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theory Equivalence
imports Identity
begin
section \<open>Homotopy\<close>
definition "homotopy A B f g \<equiv> \<Prod>x: A. f `x =\<^bsub>B x\<^esub> g `x"
definition homotopy_i (infix "~" 100)
where [implicit]: "f ~ g \<equiv> homotopy {} {} f g"
translations "f ~ g" \<leftharpoondown> "CONST homotopy A B f g"
Lemma homotopy_type [type]:
assumes
"A: U i"
"\<And>x. x: A \<Longrightarrow> B x: U i"
"f: \<Prod>x: A. B x"
"g: \<Prod>x: A. B x"
shows "f ~ g: U i"
unfolding homotopy_def
by typechk
Lemma apply_homotopy:
assumes
"A: U i" "\<And>x. x: A \<Longrightarrow> B x: U i"
"f: \<Prod>x: A. B x" "g: \<Prod>x: A. B x"
"H: f ~ g"
"x: A"
shows "H x: f x = g x"
using \<open>H:_\<close> unfolding homotopy_def
by typechk
method htpy for H::o = rule apply_homotopy[where ?H=H]
Lemma (def) homotopy_refl [refl]:
assumes
"A: U i"
"f: \<Prod>x: A. B x"
shows "f ~ f"
unfolding homotopy_def
by intros fact
Lemma (def) hsym:
assumes
"A: U i"
"\<And>x. x: A \<Longrightarrow> B x: U i"
"f: \<Prod>x: A. B x"
"g: \<Prod>x: A. B x"
"H: f ~ g"
shows "g ~ f"
unfolding homotopy_def
proof intro
fix x assuming "x: A" then have "f x = g x"
by (htpy H)
thus "g x = f x"
by (rule pathinv) fact
qed
Lemma (def) htrans:
assumes
"A: U i"
"\<And>x. x: A \<Longrightarrow> B x: U i"
"f: \<Prod>x: A. B x"
"g: \<Prod>x: A. B x"
"h: \<Prod>x: A. B x"
"H1: f ~ g"
"H2: g ~ h"
shows "f ~ h"
unfolding homotopy_def
proof intro
fix x assuming "x: A"
have *: "f x = g x" "g x = h x"
by (htpy H1, htpy H2)
show "f x = h x"
by (rule pathcomp; (rule *)?) typechk
qed
section \<open>Rewriting homotopies\<close>
calc "f ~ g" rhs g
lemmas
homotopy_sym [sym] = hsym[rotated 4] and
homotopy_trans [trans] = htrans[rotated 5]
Lemma id_funcomp_htpy:
assumes "A: U i" "B: U i" "f: A \<rightarrow> B"
shows "homotopy_refl A f: (id B) \<circ> f ~ f"
by compute
Lemma funcomp_id_htpy:
assumes "A: U i" "B: U i" "f: A \<rightarrow> B"
shows "homotopy_refl A f: f \<circ> (id A) ~ f"
by compute
Lemma funcomp_left_htpy:
assumes
"A: U i" "B: U i"
"\<And>x. x: B \<Longrightarrow> C x: U i"
"f: A \<rightarrow> B"
"g: \<Prod>x: B. C x"
"g': \<Prod>x: B. C x"
"H: g ~ g'"
shows "(g \<circ> f) ~ (g' \<circ> f)"
unfolding homotopy_def
apply (intro, compute)
apply (htpy H)
done
Lemma funcomp_right_htpy:
assumes
"A: U i" "B: U i" "C: U i"
"f: A \<rightarrow> B"
"f': A \<rightarrow> B"
"g: B \<rightarrow> C"
"H: f ~ f'"
shows "(g \<circ> f) ~ (g \<circ> f')"
unfolding homotopy_def
proof (intro, compute)
fix x assuming "x: A"
have *: "f x = f' x"
by (htpy H)
show "g (f x) = g (f' x)"
by (rewr eq: *) refl
qed
method lhtpy = rule funcomp_left_htpy[rotated 6]
method rhtpy = rule funcomp_right_htpy[rotated 6]
Lemma (def) commute_homotopy:
assumes
"A: U i" "B: U i"
"x: A" "y: A"
"p: x = y"
"f: A \<rightarrow> B" "g: A \<rightarrow> B"
"H: f ~ g"
shows "(H x) \<bullet> g[p] = f[p] \<bullet> (H y)"
using \<open>H:_\<close>
unfolding homotopy_def
apply (eq p, compute)
apply (rewr eq: pathcomp_refl, rewr eq: refl_pathcomp)
by refl
Corollary (def) commute_homotopy':
assumes
"A: U i"
"x: A"
"f: A \<rightarrow> A"
"H: f ~ (id A)"
shows "H (f x) = f [H x]"
proof -
(*FUTURE: Because we don't have very good normalization integrated into
things yet, we need to manually unfold the type of H.*)
from \<open>H: f ~ id A\<close> have [type]: "H: \<Prod>x: A. f x = x"
by (compute add: homotopy_def)
have "H (f x) \<bullet> H x = H (f x) \<bullet> (id A)[H x]"
by (rule left_whisker, rewr eq: ap_id, refl)
also have [simplified id_comp]: "H (f x) \<bullet> (id A)[H x] = f[H x] \<bullet> H x"
by (rule commute_homotopy)
finally have "?" by this
thus "H (f x) = f [H x]" by pathcomp_cancelr (fact, typechk+)
qed
section \<open>Quasi-inverse and bi-invertibility\<close>
subsection \<open>Quasi-inverses\<close>
definition "is_qinv A B f \<equiv> \<Sum>g: B \<rightarrow> A.
