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Are there results in algebraic topology -- preferably relating to homology or homotopy or phraseable in simplicial sets -- that are not true in an intuitionistic logic?

In other words, are there results that crucially rely on a law of the excluded middle, or on the axiom of choice?

Mikael Vejdemo-Johansson
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    A related question: http://mathoverflow.net/questions/50971/how-to-make-ext-and-tor-constructive – Evan Jenkins Nov 04 '13 at 22:31
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    The statement that discrete spaces have trivial cohomology! See http://golem.ph.utexas.edu/category/2013/07/cohomology_detects_failures_of.html . – Qiaochu Yuan Nov 04 '13 at 22:32
  • There are even questions which are equivalent to the continuum hypothesis, to the generalized version, and to large cardinal axioms! – Fernando Muro Nov 04 '13 at 22:40
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    I've been thinking about the same question recently! Here's a very trivial thing that goes wrong in $\mathbf{sSet}$ when you don't have AC: not all monomorphisms have the left lifting property with respect to trivial Kan fibrations. See here. (But some may argue that it's the definition of trivial Kan fibration that should be changed when AC fails.) – Zhen Lin Nov 04 '13 at 23:02
  • You can't prove that every simplicial set is cofibrant in the Kan-Quillen model structure. – Harry Gindi Dec 02 '19 at 07:40

2 Answers2

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Most of our computational tools in homological algebra break pretty badly if you don't assume surjections in $\operatorname{Set}$ split. Here's an illustration: given a group $G$ and a $G$-module $A$, we can identify (without using choice) $H^2(G; A)$ with extensions of $G$ by $A$. But $A$-valued 2-cocycles on $G$ only describe extensions $0 \to A \to E \to G \to 1$ where the surjection splits as a map of sets! In essence, removing choice adds another level to the problem of classification of extensions: first, we must classify extensions of $G$ by $A$ as sets, and then we must classify the compatible group structure to put on top of each such set extension. The same issue arises if you want to compute cohomology of Lie algebras over $\mathbb{Z}$: there are two extension problems that must be solved concurrently.

Evan Jenkins
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    What's the definition of $H^2$ you are using here? As you pointed out, previously equivalent definitions become inequivalent... – Zhen Lin Nov 04 '13 at 22:59
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    Sometimes it is sufficient to have enough projectives in the topos $Set$. In fact, removing LEM or AC means you probably want to work with analogues of sheaf cohomology. – David Roberts Nov 05 '13 at 02:06
  • The context I'm working on is, to be more exact, what happens if I do topology internal to a particular topos of sheaves. – Mikael Vejdemo-Johansson Nov 05 '13 at 21:51
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Without AC, it is impossible to prove that every set is equipotent to an ordinal (in ZF !) and it is impossible to prove that a functor is an equivalence of categories if and only if it is full faithful and essentially surjective ; so I guess that a lot of things are wrong without AC, in particular concerning combinatorial model categories because very often we have to use transfinite cardinals. For example, "Implications of large-cardinal principles in homotopical localization" or "Definable orthogonality classes in accessible categories are small" for links between large cardinal axioms and Bousfield localization.

Philippe Gaucher
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  • " it is impossible to prove that a functor is an equivalence of categories if and only if it is full faithful and essentially surjective " - hence anafunctors :-) – David Roberts Nov 05 '13 at 22:54