## July 28, 2013

### Cohomology Detects Failures of Classical Mathematics

#### Posted by Mike Shulman Ordinarily I try to avoid cross-posting between here and the Homotopy Type Theory blog, but this remark seems probably of interest to the crowd here as well. After my recent post there about how to define cohomology in HoTT, Zhen Lin kindly pointed me to a very nice paper of Andreas Blass called Cohomology detects failures of the axiom of choice. A main result of this paper is that the following two statements are equivalent (over ZF):

• The axiom of choice.
• For any discrete set $X$ and any (not necessarily abelian) group $G$, the cohomology $H^1(X;G)$ vanishes.

The fact that discrete spaces have no higher cohomology is “obvious” in classical algebraic topology, but this shows that if we want to do homotopy theory “constructively” then we need to give up that expectation. In this post I want to explain one of Blass’s proofs from a geometric point of view, and then consider what cohomology has to say about the law of excluded middle.

Here is one of Blass’s proofs that the vanishing of the cohomology of all sets implies AC. Let $p:Y\to X$ be a surjection of sets, for which we want to find a section.

Suppose first that every fiber $Y_x = p^{-1}(x)$ is isomorphic to some fixed nonempty set $S$. Of course, if we were given specified isomorphisms $\phi_x:S \cong Y_x$ for all $x$, then we could fix a particular element $s\in S$ and define $f:X\to Y$ by $f(x) = \phi_x(s)$, but we are only assuming that such isomorphisms exist. (In HoTT language, they “merely” exist.)

In this special case, from a geometric point of view, we can consider $Y$ as a locally trivial bundle over $X$ with fiber $S$, and we know that such bundles are classified by the cohomology of the base with coefficients in the automorphism group of the fiber. Thus, if we let $G = Aut(S)$, we should be able to construct a “characteristic class” of $Y$ in $H^1(X;G)$.

The way to do this depends on the definition of $H^1$. Blass points out that many of the classical definitions are not correct in the absence of choice (his criterion of “correctness” being that they give rise to long exact sequences). He takes the route of simply defining $H^1(X;G)$ to be the set of isomorphism classes of $G$-torsors over $X$.

From a category-theoretic (or HoTT) point of view, it’s more natural to define $H^1(X;G)$ to be the set of natural isomorphism classes of functors $X\to \mathbf{B}G$, where $\mathbf{B}G$ is the one-object groupoid corresponding to $G$. It’s again easy to construct our characteristic class in this case, since for $G=Aut(S)$, the groupoid $\mathbf{B}G$ is equivalent to the groupoid $\mathbf{B}'G$, whose objects are sets isomorphic to $S$ and whose morphisms are all isomorphisms between them. Then $x\mapsto Y_x$ defines a functor from $X$ to $\mathbf{B}'G$, hence also to $\mathbf{B}G$.

(In ZF, this equivalence of groupoids is only a “weak equivalence”, i.e. there is a functor $\mathbf{B}G \to \mathbf{B}'G$ that is essentially surjective and fully faithful. Thus, our “functor” $X\to \mathbf{B}G$ is only an anafunctor, so we have to use those in the definition of cohomology. In HoTT, however, this is not a problem.)

Now by assumption, $H^1(X;G)=0$, so this characteristic class is trivial. This means that our functor $Y: X\to \mathbf{B}'G$ is naturally isomorphic to the one that is constant at $S$. Such a natural transformation assigns to each $x\in X$ an isomorphism $S\cong Y_x$, and as we remarked above this is sufficient to define a section of $p$.

This handles the special case when all the fibers of $p$ are isomorphic. In order to deal with the general case, Blass uses a trick: he defines $U$ to be the set of nonempty finite sequences of elements of $Y$, with $q:U\to X$ the function which selects the first element of a sequence and then maps it down to $X$ via $p$. Then the fiber $U_x$ is a subset of $U$, but there also exists an injection $U\hookrightarrow U_x$ that adjoins some fixed element of $Y_x$ at the front of each sequence. By the Schroeder-Bernstein theorem, therefore, $U\cong U_x$, and so we can apply the previous case to get a section of $q$, and hence (by selecting first elements) a section of $p$.

It’s important, of course, that the Schroeder-Bernstein theorem does not depend on the axiom of choice. It does, however, depend on the law of excluded middle, so we might wonder, is Blass’s theorem still valid constructively?