homotopy A (fn _. A) (g \<circ>\<^bsub>A\<^esub> f) (id A) \<times> homotopy B (fn _. B) (f \<circ>\<^bsub>B\<^esub> g) (id B)"
Lemma is_qinv_type [type]:
assumes "A: U i" "B: U i" "f: A \<rightarrow> B"
shows "is_qinv A B f: U i"
unfolding is_qinv_def
by typechk
definition is_qinv_i ("is'_qinv")
where [implicit]: "is_qinv f \<equiv> Equivalence.is_qinv {} {} f"
no_translations "is_qinv f" \<leftharpoondown> "CONST Equivalence.is_qinv A B f"
Lemma (def) id_is_qinv:
assumes "A: U i"
shows "is_qinv (id A)"
unfolding is_qinv_def
proof intro
show "id A: A \<rightarrow> A" by typechk
qed (compute, intro; refl)
Lemma is_qinvI:
assumes
"A: U i" "B: U i" "f: A \<rightarrow> B"
"g: B \<rightarrow> A"
"H1: g \<circ> f ~ id A"
"H2: f \<circ> g ~ id B"
shows "is_qinv f"
unfolding is_qinv_def
proof intro
show "g: B \<rightarrow> A" by fact
show "g \<circ> f ~ id A \<and> f \<circ> g ~ id B" by (intro; fact)
qed
Lemma is_qinv_components [type]:
assumes "A: U i" "B: U i" "f: A \<rightarrow> B" "pf: is_qinv f"
shows
qinv_of_is_qinv: "fst pf: B \<rightarrow> A" and
ret_of_is_qinv: "p\<^sub>2\<^sub>1 pf: fst pf \<circ> f ~ id A" and
sec_of_is_qinv: "p\<^sub>2\<^sub>2 pf: f \<circ> fst pf ~ id B"
using assms unfolding is_qinv_def
by typechk+
Lemma (def) qinv_is_qinv:
assumes "A: U i" "B: U i" "f: A \<rightarrow> B" "pf: is_qinv f"
shows "is_qinv (fst pf)"
using \<open>pf:_\<close>[unfolded is_qinv_def] \<comment> \<open>Should be unfolded by the typechecker\<close>
apply (rule is_qinvI)
\<^item> by (fact \<open>f:_\<close>)
\<^item> by (rule sec_of_is_qinv)
\<^item> by (rule ret_of_is_qinv)
done
Lemma (def) funcomp_is_qinv:
assumes
"A: U i" "B: U i" "C: U i"
"f: A \<rightarrow> B" "g: B \<rightarrow> C"
shows "is_qinv f \<rightarrow> is_qinv g \<rightarrow> is_qinv (g \<circ> f)"
apply intros
unfolding is_qinv_def apply elims
focus vars _ _ finv _ ginv
apply intro
\<^item> by (rule funcompI[where ?f=ginv and ?g=finv])
\<^item> proof intro
show "(finv \<circ> ginv) \<circ> g \<circ> f ~ id A"
proof -
have "(finv \<circ> ginv) \<circ> g \<circ> f ~ finv \<circ> (ginv \<circ> g) \<circ> f" by compute refl
also have ".. ~ finv \<circ> id B \<circ> f" by (rhtpy, lhtpy) fact
also have ".. ~ id A" by compute fact
finally show "?" by this
qed
show "(g \<circ> f) \<circ> finv \<circ> ginv ~ id C"
proof -
have "(g \<circ> f) \<circ> finv \<circ> ginv ~ g \<circ> (f \<circ> finv) \<circ> ginv" by compute refl
also have ".. ~ g \<circ> id B \<circ> ginv" by (rhtpy, lhtpy) fact
also have ".. ~ id C" by compute fact
finally show "?" by this
qed
qed
done
done
subsection \<open>Bi-invertible maps\<close>
definition "is_biinv A B f \<equiv>
(\<Sum>g: B \<rightarrow> A. homotopy A (fn _. A) (g \<circ>\<^bsub>A\<^esub> f) (id A))
