From df9a3121d5a61477b7ff6852e34f43b6a69ad933 Mon Sep 17 00:00:00 2001 From: Sinsa Date: Tue, 19 May 2026 15:25:16 +0200 Subject: [PATCH] sitemati typo --- Tesi.log | 4 +- Tesi.pdf | Bin 334393 -> 334432 bytes Tesi.synctex.gz | Bin 188133 -> 188212 bytes Tesi.tex | 106 ++++++++++++++++++++++++------------------------ 4 files changed, 55 insertions(+), 55 deletions(-) diff --git a/Tesi.log b/Tesi.log index c9bfe22..60bfc74 100644 --- a/Tesi.log +++ b/Tesi.log @@ -1,4 +1,4 @@ -This is pdfTeX, Version 3.141592653-2.6-1.40.29 (TeX Live 2026/Arch Linux) (preloaded format=pdflatex 2026.4.6) 17 MAY 2026 18:04 +This is pdfTeX, Version 3.141592653-2.6-1.40.29 (TeX Live 2026/Arch Linux) (preloaded format=pdflatex 2026.4.6) 19 MAY 2026 15:10 entering extended mode restricted \write18 enabled. %&-line parsing enabled. @@ -885,7 +885,7 @@ exmf-dist/fonts/type1/public/amsfonts/cm/cmsy7.pfb> -Output written on Tesi.pdf (27 pages, 334393 bytes). +Output written on Tesi.pdf (27 pages, 334432 bytes). 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zE5KFr^!xjxJ@WVYmDj;dm5d79UV6vqI5c$WIqGvQMsskqAyI@Ez?hL{jEEno4Ik(= zjQxV`4^ZgN7;wLM`VHj37k2{}xzSeOeT-uM(Tb6LAtA?wSJr$!rzJq1rh7~P6L zCpaoBelLelf08K2Bcu=5M1mjNfDQz&_?@Afvemygl(r?Ll>(%^U|GGtGIbCgUMkFKLVn|9lPXvx901=0~M?2-$(4n`MVtAR$sBmt+}pL6><(Bot%M@a@z0D+h#s@ zwOBYOfof +% !TeX spellcheck = en_US \documentclass[12pt,a4paper,italian]{book} \usepackage{graphicx} % Required for inserting images @@ -49,7 +49,7 @@ \end{tikzcd}\] Then we call comma category of F and G \footnote{the name "comma category" arises from the notation (F,G) used by Lawvere}, $(F \downarrow G)$, the category with: \begin{itemize} - \item{Objects}: an object is a triple $(A,B,f)$ where $f: F(A) \to F(B)$ is a $\mathcal{C}$-morhpism + \item{Objects}: an object is a triple $(A,B,f)$ where $f: F(A) \to F(B)$ is a $\mathcal{C}$-morphism \item{Arrows}: a morphism from $(A,B,f)$ to $(A',B',f')$ is a pair $(h,k)$ of an $\mathcal{A}$-morphism $h: A \to A'$ and a $\mathcal{B}$-morphism $k: B \to B'$ such that the following diagram is commutative : % https://q.uiver.app/#q=WzAsNCxbMCwwLCJGKEEpIl0sWzIsMCwiRyhCKSJdLFswLDIsIkYoQScpIl0sWzIsMiwiRyhCJykiXSxbMCwxLCJmIl0sWzIsMywiZiciLDJdLFsxLDMsIkcoaykiXSxbMCwyLCJGKGgpIiwyXV0= \[\begin{tikzcd} @@ -96,13 +96,13 @@ \begin{theorem}\label{completenesscomma} - Let $\mathcal{A}$ and $\mathcal{B}$ be two finitely cocomplete categories and let $F:\mathcal{A} \to \mathcal{C}$, $\mathcal{B} \to \mathcal{C}$ two functors. If $F$ is a cocontinous functor - i.e. preserves small colimits - then the comma category $(F\downarrow G)$ is finitely cocomplete; morover the forgetful functors are cocontinous. \\ - Analogously if $\mathcal{A}$ and $\mathcal{B}$ are finitely complete categories and $G: \mathcal{A} \to \mathcal{C}$ is continous - i.e. preserves small limits - then $(F \downarrow G)$ is finitely complete and both forgetfoul functors are continous. + Let $\mathcal{A}$ and $\mathcal{B}$ be two finitely cocomplete categories and let $F:\mathcal{A} \to \mathcal{C}$, $\mathcal{B} \to \mathcal{C}$ two functors. If $F$ is a continuous functor - i.e. preserves small colimits - then the comma category $(F\downarrow G)$ is finitely cocomplete; moreover the forgetful functors are continuous. \\ + Analogously if $\mathcal{A}$ and $\mathcal{B}$ are finitely complete categories and $G: \mathcal{A} \to \mathcal{C}$ is continuos - i.e. preserves small limits - then $(F \downarrow G)$ is finitely complete and both forgetful functors are continuos. \end{theorem} \begin{proof} - Let's begin by proving the cocomplete case. We will start from cocompleteness of the category and then we will move on to che cocontinuity of the forgetful functors. Recall that a a category is finitely cocomplete if and only if it admits initial object and pushouts; we will construct explicitely both objects: \\ - \textbf{Initial Object}: We will construct the initial object of $(F, \downarrow G)$ by using the cocompletens of $\mathcal{A}$ and $\mathcal{B}$: indeed, let $0_\mathcal{A}$ be the initial object of $\mathcal{A}$ and $0_\mathcal{B}$ the initial object of $\mathcal{B}$. Now, since $F$ is a cocontinous functor $F(0_\mathcal{A}) = 0_\mathcal{C}$ where $0_\mathcal{C}$ is + Let's begin by proving the cocomplete case. We will start from completeness of the category and then we will move on to the cocontinuity of the forgetful functors. Recall that a a category is finitely cocomplete if and only if it admits initial object and pushouts; we will construct explicitly both objects: \\ + \textbf{Initial Object}: We will construct the initial object of $(F, \downarrow G)$ by using the completeness of $\mathcal{A}$ and $\mathcal{B}$: indeed, let $0_\mathcal{A}$ be the initial object of $\mathcal{A}$ and $0_\mathcal{B}$ the initial object of $\mathcal{B}$. Now, since $F$ is a continuous functor $F(0_\mathcal{A}) = 0_\mathcal{C}$ where $0_\mathcal{C}$ is in $\mathcal{C}$; thus there is a unique morphism $in: F(0_\mathcal{A})=0_\mathcal{C} \to G(0_\mathcal{B})$ in $\mathcal{C}$. Now, we claim $(0_\mathcal{A},0_\mathcal{B},in)$ to be initial in $(F\downarrow G)$: \begin{itemize} \item existence: let $(A,B,f)$ be an object of $(F\downarrow G)$; then there exists two unique morphisms $in_A, in_B$ respectively in $\mathcal{A}$ and $\mathcal{B}$ such that $in_A: 0_\mathcal{A} \to A $ and $in_B: 0_\mathcal{B} \to B$. Thus we have a candidate morphism $(in_A, in_B)$. In order to check that this morphism is a morphism of $(F \downarrow G)$ we need to show that the following diagram is commutative: @@ -117,11 +117,11 @@ \arrow["{Gin_B}"', from=1-3, to=3-3] \arrow["f", from=3-1, to=3-3] \end{tikzcd}\] - The diagram above is trivially commutative: since $0_\mathcal{C}$ is intial in $\mathcal{C}$ both $f\circ Fin_A: 0_\mathcal{C} \to GB$ and $Gin_B \circ in: 0_\mathcal{C} \to GB$ must be equal to $in_{GB}: 0_\mathcal{C} \to GB$. Thus $$(0_\mathcal{A}, 0_\mathcal{B}, in) \xrightarrow{(in_A, in_B)} (A,B,f)$$ is a morphism in $(F \downarrow G)$ + The diagram above is trivially commutative: since $0_\mathcal{C}$ is initial in $\mathcal{C}$ both $f\circ Fin_A: 0_\mathcal{C} \to GB$ and $Gin_B \circ in: 0_\mathcal{C} \to GB$ must be equal to $in_{GB}: 0_\mathcal{C} \to GB$. Thus $$(0_\mathcal{A}, 0_\mathcal{B}, in) \xrightarrow{(in_A, in_B)} (A,B,f)$$ is a morphism in $(F \downarrow G)$ \item uniqueness: let $(0_\mathcal{A}, 0_\mathcal{B}, in) \xrightarrow[(h,k)]{(in_A, in_B)} (A,B,f)$ two arrows in $(F \downarrow G)$; then since $0_\mathcal{A}, 0_\mathcal{B}$ are initial, then $h = in_A$ and $k = in_B$; hence $(h,k)= (in_A,in_B)$ and so $(in_A,in_B)$ is unique \end{itemize} Thus $(F \downarrow G)$ admits initial object \\ - \textbf{Pushouts}: As for the case above let construct the pushouts in $(F \downarrow G)$ by using pushout provided by cocompleteness of $\mathcal{A}, \mathcal{B}$. Take the following scenario in $(F \downarrow G)$ + \textbf{Pushouts}: As for the case above let construct the pushouts in $(F \downarrow G)$ by using pushouts provided by cocompleteness of $\mathcal{A}, \mathcal{B}$. Take the following scenario in $(F \downarrow G)$ \begin{equation}\label{push_1} % https://q.uiver.app/#q=WzAsMyxbMCwwLCIoQSxCLGYpIl0sWzIsMCwiKEEnLEInLGYnKSJdLFswLDIsIihBJycsQicnLGYnJykiXSxbMCwxLCIoaCxrKSIsMl0sWzAsMiwiKGgnLGsnKSJdXQ== \begin{tikzcd} @@ -196,7 +196,7 @@ \arrow["u"', from=3-4, to=3-6] \end{tikzcd}\] Both diagrams are commutative as consequence of the diagram (\ref{def-u}); hence projections are indeed arrows in $(F \downarrow G)$. \\ - Now we prove that the universal propery holds: + Now we prove that the universal property holds: \begin{itemize} \item existence: assume the following diagram is commutative % https://q.uiver.app/#q=WzAsNCxbMCwwLCIoQSxCLGYpIl0sWzIsMCwiKEEnLEInLGYnKSJdLFswLDIsIihBJycsQicnLGYnJykiXSxbMiwyLCIoWCxZLGcpIl0sWzAsMSwiKGgsaykiXSxbMCwyLCIoaCcsaycpIiwyXSxbMSwzLCIoeCcseScpIl0sWzIsMywiKHgnJyx5JycpIiwyXV0= @@ -286,12 +286,12 @@ \arrow["{Gu_B}", from=3-4, to=5-4] \arrow["g"', from=5-2, to=5-4] \end{tikzcd}\] - Then using (\ref{x'y'}) and (\ref{x''y''}) we can show that the external square is commuative: + Then using (\ref{x'y'}) and (\ref{x''y''}) we can show that the external square is commutative: \begin{align*} g Fx'' Fh'' = Gy'' f'' Fh' = Gy'' Gk' f = Gy'Gkf \\ g Fx' Fh = Gy' f' Fh = Gy'Gkf \end{align*} - Thus, by $P'_A$ beeing pushout in $\mathcal{C}$ there exists an unique $v: P_A' \to GY$ such that $vFp_1 = gFx'$ and $vFp_2 = gFx''$. Now , consider $gFu_A, Gu_Bu: P_A' \to GY$, notice that: + Thus, by $P'_A$ being pushout in $\mathcal{C}$ there exists an unique $v: P_A' \to GY$ such that $vFp_1 = gFx'$ and $vFp_2 = gFx''$. Now , consider $gFu_A, Gu_Bu: P_A' \to GY$, notice that: \begin{align*} gFu_AFp_1 = gFx' \\ gFu_BFp_2 = gFx'' \\ @@ -299,11 +299,11 @@ Gu_BuFp_2 = Gu_BGq_2f'' = Gy''f'' = gFx'' \\ \end{align*} So by uniqueness of $v$ we get $gFu_A = Gu_B u$ and so, since the square (\ref{uAuBarrow}) is commutative, then $(u_A,u_B)$ is an arrow of $(F \downarrow G)$ and the diagram (\ref{goalpush}) is commutative. - \item uniqueness: the uniquness of $(u_A,u_B)$ comes almost for free: indee if there was antoher arrow $(l,m)$ satisfyng commutativity in (\ref{goalpush}), then, by universal property of pushouts $l=u_A$ in $\mathcal{A}$ and $m = u_B$ in $\mathcal{B}$. Therefore we would have $(l,m) = (u_A, u_B)$ + \item uniqueness: the uniqueness of $(u_A,u_B)$ comes almost for free: indeed if there was another arrow $(l,m)$ satisfying commutativity in (\ref{goalpush}), then, by universal property of pushouts $l=u_A$ in $\mathcal{A}$ and $m = u_B$ in $\mathcal{B}$. Therefore we would have $(l,m) = (u_A, u_B)$ \end{itemize} Thus $(P_A, P_B, u)$ is the desired pushout and $(F \downarrow G)$ having initial object and pushouts is indeed finitely cocomplete. \\ - \textbf{The