $S$ the spectrum of a stricly henselian DVR and $X_0$ the special and $X_1$ the generic fibre. Why do we have $H^q(X,F) = H^q(X_0,F)$, but not $H^q(X,F) = H^q(X_1,F)$?

Answer 1

To understand this sort of question, you should begin by first understanding the case when $S$ is the open unit disk in $\mathbb C$, and $X$ is a proper family of varieties.

For example, let $X$ be the family of conics $xy - t = 0$. (Here I am writing the family of affine curves, but I really mean the corresponding family of projective curves.)

If $t$ is non-zero, the fibre $X_t$ is a smooth conic, hence homeomorphic to a $2$-sphere.

If $t = 0$, then the fibre $X_t$ is a pair of lines meeting at a point, so homeomorphic to the gluing of two $2$-spheres along a common point.

Can you see that these have different cohomology? Perhaps you can also prove that $X$ retracts onto $X_0$, and so has cohomology isomorphic to $X_0$ (and so not isomorphic to $X_1$).

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Perhaps you also have a technical question about why proper base-change implies one statement and not the other? If you confirm this, I can try to explain. But in the meantime, I suggest trying to build some intuition about the whole context by looking at simple examples of the kind described above.