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1. • CommentRowNumber1.
   • CommentAuthorDavid_Corfield
   • CommentTimeSep 28th 2018
   • (edited Sep 28th 2018)
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   Author: David_Corfield
   Format: MarkdownItexAt the end (at 42:11) of his Voevodsky Memorial [talk](https://www.youtube.com/watch?v=WfDcrN5_1wA) Pierre Deligne poses a question (for the HoTT community?). Given $\alpha: \pi_{n+k}(S^n)$, or an element in $[S^{n+k}, S^n]$ does it induce $[\alpha]: a =_n a \to a =_{n+k} a$, and do so universally? $a =_n a$ for $a:A$ stands for the iterated loop space of $A$, $\Omega^n(A)$. So is he asking whether $\alpha$ induces a map $[[S^n, A], [S^{n+k}, A]]$ (which it surely does), and so a map from $\Omega^n(A)$ to $\Omega^{n+k}(A)$? What's the challenge?

   At the end (at 42:11) of his Voevodsky Memorial talk Pierre Deligne poses a question (for the HoTT community?).

   Given $\alpha: \pi_{n+k}(S^n)$, or an element in $[S^{n+k}, S^n]$ does it induce $[\alpha]: a =_n a \to a =_{n+k} a$, and do so universally?

   $a =_n a$ for $a:A$ stands for the iterated loop space of $A$, $\Omega^n(A)$.

   So is he asking whether $\alpha$ induces a map $[[S^n, A], [S^{n+k}, A]]$ (which it surely does), and so a map from $\Omega^n(A)$ to $\Omega^{n+k}(A)$?

   What’s the challenge?

2. • CommentRowNumber2.
   • CommentAuthorAli Caglayan
   • CommentTimeSep 28th 2018
   • (edited Sep 28th 2018)
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   Author: Ali Caglayan
   Format: MarkdownItexThe maps of fundamnetal groups are truncated to the set level. I think he is asking if the truncation can be rid of. Can this be extended to a map of untruncated loop spaces.

   The maps of fundamnetal groups are truncated to the set level. I think he is asking if the truncation can be rid of. Can this be extended to a map of untruncated loop spaces.

3. • CommentRowNumber3.
   • CommentAuthorMike Shulman
   • CommentTimeSep 28th 2018
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   Author: Mike Shulman
   Format: MarkdownItexThere was some confusion at the meeting about this too. It's already an interesting question at the level Ali interpreted it; but it seems that Deligne intended his "$\pi_{n+k}(S^n)$" to refer to the homotopy groups of spheres as defined in ZFC. In other words, if we know *classically* that we have some element of a homotopy group of spheres, can we use that to define *in type theory* a corresponding map on loop spaces.

   There was some confusion at the meeting about this too. It’s already an interesting question at the level Ali interpreted it; but it seems that Deligne intended his “$\pi_{n+k}(S^n)$” to refer to the homotopy groups of spheres as defined in ZFC. In other words, if we know classically that we have some element of a homotopy group of spheres, can we use that to define in type theory a corresponding map on loop spaces.

4. • CommentRowNumber4.
   • CommentAuthorDavid_Corfield
   • CommentTimeSep 30th 2018
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   Author: David_Corfield
   Format: MarkdownItexAnd what can we say to the question as posed in #2 and #3? Is there something one can 'reasonably' ask for, that can't be delivered?

   And what can we say to the question as posed in #2 and #3? Is there something one can ’reasonably’ ask for, that can’t be delivered?

5. • CommentRowNumber5.
   • CommentAuthorMike Shulman
   • CommentTimeSep 30th 2018
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   Author: Mike Shulman
   Format: MarkdownItexIf it's "reasonable" to ask for a map $\pi_{n+k}(S^n) \to \Omega^n X \to \Omega^{n+k} X$, then yes; I don't think we can deliver that. But even in ZFC such a map doesn't really exist canonically. You can choose a representative of any $\alpha \in \pi_{n+k}(S^n)$, which in turn induces a map $\Omega^n X \to \Omega^{n+k} X$, and equivalent representatives of $\alpha$ induce homotopic maps, but not *uniquely* homotopic ones. So the axiom of choice can give you a particular map, but it's not canonically determined.

   If it’s “reasonable” to ask for a map $\pi_{n+k}(S^n) \to \Omega^n X \to \Omega^{n+k} X$, then yes; I don’t think we can deliver that. But even in ZFC such a map doesn’t really exist canonically. You can choose a representative of any $\alpha \in \pi_{n+k}(S^n)$, which in turn induces a map $\Omega^n X \to \Omega^{n+k} X$, and equivalent representatives of $\alpha$ induce homotopic maps, but not uniquely homotopic ones. So the axiom of choice can give you a particular map, but it’s not canonically determined.