forgetfoul functors are cocontinous}: the cocontinuity of $\Pi_1, \Pi_2$ arises immediatly from the structure of intial object and pushouts of $(F \downarrow G)$. Indeed take the inital object of $(F \downarrow G)$, $(0_A, 0_B, in)$; then $\Pi_1((0_A, 0_B, in))=0_A$ and $\Pi_2((0\_A, 0\_B, in)) = 0_B$ are respectively the initial object of $\mathcal{A}$ and $\mathcal{B}$. Since the same happens for the pushouts of $(F \downarrow G)$ then both $\Pi_1$ and $\Pi_2$ preserves finite colimits. \\ - Now, let's move on to the finitely complete case; so suppose that both $\mathcal{A}$ and $\mathcal{B}$ are finitely complete and that $G$ is finitely continous. The structure of the proof is completely analogue \footnote{Notice that we can also argue that, beeing $(F \downarrow G)^{op} \cong (F^{op} \downarrow G^{op})$, the complete case comes for free from the cocomplete case} and so we will omit some details. \\ + \textbf{The forgetful functors are continuous}: the cocontinuity of $\Pi_1, \Pi_2$ arises immediately from the structure of initial object and pushouts of $(F \downarrow G)$. Indeed take the initial object of $(F \downarrow G)$, $(0_A, 0_B, in)$; then $\Pi_1((0_A, 0_B, in))=0_A$ and $\Pi_2((0\_A, 0\_B, in)) = 0_B$ are respectively the initial object of $\mathcal{A}$ and $\mathcal{B}$. Since the same happens for the pushouts of $(F \downarrow G)$ then both $\Pi_1$ and $\Pi_2$ preserves finite colimits. \\ + Now, let's move on to the finitely complete case; so suppose that both $\mathcal{A}$ and $\mathcal{B}$ are finitely complete and that $G$ is finitely continuos. The structure of the proof is completely analog \footnote{Notice that we can also argue that, being $(F \downarrow G)^{op} \cong (F^{op} \downarrow G^{op})$, the complete case comes for free from the cocomplete case} and so we will omit some details. \\ \textbf{Terminal Objects}: following the structure of the cocomplete proof we claim $(1_\mathcal{A}, 1_\mathcal{B}, un)$, where $un: F(1_\mathcal{A}) \to 1_\mathcal{C} = G(1_\mathcal{B})$ is the unique morphism granted by $G(1_\mathcal{B})$ be terminal in $\mathcal{C}$, to be the terminal object in $(F \downarrow G)$ \begin{itemize} \item existence: let $(A,B,f)$ be an object of $(F\downarrow G)$; then there exists two unique morphisms $un_A, un_B$ respectively in $\mathcal{A}$ and $\mathcal{B}$ such that $un_A: A \to 1_\mathcal{A} $ and $un_B: B \to 1_\mathcal{B}$. Thus we have a candidate morphism $(un_A, un_B)$. In order to check that this morphism is a morphism of $(F \downarrow G)$ we need to show that the following diagram is commutative: @@ -317,7 +317,7 @@ \arrow["f", from=3-1, to=3-3] \end{tikzcd}\] The diagram above is trivially commutative since $G1_\mathcal{B} = 1_\mathcal{C}$ is terminal in $\mathcal{C}$ - \item uniqueness: the uniquness is directly inherited by uniness of $un_A$ and $un_B$ in $\mathcal{A}$ and $\mathcal{B}$ + \item uniqueness: the uniqueness is directly inherited by uniqueness of $un_A$ and $un_B$ in $\mathcal{A}$ and $\mathcal{B}$ \end{itemize} \textbf{Pullbacks}: As for the pushouts we claim the pullback of the following diagram: % https://q.uiver.app/#q=WzAsMyxbMCwyLCIoQScsQicsZikiXSxbMiwyLCIoQScnLEInJyxmJycpIl0sWzIsMCwiKEEsQixmKSJdLFswLDEsIihoJyxrJykiXSxbMiwxLCIoaCxrKSIsMl1d @@ -345,12 +345,12 @@ \arrow["Gk"', from=2-4, to=4-4] \arrow["{Gk'}", from=4-2, to=4-4] \end{tikzcd}\] - From the commutativity of this diagram we deduce that $(q_i, p_i)$ are actuall arrows of $(F \downarrow G)$, indeed: + From the commutativity of this diagram we deduce that $(q_i, p_i)$ are actual arrows of $(F \downarrow G)$, indeed: \begin{align} Gq_2u = f'Fp_2 \\ Gq_1u = fFp_1 \end{align} - Now, we cheeck that the universal propery holds + Now, we check that the universal property holds \begin{itemize} \item Existence: assume the external square to be commutative: \begin{equation}\label{pull} @@ -395,7 +395,7 @@ \end{tikzcd} \end{equation} Now, consider the couple $(u_A,u_B)$; by commutativity of (\ref{pullstart}) if $(u_A,u_B): (X,Y,g) \to (P_A,P_B, u)$ is an actual arrow of $(F \downarrow G)$ then it complete the diagram (\ref{pull}) to a commutative diagram. \\ - We procede as in the case of pushouts; consider the following square: + We proceed as in the case of pushouts; consider the following square: % https://q.uiver.app/#q=WzAsOCxbMCwwLCJGWCJdLFsxLDAsIkdZIl0sWzEsMSwiR3VfQiBcXGNvbmcgUF9CJyJdLFswLDEsIkZQX0EiXSxbMSwyLCJHQiciXSxbMiwyLCJHQicnIl0sWzIsMSwiR0IiXSxbMCwyLCJGQSciXSxbMCwxLCJnIl0sWzEsMl0sWzAsM10sWzMsMiwidSIsMl0sWzIsNCwiR3FfMiJdLFs0LDUsIkdrJyIsMl0sWzIsNiwiR3FfMSIsMl0sWzYsNSwiR2siXSxbMCwyLCJcXGV4aXN0cyAhIHYgIiwxXSxbMSw2LCJHeSIsMCx7ImN1cnZlIjotMn1dLFs3LDQsImYnIiwyXSxbMCw3LCJGeCciLDIseyJjdXJ2ZSI6NH1dXQ== \[\begin{tikzcd} FX & GY & \\ @@ -428,23 +428,23 @@ Thus by uniqueness of $v: FX \to P_B'$ then $uFu_A = Gu_B f$ and so $(u_A,u_B): (X,Y,g) \to (P_A,P_B,u)$ is an arrow of $(F \downarrow G)$ and complete the diagram (\ref{pull}) to a commutative diagram \item uniqueness: uniqueness of $u_A$ and $u_B$ in $\mathcal{A}$ and $\mathcal{B}$ yields to uniqueness of $(u_A,u_B)$ in $(F \downarrow G)$ \end{itemize} - \textbf{Forgetful functors are continous}: the proof is analogue to the cocomplete case, since both $\Pi_1,\Pi_2$ preserves terminal object and pullbacks. \\ - We proved that under the assumption of $\mathcal{A}$ and $\mathcal{B}$ beeing cocomplete and $F$ beeing cocontinous the comma category $(F \downarrow G)$ is cocomplete and that $\Pi_i$ is cocontinous; morevoer we proved the dual result with complteness assumption, therefore the statement is proved and we are done. + \textbf{Forgetful functors are continuos}: the proof is analogous to the cocomplete case, since both $\Pi_1,\Pi_2$ preserves terminal object and pullbacks. \\ + We proved that under the assumption of $\mathcal{A}$ and $\mathcal{B}$ being cocomplete and $F$ being cocontinuos the comma category $(F \downarrow G)$ is cocomplete and that $\Pi_i$ is cocontinuos; moreover we proved the dual result with completeness assumption, therefore the statement is proved and we are done. \end{proof} \begin{definition}[Freyd Cover] - Let $\mathcal{C}$ be a small category (serve veramente sia piccola?) with a terminal odject $1$, then we call its \textit{Freyd Cover}, $Cov(\mathcal{C})$, the comma category $Id \downarrow \mathcal{C}(1,\text{-})$ where $Id: Set \to Set$ is the identical functor + Let $\mathcal{C}$ be a small category (serve veramente sia piccola?) with a terminal object $1$, then we call its \textit{Freyd Cover}, $Cov(\mathcal{C})$, the