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1. • CommentRowNumber1.
   • CommentAuthorDavid_Corfield
   • CommentTimeMay 9th 2016
   • (edited May 9th 2016)
   • PermaLink
   Author: David_Corfield
   Format: MarkdownItexI've been having a chat with someone worried about the weakness of identity in HoTT. Say you define a modified noun, like handsome man, by dependent sum with a mere proposition handsome(m). Then despite any two warrants, $p$ and $q$, for handsome(John) being equal, if I now have a further dependency, say, events of handsome men, then I have only an isomorphism between events(John, p) and events(John, q) when I should have equality. Some thoughts in response. Presumably it should be the case that the type is properly dependent on the modified noun. I mean, one might take Events(x) to be dependent on men in general, then restrict to Events for handsome men. Since the events don't depend on the handsomeness of the man, that might sound reasonable. Take 'Confessions of opium eaters'. Say I have, $x: Human \vdash confessions(x): Type$ $x: Human \vdash opium-eating(x): Type$ Then I form, $x: man \vdash ||opium-eating(x)||: Prop$ and the dependent sum $opium-eater := \sum_{x:Human} ||opium-eating(x)||$ Then, yes, I may form the new variant on confessions, let's say confessions*, which is dependent on opium-eaters: $(x, a): \sum_{x: Human} ||opium-eating|| \vdash confessions*(x, a) := confessions(x)$ But perhaps we do have cases where the dependency is on the modification too. Biology comes to mind: the children born of a woman. $ x: Human \vdash female(x): Prop$ $Woman := \sum_{x: Human} female(x)$ $(x, t) : Woman \vdash BornChild(x, t): Type$ Given (Eve, p), (Eve, q) : Woman, we know that p = q: female(Eve), but then BornChild(Eve, p) is only known to be isomorphic to BornChild(Eve, q), is it? Do we have even (Eve, p) = (Eve, q)? Oh, is this where that [heterogeneous equality](https://homotopytypetheory.org/2012/11/21/on-heterogeneous-equality/) shows up? So here an identity 'over' the identity on Eve? Say we have b: (p = q), do I then get my 'identity over' to provide an isomorphism from BornChild(Eve, p) to BornChild(Eve, q)? But I want these to be the same children. Should I somehow say that BornChild(x, t) is a subtype of Human, and use the latter to provide an identity?

   I’ve been having a chat with someone worried about the weakness of identity in HoTT. Say you define a modified noun, like handsome man, by dependent sum with a mere proposition handsome(m). Then despite any two warrants, $p$ and $q$, for handsome(John) being equal, if I now have a further dependency, say, events of handsome men, then I have only an isomorphism between events(John, p) and events(John, q) when I should have equality.

   Some thoughts in response. Presumably it should be the case that the type is properly dependent on the modified noun. I mean, one might take Events(x) to be dependent on men in general, then restrict to Events for handsome men. Since the events don’t depend on the handsomeness of the man, that might sound reasonable.

   Take ’Confessions of opium eaters’.

   Say I have,

   $x: Human \vdash confessions(x): Type$

   $x: Human \vdash opium-eating(x): Type$

   Then I form,

   $x: man \vdash ||opium-eating(x)||: Prop$

   and the dependent sum

   $opium-eater := \sum_{x:Human} ||opium-eating(x)||$

   Then, yes, I may form the new variant on confessions, let’s say confessions*, which is dependent on opium-eaters:

   $(x, a): \sum_{x: Human} ||opium-eating|| \vdash confessions*(x, a) := confessions(x)$

   But perhaps we do have cases where the dependency is on the modification too.

   Biology comes to mind: the children born of a woman.

   $x: Human \vdash female(x): Prop$

   $Woman := \sum_{x: Human} female(x)$

   $(x, t) : Woman \vdash BornChild(x, t): Type$

   Given (Eve, p), (Eve, q) : Woman, we know that p = q: female(Eve), but then BornChild(Eve, p) is only known to be isomorphic to BornChild(Eve, q), is it?

   Do we have even (Eve, p) = (Eve, q)? Oh, is this where that heterogeneous equality shows up? So here an identity ’over’ the identity on Eve?

   Say we have b: (p = q), do I then get my ’identity over’ to provide an isomorphism from BornChild(Eve, p) to BornChild(Eve, q)?

   But I want these to be the same children. Should I somehow say that BornChild(x, t) is a subtype of Human, and use the latter to provide an identity?

2. • CommentRowNumber2.
   • CommentAuthorMike Shulman
   • CommentTimeMay 9th 2016
   • PermaLink
   Author: Mike Shulman
   Format: MarkdownItex> I have only an isomorphism between events(John, p) and events(John, q) when I should have equality. But isomorphism is equal to equality, by univalence!

   I have only an isomorphism between events(John, p) and events(John, q) when I should have equality.

   But isomorphism is equal to equality, by univalence!

3. • CommentRowNumber3.
   • CommentAuthorDavid_Corfield
   • CommentTimeMay 9th 2016
   • PermaLink
   Author: David_Corfield
   Format: MarkdownItexSo perhaps the worry is that both are to be thought of as subsets of some type Events, and one wants these to be exactly the same subsets.

   So perhaps the worry is that both are to be thought of as subsets of some type Events, and one wants these to be exactly the same subsets.

4. • CommentRowNumber4.
   • CommentAuthorMike Shulman
   • CommentTimeMay 9th 2016
   • PermaLink
   Author: Mike Shulman
   Format: MarkdownItexWhere "the same" means... isomorphic? In type theory a subset of a type $E$ is either a monomorphic function $S \hookrightarrow E$ or a characteristic function $\chi_S : E\to Prop$. In the first case, two subsets are the same (by univalence) if we have $S\cong S'$ commuting with the injections to $E$, and in the second case they are the same (by function extensionality and univalence) if we have $\chi_S(e) \leftrightarrow \chi_{S'}(e)$ for all $e:E$.

   Where “the same” means… isomorphic?

   In type theory a subset of a type $E$ is either a monomorphic function $S \hookrightarrow E$ or a characteristic function $\chi_S : E\to Prop$. In the first case, two subsets are the same (by univalence) if we have $S\cong S'$ commuting with the injections to $E$, and in the second case they are the same (by function extensionality and univalence) if we have $\chi_S(e) \leftrightarrow \chi_{S'}(e)$ for all $e:E$.