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1. • CommentRowNumber1.
   • CommentAuthorMichael_Bachtold
   • CommentTimeMay 13th 2015
   • (edited May 13th 2015)
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   Author: Michael_Bachtold
   Format: MarkdownItexFollowing [this discussion](http://nforum.mathforge.org/discussion/6583/differential-operators-as-cokleisli-morphisms-for-the-jet-comonad/?Focus=53301#Comment_53301), I take the opportunity to articulate some confusions I have, regarding the notions of path and identity in the context of cohesion. Contrary to the situation in "bare" homotopy type theory, it seems to me that "path" and "identity" no longer mean the same for cohesive types. It's maybe easiest to ask some concrete questions (apologies if these are completely misguided): 1. If I consider a manifold as a cohesive type, for concreteness the unit circle $S^1$. Is it right that there are no non-trivial identities between points of $S^1$? So $S^1$ is an h-set? If I had considered the real line $R$, the same applies and moreover $R\neq *$? 2. Is it right that there are plenty of non-trivial paths in $R$, which it's safe to think of as maps $[0,1]\to R$? 3. If I consider the [[shape modality]] of $R$, $\int R$. Am I right that this is still an h-set, which is neither isomorphic to $R$ nor to a point? I concluded this from what David wrote: >As with the shape modality, identifying path connected points in a space, one retains the information of different identifications, and identifications between identifications, etc. So in particular, since I retain the information of _different_ identifications (or paths), there should be a way to distinguish between a constant path and a non-trivial path homotopic to it. This confuses me, since it seems to contradict univalence. On the other hand, if I think of paths and identites as different notions, and substitute path for identification in the above it seems to make sense. On the other hand Urs wrote: >think of $\Pi_{inf}X \coloneqq \Im(X)$ as the result of forming a new groupoid from the groupoid $X$ by adding in further isomorphisms between all objects (points of $X$) that are infinitesimal neighbours. These isomorphisms are hence infinitesimal paths. Here you say isomorphism are paths, which makes me think that paths are identities after all...

   Following this discussion, I take the opportunity to articulate some confusions I have, regarding the notions of path and identity in the context of cohesion. Contrary to the situation in “bare” homotopy type theory, it seems to me that “path” and “identity” no longer mean the same for cohesive types.

   It’s maybe easiest to ask some concrete questions (apologies if these are completely misguided):

   1. If I consider a manifold as a cohesive type, for concreteness the unit circle $S^1$. Is it right that there are no non-trivial identities between points of $S^1$? So $S^1$ is an h-set? If I had considered the real line $R$, the same applies and moreover $R\neq *$?

   2. Is it right that there are plenty of non-trivial paths in $R$, which it’s safe to think of as maps $[0,1]\to R$?

   3. If I consider the shape modality of $R$, $\int R$. Am I right that this is still an h-set, which is neither isomorphic to $R$ nor to a point?

   I concluded this from what David wrote:

   As with the shape modality, identifying path connected points in a space, one retains the information of different identifications, and identifications between identifications, etc.

   So in particular, since I retain the information of different identifications (or paths), there should be a way to distinguish between a constant path and a non-trivial path homotopic to it. This confuses me, since it seems to contradict univalence. On the other hand, if I think of paths and identites as different notions, and substitute path for identification in the above it seems to make sense.

   On the other hand Urs wrote:

   think of $\Pi_{inf}X \coloneqq \Im(X)$ as the result of forming a new groupoid from the groupoid $X$ by adding in further isomorphisms between all objects (points of $X$) that are infinitesimal neighbours. These isomorphisms are hence infinitesimal paths.

   Here you say isomorphism are paths, which makes me think that paths are identities after all…

2. • CommentRowNumber2.
   • CommentAuthorUrs
   • CommentTimeMay 13th 2015
   • (edited May 13th 2015)
   • PermaLink
   Author: Urs
   Format: MarkdownItexOne issue, beyond my control, is that it is a bad idea to speak in bare homotopy type theory of paths. This comes from thinking, by default, of bare homotopy types as being the shape of topological spaces, which is good for some purposes but is a conceptual trick that in the hands of the unwary may backfire. Also, there is long established terminology for the non-paths in bare homotopy types, namely they are the morphisms of an $\infty$-groupoid. And it being an $\infty$-groupoid, they are all (higher) isomorphisms. With that out of the way, here are the replies to your questions: 1. The circle regarded as $S^1 \in SmoothMfd \hookrightarrow Smooth \infty Grpd$ is a 0-truncated cohesive homotopy type, in that, indeed, it does not have any nontrivial morphisms inside it, it is an h-set, yes, for emphasis: a _cohesive h-set_. Same for $\mathbb{R}^1 \in SmoothMfd \hookrightarrow Smooth \infty Grpd$ and $\mathbb{R}^1 \neq \ast$ in $Smooth \infty Grpd$, yes. 2. Yes. 3. Not quite. So $\int S^1 \simeq B \mathbb{Z}$ in $Smooth \infty Grpd$, where on the right hand side we have the homotopy type of the circle , which is the one that all presently available HoTT literature means when it says "circle". So $\int S^1$ is a cohesive homotopy 1-type that happens to be a geometrically discrete homotopy 1-type. Notice that the _morphisms_ in $\int S^1$ are the _geometric paths_ in $S^1$. Generally $\int$ may be thought of as producing fundamental path $\infty$-groupoids. Generally, for $X \in SmoothMfd \hookrightarrow Smooth \infty Grpd$, then $\int X$ is the geometrically discrete homotopy type which is presented by the topological space underlying $X$. Let me know if this clears up the remainder of what you wrote.

