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1. • CommentRowNumber1.
   • CommentAuthorNima
   • CommentTimeAug 11th 2021
   • (edited Aug 11th 2021)
   • PermaLink
   Author: Nima
   Format: MarkdownItexI think I'm missing something here. Say we have an equivalence $A \simeq B$ and a polymorphic predicate: $P : \Pi_{A:U} A \rightarrow U$ Then should we not expect the following equivalence? $\Sigma_{a:A} {P (A, a)} \simeq \Sigma_{b:B} {P (B, b)}$ It seems to me like this should be supported through substitute of like for like, but I can't figure out how to prove this through the language of HoTT, even when using Univalence. Is my thinking correct here or am I missing something?

   I think I’m missing something here.

   Say we have an equivalence

   $A \simeq B$

   and a polymorphic predicate:

   $P : \Pi_{A:U} A \rightarrow U$

   Then should we not expect the following equivalence?

   $\Sigma_{a:A} {P (A, a)} \simeq \Sigma_{b:B} {P (B, b)}$

   It seems to me like this should be supported through substitute of like for like, but I can’t figure out how to prove this through the language of HoTT, even when using Univalence. Is my thinking correct here or am I missing something?

2. • CommentRowNumber2.
   • CommentAuthorDavid_Corfield
   • CommentTimeAug 12th 2021
   • (edited Aug 12th 2021)
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   Author: David_Corfield
   Format: MarkdownItexSo given an equivalence $f: A \simeq B$, does it induce an equivalence $\Sigma_{a:A} {P (A, a)} \simeq \Sigma_{b:B} {P (B, b)}$? Doesn't this concern that dependent map of Lemma 2.3.4 of the HoTT book? In the terms here: $$ apd_P: \prod_{f:A =B}(f_{\ast} P(A) =_{B \to U} P(B)). $$

   So given an equivalence $f: A \simeq B$, does it induce an equivalence $\Sigma_{a:A} {P (A, a)} \simeq \Sigma_{b:B} {P (B, b)}$?

   Doesn’t this concern that dependent map of Lemma 2.3.4 of the HoTT book? In the terms here:

   $$apd_P: \prod_{f:A =B}(f_{\ast} P(A) =_{B \to U} P(B)).$$
3. • CommentRowNumber3.
   • CommentAuthorDavid_Corfield
   • CommentTimeAug 12th 2021
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   Author: David_Corfield
   Format: MarkdownItexSo you have equivalences between $P(A)(a)$ and $f_\ast P(A)(f(a))$ and $P(B)(f(a))$.

   So you have equivalences between $P(A)(a)$ and $f_\ast P(A)(f(a))$ and $P(B)(f(a))$.

4. • CommentRowNumber4.
   • CommentAuthorUlrik
   • CommentTimeAug 12th 2021
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   Author: Ulrik
   Format: MarkdownItexYes, it follows that $P(B) \circ f =_{A \to U} P(A)$ (transporting on the other side) and hence for every $a : A$, $P(B)(f(a)) \simeq P(A)(a)$. This in turn induces the desired equivalence.

   Yes, it follows that $P(B) \circ f =_{A \to U} P(A)$ (transporting on the other side) and hence for every $a : A$, $P(B)(f(a)) \simeq P(A)(a)$. This in turn induces the desired equivalence.

5. • CommentRowNumber5.
   • CommentAuthorNima
   • CommentTimeAug 15th 2021
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   Author: Nima
   Format: MarkdownItexThanks David and Ulrik. I think I follow.

   Thanks David and Ulrik. I think I follow.