\<times> (\<Sum>g: B \<rightarrow> A. homotopy B (fn _. B) (f \<circ>\<^bsub>B\<^esub> g) (id B))"
Lemma is_biinv_type [type]:
assumes "A: U i" "B: U i" "f: A \<rightarrow> B"
shows "is_biinv A B f: U i"
unfolding is_biinv_def by typechk
definition is_biinv_i ("is'_biinv")
where [implicit]: "is_biinv f \<equiv> Equivalence.is_biinv {} {} f"
translations "is_biinv f" \<leftharpoondown> "CONST Equivalence.is_biinv A B f"
Lemma is_biinvI:
assumes
"A: U i" "B: U i" "f: A \<rightarrow> B"
"g: B \<rightarrow> A" "h: B \<rightarrow> A"
"H1: g \<circ> f ~ id A" "H2: f \<circ> h ~ id B"
shows "is_biinv f"
unfolding is_biinv_def
proof intro
show "<g, H1>: \<Sum>g: B \<rightarrow> A. g \<circ> f ~ id A" by typechk
show "<h, H2>: \<Sum>g: B \<rightarrow> A. f \<circ> g ~ id B" by typechk
qed
Lemma is_biinv_components [type]:
assumes "A: U i" "B: U i" "f: A \<rightarrow> B" "pf: is_biinv f"
shows
section_of_is_biinv: "p\<^sub>1\<^sub>1 pf: B \<rightarrow> A" and
retraction_of_is_biinv: "p\<^sub>2\<^sub>1 pf: B \<rightarrow> A" and
ret_of_is_biinv: "p\<^sub>1\<^sub>2 pf: p\<^sub>1\<^sub>1 pf \<circ> f ~ id A" and
sec_of_is_biinv: "p\<^sub>2\<^sub>2 pf: f \<circ> p\<^sub>2\<^sub>1 pf ~ id B"
using assms unfolding is_biinv_def
by typechk+
Lemma (def) is_biinv_if_is_qinv:
assumes "A: U i" "B: U i" "f: A \<rightarrow> B"
shows "is_qinv f \<rightarrow> is_biinv f"
apply intros
unfolding is_qinv_def is_biinv_def
by (rule distribute_Sig)
Lemma (def) is_qinv_if_is_biinv:
assumes "A: U i" "B: U i" "f: A \<rightarrow> B"
shows "is_biinv f \<rightarrow> is_qinv f"
apply intro
unfolding is_biinv_def apply elims
focus vars _ _ _ g H1 h H2
apply (rule is_qinvI)
\<^item> by (fact \<open>g: _\<close>)
\<^item> by (fact \<open>H1: _\<close>)
\<^item> proof -
have "g ~ g \<circ> (id B)" by compute refl
also have ".. ~ g \<circ> f \<circ> h" by rhtpy (rule \<open>H2:_\<close>[symmetric])
also have ".. ~ (id A) \<circ> h" by (comp funcomp_assoc[symmetric]) (lhtpy, fact)
also have ".. ~ h" by compute refl
finally have "g ~ h" by this
then have "f \<circ> g ~ f \<circ> h" by (rhtpy, this)
also note \<open>H2:_\<close>
finally show "f \<circ> g ~ id B" by this
qed
done
done
Lemma (def) id_is_biinv:
"A: U i \<Longrightarrow> is_biinv (id A)"
by (rule is_biinv_if_is_qinv) (rule id_is_qinv)
Lemma (def) funcomp_is_biinv:
assumes
"A: U i" "B: U i" "C: U i"
"f: A \<rightarrow> B" "g: B \<rightarrow> C"
shows "is_biinv f \<rightarrow> is_biinv g \<rightarrow> is_biinv (g \<circ> f)"
apply intros
focus vars pf pg
by (rule is_biinv_if_is_qinv)
(rule funcomp_is_qinv; rule is_qinv_if_is_biinv, fact)
done
section \<open>Equivalence\<close>
text \<open>
Following the HoTT book, we first define equivalence in terms of
bi-invertibility.