   One issue, beyond my control, is that it is a bad idea to speak in bare homotopy type theory of paths. This comes from thinking, by default, of bare homotopy types as being the shape of topological spaces, which is good for some purposes but is a conceptual trick that in the hands of the unwary may backfire.

   Also, there is long established terminology for the non-paths in bare homotopy types, namely they are the morphisms of an $\infty$-groupoid. And it being an $\infty$-groupoid, they are all (higher) isomorphisms.

   With that out of the way, here are the replies to your questions:

   1. The circle regarded as $S^1 \in SmoothMfd \hookrightarrow Smooth \infty Grpd$ is a 0-truncated cohesive homotopy type, in that, indeed, it does not have any nontrivial morphisms inside it, it is an h-set, yes, for emphasis: a cohesive h-set.

      Same for $\mathbb{R}^1 \in SmoothMfd \hookrightarrow Smooth \infty Grpd$ and $\mathbb{R}^1 \neq \ast$ in $Smooth \infty Grpd$, yes.

   2. Yes.

   3. Not quite. So $\int S^1 \simeq B \mathbb{Z}$ in $Smooth \infty Grpd$, where on the right hand side we have the homotopy type of the circle , which is the one that all presently available HoTT literature means when it says “circle”. So $\int S^1$ is a cohesive homotopy 1-type that happens to be a geometrically discrete homotopy 1-type.

      Notice that the morphisms in $\int S^1$ are the geometric paths in $S^1$. Generally $\int$ may be thought of as producing fundamental path $\infty$-groupoids.

      Generally, for $X \in SmoothMfd \hookrightarrow Smooth \infty Grpd$, then $\int X$ is the geometrically discrete homotopy type which is presented by the topological space underlying $X$.

   Let me know if this clears up the remainder of what you wrote.

3. • CommentRowNumber3.
   • CommentAuthorMike Shulman
   • CommentTimeMay 13th 2015
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   Author: Mike Shulman
   Format: MarkdownItexRight. A general cohesive homotopy type may have both *paths* (in the topological sense) and *identifications*, and two points may be connected by a path but not identified. The shape modality *makes paths into new identifications* (but also keeps the old identifications around). It does, however, happen to be the case that the shape of the (Dedekind) *real numbers* is a point (and in particular an h-set), because the real numbers are *topologically* (cohesively) contractible, in the sense that any two points are connected by a path (a map $[0,1]\to \mathbb{R}$), any two paths (in this sense) are homotopic (by a map $[0,1]\times [0,1]\to \mathbb{R}$), etc., so we have $ʃ \mathbb{R} = \ast$. (Note that the shape modality ʃ is an [esh](http://en.wikipedia.org/wiki/Esh_%28letter%29), `ʃ`, not an integral sign.)

   Right. A general cohesive homotopy type may have both paths (in the topological sense) and identifications, and two points may be connected by a path but not identified. The shape modality makes paths into new identifications (but also keeps the old identifications around).

   It does, however, happen to be the case that the shape of the (Dedekind) real numbers is a point (and in particular an h-set), because the real numbers are topologically (cohesively) contractible, in the sense that any two points are connected by a path (a map $[0,1]\to \mathbb{R}$), any two paths (in this sense) are homotopic (by a map $[0,1]\times [0,1]\to \mathbb{R}$), etc., so we have $ʃ \mathbb{R} = \ast$.

   (Note that the shape modality ʃ is an esh, ʃ, not an integral sign.)

4. • CommentRowNumber4.
   • CommentAuthorMichael_Bachtold
   • CommentTimeMay 14th 2015
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   Author: Michael_Bachtold
   Format: MarkdownItexThanks Urs and Mike, that clarifies some things!

   Thanks Urs and Mike, that clarifies some things!