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    • CommentRowNumber1.
    • CommentAuthorNima
    • CommentTimeAug 11th 2021
    • (edited Aug 11th 2021)

    I think I’m missing something here.

    Say we have an equivalence

    ABA \simeq B

    and a polymorphic predicate:

    P:Π A:UAUP : \Pi_{A:U} A \rightarrow U

    Then should we not expect the following equivalence?

    Σ a:AP(A,a)Σ b:BP(B,b)\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?

    • CommentRowNumber2.
    • CommentAuthorDavid_Corfield
    • CommentTimeAug 12th 2021
    • (edited Aug 12th 2021)

    So given an equivalence f:ABf: A \simeq B, does it induce an equivalence Σ a:AP(A,a)Σ b:BP(B,b)\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: f:A=B(f *P(A)= BUP(B)). apd_P: \prod_{f:A =B}(f_{\ast} P(A) =_{B \to U} P(B)).
    • CommentRowNumber3.
    • CommentAuthorDavid_Corfield
    • CommentTimeAug 12th 2021

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

    • CommentRowNumber4.
    • CommentAuthorUlrik
    • CommentTimeAug 12th 2021

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

    • CommentRowNumber5.
    • CommentAuthorNima
    • CommentTimeAug 15th 2021

    Thanks David and Ulrik. I think I follow.