\<close>
definition equivalence (infix "\<simeq>" 110)
where "A \<simeq> B \<equiv> \<Sum>f: A \<rightarrow> B. Equivalence.is_biinv A B f"
Lemma equivalence_type [type]:
assumes "A: U i" "B: U i"
shows "A \<simeq> B: U i"
unfolding equivalence_def by typechk
Lemma (def) equivalence_refl:
assumes "A: U i"
shows "A \<simeq> A"
unfolding equivalence_def
proof intro
show "is_biinv (id A)" by (rule is_biinv_if_is_qinv) (rule id_is_qinv)
qed typechk
Lemma (def) equivalence_symmetric:
assumes "A: U i" "B: U i"
shows "A \<simeq> B \<rightarrow> B \<simeq> A"
apply intros
unfolding equivalence_def
apply elim
apply (dest (4) is_qinv_if_is_biinv)
apply intro
\<^item> by (rule qinv_of_is_qinv) facts
\<^item> by (rule is_biinv_if_is_qinv) (rule qinv_is_qinv)
done
Lemma (def) equivalence_transitive:
assumes "A: U i" "B: U i" "C: U i"
shows "A \<simeq> B \<rightarrow> B \<simeq> C \<rightarrow> A \<simeq> C"
proof intros
fix AB BC assume *: "AB: A \<simeq> B" "BC: B \<simeq> C"
then have
"fst AB: A \<rightarrow> B" and 1: "snd AB: is_biinv (fst AB)"
"fst BC: B \<rightarrow> C" and 2: "snd BC: is_biinv (fst BC)"
unfolding equivalence_def by typechk+
then have "fst BC \<circ> fst AB: A \<rightarrow> C" by typechk
moreover have "is_biinv (fst BC \<circ> fst AB)"
using * unfolding equivalence_def by (rule funcomp_is_biinv 1 2) facts
ultimately show "A \<simeq> C"
unfolding equivalence_def by intro facts
qed
text \<open>
Equal types are equivalent. We give two proofs: the first by induction, and
the second by following the HoTT book and showing that transport is an
equivalence.
\<close>
Lemma
assumes "A: U i" "B: U i" "p: A =\<^bsub>U i\<^esub> B"
shows "A \<simeq> B"
by (eq p) (rule equivalence_refl)
text \<open>
The following proof is wordy because (1) typechecker normalization is still
rudimentary, and (2) we don't yet have universe level inference.
\<close>
Lemma (def) equiv_if_equal:
assumes
"A: U i" "B: U i" "p: A =\<^bsub>U i\<^esub> B"
shows "<transp (id (U i)) p, ?isequiv>: A \<simeq> B"
unfolding equivalence_def
apply intro defer
\<^item> apply (eq p)
\<^enum> vars A B
apply (comp at A in "A \<rightarrow> B" id_comp[symmetric])
using [[solve_side_conds=1]]
apply (comp at B in "_ \<rightarrow> B" id_comp[symmetric])
apply (rule transport, rule Ui_in_USi)
by (rule lift_universe_codomain, rule Ui_in_USi)
\<^enum> vars A
using [[solve_side_conds=1]]
apply (comp transport_comp)
\<circ> by (rule Ui_in_USi)
\<circ> by compute (rule U_lift)
\<circ> by compute (rule id_is_biinv)
done
done
\<^item> \<comment> \<open>Similar proof as in the first subgoal above\<close>
apply (comp at A in "A \<rightarrow> B" id_comp[symmetric])
using [[solve_side_conds=1]]
apply (comp at B in "_ \<rightarrow> B" id_comp[symmetric])
apply (rule transport, rule Ui_in_USi)
by (rule lift_universe_codomain, rule Ui_in_USi)
done
end
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