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Yesterday, I read the following question : Finding a space such that $X\cong X+2$ and $X\not\cong X+1$, and found the result very astounding.

Does it still hold in homotopy type theory? As a type theorist, I don't really understand the answer…

I tried to redo the proof, translating it in type-theoretic words, but and I don't seem to find a description of Stone-Čech compactification easily translatable to HoTT.

Is there an analogous of SČ-compactification in homotopy type theory (or at least, a description of $\beta\mathbb N$)?

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Kevin Quirin
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  • Correct me if I'm wrong, but in HoTT a type is simply a space, and equivalence of types is simply homotopy equivalence of spaces, right? So what makes you think that result holds "in topology" but not "in HoTT"? – Najib Idrissi Dec 17 '15 at 08:49
  • @NajibIdrissi Actually, a type is more a infinity-groupoid than a space, so some results true in usual topology might be false in HoTT, or at least need a generalization – Kevin Quirin Dec 17 '15 at 10:09
  • I've always been taught that an $\infty$-groupoid is "the same thing" as a space (there's a Quillen equivalence, so from the point of view of homotopy there's not much difference). Can you give an example of a result (of a similar nature as this one) holding in topology but not in HoTT? – Najib Idrissi Dec 17 '15 at 10:12
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    @KevinQuirin - You wrote "$X+1\not\simeq1$ and $X+2\simeq X$", but I suspect you mean "$X+1\not\simeq X$ and $X+2\simeq X$". – Pierre-Yves Gaillard Dec 17 '15 at 10:17
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    @Najib See https://groups.google.com/forum/#!topic/HomotopyTypeTheory/Wf-H6rH155g for several thoughtful comments on the "space vs. homotopy type" issue. – Ingo Blechschmidt Dec 17 '15 at 11:38
  • @IngoBlechschmidt Ah, I see the problem. I assumed that the question linked at the top was using $\cong$ to denote homotopy equivalence (a rather stupid assumption to be honest, I was probably confused by the context), and that they had constructed a space $X$ such that $X$ was htpy.eq. to $X+2$ but not $X+1$. My mistake. My point (and it's now in Mike Shulman's answer) was that this question is nothing special about HoTT, it's simply a question about spaces and homotopy equivalences of spaces. – Najib Idrissi Dec 17 '15 at 16:21

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The question you linked to is, if I understand it correctly, asking about homeomorphism of topological spaces (that's what the symbol $\cong$ usually means). Types in HoTT behave like $\infty$-groupoids, which can (classically) be presented by topological spaces, but only up to homotopy equivalence. So the first question is whether for the example spaces listed, it might still be the case that $X+1\simeq X$ even though $X+1\not\cong X$. I don't know. However, if it does happen that $X+1\not\simeq X$ as well, then it will automatically be true in the "standard" model of HoTT (where types are interpreted by classical $\infty$-groupoids) that such a type exists. However, that is a very different thing from saying that such a type can be constructed inside the formal system of HoTT. In particular, the Stone-Cech compactification is an operation on topological spaces, not on $\infty$-groupoids, so there is no obvious way to give it meaning inside HoTT.

Mike Shulman
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  • My bad, I thought at first that the original question was about homotopy equivalence, and then realised my mistake. However, I don't seem to find a reason for such a type not to exist, and my only hint so far is to check whether the homeomorphisms presented can be changed into homotopy equivalences – Kevin Quirin Dec 17 '15 at 17:46
  • @KevinQuirin A homeomorphism is a particular case of homotopy equivalence. What you want to show is that there is no homotopy equivalence $X+1 \simeq X$ (so far all you know is that there's no homeomorphism), not that the homeomorphism $X \cong X+2$ "can be changed into a homotopy equivalence" (which is obvious since it's already one). – Najib Idrissi Dec 18 '15 at 09:42