comma category $Id \downarrow \mathcal{C}(1,\text{-})$ where $Id: Set \to Set$ is the identical functor \end{definition} \begin{definition}[Projection Functor of Freyd Cover] Recall that there are two canonical functors $\Pi_1,\Pi_2$ associated with a comma category. Let $\mathcal{C}$ be a category with terminal object and let $Cov(\mathcal{C})$ be its Frey Cover. We call the functor $\Pi_2: Cov(\mathcal{C}) \to \mathcal{C}$ the projection functor of the Freyd Cover. \end{definition} \begin{theorem}\label{cover of bicartesian} - Let $\mathcal{C}$ be a bicartesian category - i.e. a category with terminal object, initial object and binary products and coproducs - then $Cov(\mathcal{C})$ is bicartesian. Moreover $\Pi_2 : Cov(\mathcal{C}) \to \mathcal{C}$ is a cartesian closed functor + Let $\mathcal{C}$ be a bicartesian category - i.e. a category with terminal object, initial object and binary products and coproducts - then $Cov(\mathcal{C})$ is bicartesian. Moreover $\Pi_2 : Cov(\mathcal{C}) \to \mathcal{C}$ is a cartesian closed functor \end{theorem} \begin{proof} - The proof follows a similar structure to the proof of \ref{completenesscomma}, so we will construct initial and terminal objects, binary products and copruducts and then we will prove that $\Pi_2$ is closed with respect to this constructions. \\ - \textbf{Initial and Terminal Object}: the initial object of $Cov(\mathcal{C})$ is $(0_{Set}, 0_\mathcal{C}, in)$ and the terminal object is $(1_{Set}, 1_\mathcal{C}, un)$. The construction is the same of the construction in \ref{completenesscomma} and therefore the proof is ommitted \\ - \textbf{Binary Products}: Recall that $\mathcal{C}(1_\mathcal{C},-): \mathcal{C} \to Set$ is a continous functor - i.e. preserves limits - and so in particular it preserves binary products. Therefore if $A,B$ are objects of $\mathcal{C}$ we have a set isomorphism $\sigma$: + The proof follows a similar structure to the proof of \ref{completenesscomma}, so we will construct initial and terminal objects, binary products and coproducts and then we will prove that $\Pi_2$ is closed with respect to this constructions. \\ + \textbf{Initial and Terminal Object}: the initial object of $Cov(\mathcal{C})$ is $(0_{Set}, 0_\mathcal{C}, in)$ and the terminal object is $(1_{Set}, 1_\mathcal{C}, un)$. The construction is the same of the construction in \ref{completenesscomma} and therefore the proof is omitted \\ + \textbf{Binary Products}: Recall that $\mathcal{C}(1_\mathcal{C},-): \mathcal{C} \to Set$ is a continuos functor - i.e. preserves limits - and so in particular it preserves binary products. Therefore if $A,B$ are objects of $\mathcal{C}$ we have a set isomorphism $\sigma$: % https://q.uiver.app/#q=WzAsNCxbMSwwLCJcXG1hdGhjYWx7Q30oMV9cXG1hdGhjYWx7Q30sQSBcXHRpbWVzIEIpIl0sWzAsMCwiXFxtYXRoY2Fse0N9KDFfXFxtYXRoY2Fse0N9LCBBKSBcXHRpbWVzIFxcbWF0aGNhbHtDfSgxX1xcbWF0aGNhbHtDfSxCKSJdLFswLDEsIihoLCBrKSJdLFsxLDEsIjwgaCxrID46IDFfXFxtYXRoY2Fse0N9IFxcdG8gQSBcXHRpbWVzIEIgIl0sWzEsMF0sWzIsMywiIiwwLHsic3R5bGUiOnsidGFpbCI6eyJuYW1lIjoibWFwcyB0byJ9fX1dXQ== \[\begin{tikzcd} {\mathcal{C}(1_\mathcal{C}, A) \times \mathcal{C}(1_\mathcal{C},B)} & {\mathcal{C}(1_\mathcal{C},A \times B)} \\ @@ -468,7 +468,7 @@ \arrow["g"', from=3-2, to=3-3] \end{tikzcd}\] and