# Intuitionistic algebraic topology?

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?

• A related question: mathoverflow.net/questions/50971/… – Evan Jenkins Nov 4 '13 at 22:31
• The statement that discrete spaces have trivial cohomology! See golem.ph.utexas.edu/category/2013/07/… . – Qiaochu Yuan Nov 4 '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 4 '13 at 22:40
• 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 4 '13 at 23:02
• You can't prove that every simplicial set is cofibrant in the Kan-Quillen model structure. – Harry Gindi Dec 2 '19 at 7:40

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.
• What's the definition of $H^2$ you are using here? As you pointed out, previously equivalent definitions become inequivalent... – Zhen Lin Nov 4 '13 at 22:59
• 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 5 '13 at 2:06