Now by a classical result of Bishop and Diaconescu, the axiom of choice implies the law of excluded middle. I will explain the proof in a moment; the crucial thing to note is that it only uses AC to obtain a section of a surjection all of whose fibers are isomorphic. Since the first part of Blass’s argument (when all fibers are isomorphic) seems perfectly constructive, it will therefore also imply LEM, so that the second part of the argument can freely use Schroeder-Bernstein.

Why does AC imply LEM? Suppose $P$ is a proposition for which we want to show “$P$ or not $P$”. Let $\Sigma P$ be the quotient of $2 = \{N,S\}$ by the equivalence relation which identifies $N\sim S$ exactly when $P$ holds. (In reproducing this proof in HoTT, Spitters and Rijke have observed that $\Sigma P$ is exactly the homotopy-theoretic suspension of $P$, hence the notation.) Then $p:2 \to \Sigma P$ is a surjection. If it has a section $f:\Sigma P \to 2$, then we can consider $f([N])$ and $f([S])$, each of which is either $N$ or $S$. This gives four cases, two of which easily imply $P$, and the other two of which easily imply “not $P$”.

Now what are the fibers of $p$? Every element of $\Sigma P$ is either $[N]$ or $[S]$ (though this is not necessarily an exclusive-or), so it suffices to show that the fibers over $[N]$ and $[S]$ are isomorphic. The fiber over $[N]$ always contains $N$, and it contains $S$ iff $P$. Similarly, the fiber over $[S]$ always contains $S$, and it contains $N$ iff $P$. Thus, the two are isomorphic by an isomorphism that interchanges $N$ and $S$.

Thus we recover, even constructively, Blass’s theorem that the vanishing of $H^1$ of sets is equivalent to the axiom of choice. Blass’s original theorem was actually sharper: the vanishing of $H^1(X;G)$ for all $G$ and a particular set $X$ is equivalent to projectivity of $X$. Can this sharper version also be obtained constructively? I don’t know.

Regardless, I think it’s quite cute that we can connect geometric ideas like characteristic classes of fiber bundles in cohomology with logical ideas like the axiom of choice. In some sense, this is what topos theory is supposed to be all about.

Posted at July 28, 2013 5:43 AM UTC

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### Re: Cohomology Detects Failures of Classical Mathematics

Nice! Are there structures other than discrete sets whose cohomology fails to behave as expected when the axiom of choice doesn’t hold? For example, algebraic gadgets of some sort?

I would imagine so, but my memory only conjures forth the Whitehead problem, “is every abelian group $A$ with $Ext^1(A,\mathbb{Z}) = 0$ free?” This sounds vaguely choice-ish, since it’s about surjections and whether they split: it’s equivalent to “if every surjection $f: B \to A$ with kernel $\mathbb{Z}$ splits, is $A$ free?” But it’s independent of ZFC and related to more subtle axioms of set theory.

So, now I want to know quite generally: “What axioms of set theory does cohomology detect?”

Cohomology can be regarded as the art of counting holes, so we should use it to locate holes in the foundations of mathematics. Posted by: John Baez on July 29, 2013 4:39 AM | Permalink | Reply to this

### Re: Cohomology Detects Failures of Classical Mathematics

Perhaps the cohomology of a classifying topos of some of the set theory axioms could do it…

Posted by: David Roberts on July 29, 2013 6:24 AM | Permalink | Reply to this

### Re: Cohomology Detects Failures of Classical Mathematics

Talking of Whitehead, shouldn’t his theorem be relevant to this post. As Mike once said

Now I think we can think of Whitehead’s theorem as a “classicality” axiom which can fail in $(\infty,1)$-topoi, akin to the law of excluded middle and the axiom of choice (which can already fail in 1-topoi).

Anything cohomological there?

Posted by: David Corfield on July 29, 2013 7:35 AM | Permalink | Reply to this

### Re: Cohomology Detects Failures of Classical Mathematics

Yeah, I’ve also been wondering whether the failure of Whitehead’s theorem could be detected cohomologically. However, it seems a bit unlikely, because cohomology is basically the homotopy groups of mapping spaces, while the failure of Whitehead’s theorem means precisely that there are things that can’t be detected by homotopy groups. (-: However, it’s conceivable that there’s some way to massage a failure of Whitehead’s theorem into some other invariant, that could be detected by homotopy groups.

