Does cohomology and base change hold if supported at a point?

I have a flat, quasicompact, and separated map $$p : X \to \mathbb{A}^1$$ and I know that $$R^i p_* \mathcal{O}_X$$ vanishes everywhere except possibly $$0 \in \mathbb{A}^1$$.

Q1: Does "cohomology and base change" automatically hold here? I.e., is the natural map $$(R^i p_* \mathcal{O}_X)_s \otimes_{\mathcal{O}_{\mathbb{A}^1, s}} k(s) \to H^i(X_s, \mathcal{O}_{X, s})$$ an isomorphism for all $$s \in \mathbb{A}^1$$? This is clear for all nonzero $$s$$ and I believe it should follow for $$0 \in \mathbb{A}^1$$ automatically.

Q2: Does it then follow that $$R^i p_* \mathcal{O}_X = 0$$? Theorem 1.1 in Conrad's notes would then imply that $$R^i p_* \mathcal{O}_X$$ was locally free, hence constant rank 0, hence 0.

Q3: If $$X \overset{v}{\to} Y \to \mathbb{A}^1$$ are proper maps and both $$X$$ and $$Y$$ are flat over $$\mathbb{A}^1$$ ($$v$$ is proper birational, n.n. flat), then does cohomology of $$\mathcal{O}_X$$ along $$v$$ commute with base change in $$\mathbb{A}^1$$? The versions of Cohomology and Base Change I've seen have one map which must be both proper and flat, as opposed to a composite being flat and the first map proper. This is my original situation, and I got to the above by localizing in $$Y$$.

I've been trying to do this descending induction, where $$R^N p_* \mathcal{O}_X$$ is zero for $$N \gg 0$$ for dimensional reasons, so $$R^{N-1} p_* \mathcal{O}_X$$ commutes with base change, but I also need $$R^{N-2} p_* \mathcal{O}_X$$ to commute with base change to conclude $$R^{N-1}p_* \mathcal{O}_X$$ was locally free, hence zero, and continue.

• It seems to me that you are assuming your hypothesis also for $i=0$, which is impossible — $p_*\mathscr{O}_X$ contains $\mathscr{O}_{\mathbb{A}^1}$.
– abx
Sep 10, 2020 at 5:18
• What happens for the projection from $\mathbb{A}^3\setminus\{0\}$ to $\mathbb{A}^1$? Sep 10, 2020 at 6:59
• @abx I understand, but I'd like to at least get it for $i \geq 2$ if possible -- I'll take what I can get. Sep 10, 2020 at 17:04
• @PiotrAchinger Thank you for bringing up this example. I thought I checked it but I made a mistake. I don't think that arises in my situation because I got there by (Zariski) localizing a genuinely proper map $X \to Y$ in $Y$ and then I have an affine $Y$ over $\mathbb{A}^1$ (as in Q3). I don't believe $\mathbb{A}^3 \setminus 0$ can factor as a projective map to an affine over $\mathbb{A}^1$ because the map to an affine would have to factor through $\mathbb{A}^3$ and then the inclusion $\mathbb{A}^3 \setminus 0 \subseteq \mathbb{A}^3$ would have to be closed. I'll rewrite the question better. Sep 10, 2020 at 17:09