# On the ordered set of real numbers, does sheaf+cosheaf imply constant?

I have a technical question about unbounded chain complexes. I couldn't think of a descriptive title for it.

Let $P$ be a chain complex of contravariant functors on $\mathbf{R}$ (the real numbers regarded as an ordered set), valued in the category of abelian groups.

Suppose that for every real number $s$, both of the maps $$P(s) \to \varprojlim_{t < s} P(t), \qquad P(s) \leftarrow \varinjlim_{t > s} P(t)$$ are quasi-isomorphisms, where the lims indicate homotopy limit and colimit. Does it follow that all of the maps $P(s_1) \to P(s_2)$ are quasi-isomorphisms?

When the $P(s)$ have homology groups that are uniformly bounded above, this is a result of Kashiwara. It plays a key role in the foundations of microlocal sheaf theory, especially for sheaves that are not necessarily constructible.

Dima Tamarkin showed me how to remove the boundedness assumption, if you replace abelian groups by $k$-vector spaces for a field $k$. In that case the abelian category of presheaves of vector spaces on $\mathbf{R}$ has homological dimension $1$, so the derived category has a simple structure.

I'm interested in the case where $P$ takes values in the $\infty$-category of spectra, where you can formulate the same question, but already I don't know what to expect for abelian groups.

• Thanks Dmitri. I think I haven't understood your criterion for isomorphism of spectra yet, in the second paragraph. But before I dig in let me check something: in the third paragraph you take a supremum over a set of numbers $r$ that obey $r > q \geq t$, which implies $r > t$. Then in parentheses you argue that this set of numbers is nonempty because it contains $r = t$. Is that a problem? – David Treumann Mar 15 '16 at 15:49