# Artin vanishing for Stein manifolds and restriction maps

In the setting of complex Stein manifolds $$X$$ of complex dimension $$d$$, the theorem of Andreotti--Frankel implies the vanishing of the singular cohomology group $$H^i(X,\mathbb Z)=0$$ for $$i>d$$. With complex coefficients, a simple argument for this is to compute the cohomology in terms of the cohomology of the de Rham complex. Their theorem gives a more precise Morse-theoretic statement.

Now let $$U\subset X$$ be another Stein manifold open in $$X$$, and assume that the map $$\mathcal O(X)\to \mathcal O(U)$$ has dense image. (This condition is not automatic, and necessary for the following.)

Conjecture. In top degree, the map $$H^d(X,\mathbb Z)\to H^d(U,\mathbb Z)$$ is surjective.

Is this known? The same result should also be true with constructible coefficients. With $$\mathbb C$$-coefficients, it follows from the comparison with de Rham cohomology (at least when cohomology groups are finite-dimensional, which I'm happy to assume). Is there some argument using Morse theory?

The motivation for the question is that the analogue in rigid-analytic geometry is true (but I found it quite surprising); it is essentially equivalent to a version of Artin vanishing for affine schemes over absolutely integrally closed valuation rings stated by Gabber in Oberwolfach last year.

• @JasonStarr An example would be a small open ball inside affine space Sep 2, 2021 at 11:21
• What topology are you using? For the open unit disk in the complex plane, the holomorphic restriction of $1/(1-z)$ is certainly not approximated by restrictions of entire holomorphic functions in the uniform topology (since these are bounded on the closure of the disk). Sep 2, 2021 at 11:40
• I'm using the Frechet topology on $\mathcal O(X)$ and $\mathcal O(U)$ (of uniform convergence on compact subsets), as I believe is the standard choice. (Sorry for not specifying.) Sep 2, 2021 at 11:42
• Interesting question. The inclusion of $\mathbb{C}^*\subset \mathbb{C}$ is an open immersion of Stein manifolds where the map on $H^d$ is not surjective, but I'm not sure about the density condition for function spaces. Sep 2, 2021 at 12:53
• It doesn't have dense image: One argument is to observe that the residue at $0$ of a function on $\mathbb C^\times$ gives a nonzero continuous functional which vanishes on the image of $\mathcal{O}(\mathbb C)$. The same argument applies more generally to hyperplane complements. Sep 2, 2021 at 12:56