# Exactness is often an open condition. How often?

Let $\mathcal C$ denote an abelian category. Consider a formal $\mathbb Z$-graded object $V_\bullet$ in $\mathcal C$, by which I simply mean a $\mathbb Z$-indexed list $n\mapsto V_n$ of objects. A differential on $V_\bullet$ is a sequence of maps $\partial : V_n \to V_{n-1}$ such that $\partial^2 = 0$; for example, the sequence of all $0$s works. A differential is exact if for every $n$, $\ker \partial|_{V_n} = \partial(V_{n+1})$. Clearly, for fixed $V_\bullet \in \mathcal C$, the collection of differentials is the intersection of some quadrics in some abelian group. I have less of an idea of what type of space the exact differentials constitute, which is one of the motivations for my question.

Thus, fix a $\mathbb Z$-graded object $V_\bullet \in \mathcal C$. Suppose that you fix also an exact differential $\partial_0$. There are many situations where one is interested in "varying" $\partial_0$ to some new differential $\partial$. One situation that arises is the "formal deformation": we can formally extend $\mathcal C$ to a new category $\mathcal C[\![\hbar]\!]$ by tensoring every hom set with $\mathbb Z[\![\hbar]\!]$, and then consider differentials $\partial$ on $V_\bullet[\![\hbar]\!]$ such that $\partial = \partial_0 \mod\hbar$. I can also imagine more algebrogeometric contexts: I expect that there is a natural way to give the hom spaces in an abelian category, and thus the space of differentials, the structure of affine algebraic varieties; then I could probe this space by, say, local Artinian rings. I could also, of course, probe non-infinitesimal neighborhoods.

In any case, one should expect that if $\partial_0$ is exact that $\partial$ is "close enough" to $\partial_0$, then $\partial$ is also exact. I know how to prove a special case of this. Say that $\mathcal C$ has the Axiom of Choice if every epimorphism in $\mathcal C$ splits. Then, since $\mathcal C$ is abelian, it must also have the Axiom of "co-Choice", and one can always choose a deformation retraction of any complex onto its homology, for example, and also one can always choose an isomorphism between a (Hausdorff exhaustive) filtered object and its associated graded. Under these conditions, I can prove that formal deformations of exact differentials remain exact. Another somewhat trivial example is when $V_\bullet = (V_0,V_1)$ is two terms long. Then an exact differential is an isomorphism $\partial_0 : V_1 \overset\sim\to V_0$. Consider the situation when $\mathcal C$ is, say, the category of finite-dimensional vector spaces. Then the space of isomorphisms is naturally a Zariski-open (and hence analytically open, if we are over $\mathbb R$) subset of the space of all maps.

Question: Is there a way to give the the space of differentials on $V_\bullet$ a natural (Zariski, say, but I'm agnostic) topology, perhaps under some conditions on $\mathcal C$, and if so, how? Under what conditions does the slogan "exactness is an open condition" hold? As a special case, does it hold for "infinitesimal" (either in the local-Artinian or $\hbar$ sense) deformations in the absence of Choice?

A further version of my question recognizes the fact that many categories are naturally enriched in topological abelian groups; categories of Banach spaces, for example. Then one can also ask: when is exactness an open condition for "analytic" topologies?

Lest you say that exactness is always an open condition, let me end with a trivial parting comment. I could, if I so chose, give every space the indiscrete topology. For this topology, exactness certainly is not open, as deforming all the way to $\partial = 0$ would be a "small" deformation.

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The set of complexes of vector spaces of given finite dimension vector "are" an affine space, and the dimension of their cohomology groups is a lower (upper? I never know which) semi-continuous function. Since it is possitive, that gives your continuity in that case. – Mariano Suárez-Alvarez Feb 10 '13 at 6:07
Regarding your parting comment, I think it is reasonable to require that the zero element be closed in any topologized hom space. – S. Carnahan Feb 10 '13 at 9:13
Mariano, I think you mean affine variety, with equations $d^2=0$. – Allen Knutson Feb 10 '13 at 11:25
"I expect that there is a natural way to give the hom spaces in an abelian category, and thus the space of differentials, the structure of affine algebraic varieties;" I am not sure if this works without any assumption on the abelian category. – Martin Brandenburg Feb 10 '13 at 15:13
@Mariano: Right, for finite-dimensional vector spaces I know how to do everything I've asked. Already for infinite-dimensional vector spaces, though, I know how to prove semicontinuity for "infinitesimal" open neighborhoods, but I don't know enough about infinite-dimensional vector spaces to say things finitely... – Theo Johnson-Freyd Feb 12 '13 at 5:11