Posted by: Mike Shulman on July 30, 2013 6:54 PM | Permalink | Reply to this

### Re: Cohomology Detects Failures of Classical Mathematics

Good question! Here’s another example that came up recently in another thread. Classically, $Ext^n(A,B)$ for abelian groups $A$ and $B$ vanishes when $n\gt 1$, but it seems unlikely to me that this will be the case without choice. Of course there is the question of the correct definition of $Ext^n(A,B)$. My current inclination is that the correct definition in homotopy type theory will be $Ext^n(A,B) = \pi_{-n} Hom_{H\mathbb{Z}}(H A, H B).$ That is, the $Ext$ groups are the (cohomologically reindexed) homotopy groups of a mapping spectrum of $H\mathbb{Z}$-module spectra. (Here $H A$, for an abelian group $A$, denotes the Eilenberg-Mac Lane spectrum with $(H A)_n = K(A,n)$.) This is correct in classical mathematics, although many people might find it somewhat perverse as a definition.

To identify it with the more common definitions, we use the fact that $H\mathbb{Z}$-modules are equivalent to unbounded chain complexes of abelian groups. Then the facts that free abelian groups are projective, and subgroups of free abelian groups are free abelian, imply that every abelian group has a 2-term projective resolution, causing $Ext^n(A,B)=0$ for $n\gt 1$. However, it seems unlikely to me that any of these facts is true constructively.

There is a slight obstacle in that we don’t currently know how to define $H\mathbb{Z}$-modules or their mapping spaces in homotopy type theory—they have the same sort of infinity problems as simplicial types, $(\infty,1)$-categories, etc. We can, however, define some closely related groups $Ext^n_{\mathbb{S}}(A,B) = \pi_{-n} Hom(H A, H B)$ by taking the ordinary mapping spectrum (ignoring the $H\mathbb{Z}$-module structures that we don’t know how to define). These groups are called “$Ext$ over the sphere spectrum”.

There are equivalent definitions by classification. Namely, $Ext^n_{\mathbb{S}}(A,B)$ classifies fiber sequences of spectra $\Sigma^{n-1} (H B) \to E \to H A$ while $Ext^n_{H\mathbb{Z}}(A,B)$ (the usual $Ext$) classifies analogous fiber sequences of $H\mathbb{Z}$-modules. When $n=1$, this ought to reproduce the usual definition of $Ext^1(A,B)$ in terms of extensions of abelian groups, but probably recovering the version of $Ext^n(A,B)$ for $n\gt 1$ in terms of “iterated extensions” requires identifying $H\mathbb{Z}$-modules with chain complexes.

Posted by: Mike Shulman on July 30, 2013 6:48 PM | Permalink | Reply to this

### Re: Cohomology Detects Failures of Classical Mathematics

This is a very naive question, but I vaguely remember that, in the infinite dimensional context, Drinfeld has suggested that the usual definition of projective is “wrong”, and that is the source of a lot of the set-theoretic oddities. Instead, one should use flat and Mittag-Leffler. Would changing the definition in this way have any impact on the result here (presumably defining exts in terms of resolutions of this new definition of projective)?

Posted by: Aaron Bergman on July 31, 2013 1:39 AM | Permalink | Reply to this

### Re: Cohomology Detects Failures of Classical Mathematics

I’m not sure exactly what you mean by “infinite-dimensional”, but my instinct is no. When you say “the result here” what exactly are you referring to? Blass’s theorem that cohomology detects failures of LEM and AC, or the discussion in the other comments of Ext?

One of the problems with defining (co)homological concepts in terms of resolutions constructively is that such resolutions may not exist. For instance, every abelian group still has a resolution by free ones, but free abelian groups are no longer necessarily projective, so this resolution doesn’t have the properties you want to get a good definition of cohomology. Can you show constructively that everything has a resolution by flat and Mittag-Leffler things? Do they have the behavior you want for cohomology? (In particular, does such a definition give you long exact sequences?)

Posted by: Mike Shulman on July 31, 2013 4:00 AM | Permalink | Reply to this

### Re: Cohomology Detects Failures of Classical Mathematics

I’ll admit that I haven’t had a chance to read the post in detail, and this stuff is fairly far out of my (former) areas of expertise – hence the disclaimer – but I was thinking back to this old mathoverflow answer which refers to homological dimension depending on the continuum hypothesis. This is a far cry from a dependence on the continuum hypothesis, but it did make me a bit curious.

Posted by: Aaron on July 31, 2013 4:41 AM | Permalink | Reply to this

### Re: Cohomology Detects Failures of Classical Mathematics

Interesting! I wonder whether free modules are constructively flat and Mittag-Leffler and whether they give a useful theory of resolutions. I don’t have time to work it out myself, though.

Posted by: Mike Shulman on July 31, 2013 5:52 PM | Permalink | Reply to this

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