Converses to Cartan's Theorem B

Question

Here is a phrasing of some Cartan Theorem B statements:

Consider the following conditions:

1. $X$ is a {Stein manifold, affine scheme, coherent analytic subvariety of $\mathbb{R}^n$, contractible subset of $\mathbb{R}^n$}.
2. $\mathcal{F}$ is a {coherent, quasi-coherent, coherent, locally constant} sheaf (of $\mathcal{O}_X$-modules) on $X$.
3. The sheaf cohomology $\operatorname{H}^p(X,\mathcal{F})$ is zero for $p\geqslant1$.

Then (1) + (2) $\implies$ (3).

Here this statement applies to four geometric contexts: {complex-analytic, complex-algebraic, real-analytic, smooth}. In all of these, a "sharpness" statement also holds:

Given arbitrary $X$, if every sheaf $\mathcal{F}$ (of $\mathcal{O}_X$-modules) that satisfies (2) also satisfies (3), then $X$ satisfies (1).

(For the two complex contexts, this is "classical"; for the real-analytic case, I believe this should be covered in Cartan's paper "Variétés analytiques réelles et variétés analytiques complexes" but I should check this; for the smooth case, this is quasi-folklore.)

Question. For the four contexts above, is the remaining implication known to be true? i.e. is the statement

Let $X$ satisfy (1); if an arbitrary sheaf $\mathcal{F}$ (of $\mathcal{O}_X$-modules) satisfies (3) then it satisfies (2).

true?

I assume this should be false, at least in the smooth case, given the existence of e.g. injective sheaves that are not locally constant, and I think in the complex-algebraic case it's also false (thanks to skyscraper sheaves) but I'm not so sure about the other cases.

Ideally an answer would consist of four "yes/no" answers with some references, but I'd be happier with some partial answers than with no answers at all :-)

Answer 1

For (4), there are finite CW complexes $X$ which are not contractible, but such that $H^q(X, \mathcal{F})$ vanishes for all (finite rank) locally constant sheaves $\mathcal{F}$. Specifically, take $X$ to have one vertex $v$; four edges $a$, $b$, $c$, $d$ from $v \to v$, and four pentagonal $2$-cells with boundaries $b^{-1} a^{-1} b^2 a$, $c^{-1} b^{-1} c^2 b$, $d^{-1} c^{-1} d^2 c$ and $a^{-1} d^{-1} a^2 d$. We easily compute that this complex is connected and has vanishing cohomology, so the trivial locally constant sheaf on $X$ has vanishing cohomology. The fundamental group of $X$ is Higman's group.

Locally constant sheaves (of rank $N$, over $k$) come from maps $\pi_1(X) \to \text{GL}_N(k)$, and Higman's group has no non-constant maps to $\text{GL}_N(k)$ for any $k$ or $N$ (see Remark 2 in Tao's blogpost), so the only locally constant sheaves on $X$ are trivial ones, and thus have vanishing cohomology.

But the fundamental group (Higman's group) is nontrivial, so this is not contractible.

I think you should be able to thicken $X$ to a $5$-fold by standard tricks, but I didn't think through the details of this.

---

If, on the other hand, "local system" doesn't mean "finite rank", then $\tilde{H}^q(X, \mathcal{F})=0$ for all $q$, where $X$ is a CW complex, implies $X$ contractible. Suppose that $X$ is such a CW complex. Since $\tilde{H}^0$ with constant coefficients vanishes, $X$ is connected.

I next claim that $X$ is simply connected (this is the only place that I'll use a non-trivial local system). Let $G = \pi_1(X)$. Let $\mathbb{Z}G$ be the free abelian group on $G$, so an element of $\mathbb{Z}G$ is of the form $\sum a_g g$, where the sum is finite. Let $U \subset \mathbb{Z}G$ be $\{ \sum a_g g : \sum a_g = 0 \}$, so we have a short exact sequence of $G$-modules $$0 \to U \to \mathbb{Z}G \to \mathbb{Z} \to 0$$ where the last map sends $\sum a_g g$ to $\sum a_g$. This induces a short exact sequence of locally constant sheaves, which I'll denote by the underlined versions of the $G$-modules.

Since $H^1(X, \underline{U})$ vanishes, $H^0(X, \underline{\mathbb{Z}G}) \to H^0(X, \mathbb{Z}) = \mathbb{Z}$ must be surjective. Now, $H^0(X, \underline{\mathbb{Z}G})$ is the invariants $$(\mathbb{Z}G)^G = \begin{cases} \mathbb{Z} \sum_{g \in G} g & G \ \text{finite} \\ 0 & G \ \text{infinite} \end{cases}.$$ So the image of $H^0(X, \underline{\mathbb{Z}G})$ in $H^0(X, \mathbb{Z})$ is $$\begin{cases} |G| \mathbb{Z} & G \ \text{finite} \\ 0 & G \ \text{infinite} \end{cases}.$$ We see that the only way for the map to be surjectiveis if $|G|=1$, so $X$ is simply connected, as claimed.

Now, using the vanishing of $H^q(X, \underline{Z})$ for higher $q$, we see that all the reduced cohomology groups of $X$ vanish. So, by the universal coefficient theorem, all of the homology groups vanish. So the map $X \to \text{point}$ induces an isomorphism on all homology groups, and $X$ is simply connected. From Corollary 4.33 in Hatcher, that means that the map $X \to \text{point}$ is a homotopy equivalence, so $X$ is contractible.

PS: The reason that I kept saying that $X$ was a CW complex is because that is the assumption in Hatcher.