we claim the projections of $(X\times Y, A \times B, \sigma \circ (f \times g))$ to be $(\pi_X,\pi_A):(X\times Y, A \times B, \sigma \circ (f \times g)) \to (X, A,f)$ and $(\pi_Y,\pi_B):(X\times Y, A \times B, \sigma \circ (f \times g)) \to (Y, B,f)$. \\ - We start by checking that our candidate projections are arrows of $Cov(\mathcal{C})$ and then we will check that the universal property holds. In order to check that our projection are actual arrows we shall prove that $ \mathcal{C}(1_\mathcal{C},\pi_A) \sigma (f \times g) = f \pi_X$ and that $\mathcal{C}(1_\mathcal{C}, \pi_B) \sigma (f \times g) = \pi_B g$ but this two identities follow directly from the fact that $\mathcal{C}(1_\mathcal{C},-)$ preserves products and that products are unique up to ismorphism, thus $\pi_1 = \mathcal{C}(\_\mathcal{C},-)\pi_A$ and $\pi_2 = \mathcal{C}(\_\mathcal{C},-)\pi_B$ and so the indentity are trivial. Let's move on to the universal property. + We start by checking that our candidate projections are arrows of $Cov(\mathcal{C})$ and then we will check that the universal property holds. In order to check that our projection are actual arrows we shall prove that $ \mathcal{C}(1_\mathcal{C},\pi_A) \sigma (f \times g) = f \pi_X$ and that $\mathcal{C}(1_\mathcal{C}, \pi_B) \sigma (f \times g) = \pi_B g$ but this two identities follow directly from the fact that $\mathcal{C}(1_\mathcal{C},-)$ preserves products and that products are unique up to isomorphism, thus $\pi_1 = \mathcal{C}(\_\mathcal{C},-)\pi_A$ and $\pi_2 = \mathcal{C}(\_\mathcal{C},-)\pi_B$ and so the identity are trivial. Let's move on to the universal property. \begin{itemize} \item existence: let consider the following situation: \begin{equation}\label{prodgoal} @@ -483,7 +483,7 @@ \arrow["{(l',m')}"', from=3-2, to=3-3] \end{tikzcd} \end{equation} - then, since $X \times Y$ and $A \times B$ are product rispectively in $Set$ and $\mathcal{C}$ we have the following commutative diagrams: + then, since $X \times Y$ and $A \times B$ are product respectively in $Set$ and $\mathcal{C}$ we have the following commutative diagrams: % https://q.uiver.app/#q=WzAsOCxbMSwwLCJYXFx0aW1lcyBZIl0sWzAsMSwiWCJdLFsyLDEsIlkiXSxbMSwxLCJaIl0sWzMsMSwiQSJdLFs0LDAsIkEgXFx0aW1lcyBCIl0sWzUsMSwiQiJdLFs0LDEsIkMiXSxbMCwxLCJcXHBpX1giLDJdLFswLDIsIlxccGlfWSJdLFszLDEsImwiXSxbMywyLCJsJyIsMl0sWzMsMCwiXFxleGlzdHMgISBcXGJhcntsfSIsMV0sWzUsNCwiXFxwaV9BIiwyXSxbNSw2LCJcXHBpX0IiXSxbNyw0LCJtIl0sWzcsNiwibSciLDJdLFs3LDUsIlxcZXhpc3RzICEgXFxiYXJ7bX0iLDFdXQ== \[\begin{tikzcd} & {X\times Y} &&& {A \times B} & \\ @@ -538,17 +538,17 @@ \arrow["{(\iota_{2})_\mathcal{C}}"', from=3-3, to=2-4] \end{tikzcd}\] The proof is dual to the proof of binary product and so we omit the calculations. \\ - \textbf{Projection is cartesian closed}: the proof of $\Pi_2$ beeing cartesian closed functor is identical to the proof of $\Pi_1$ beeing continous in \ref{completenesscomma} + \textbf{Projection is cartesian closed}: the proof of $\Pi_2$ being cartesian closed functor is identical to the proof of $\Pi_1$ being continuos in \ref{completenesscomma} \end{proof} \begin{remark} Notice that this theorem can be generalized for arbitrary comma categories with appropriate assumptions. In fact the construction process of products and coproduct of this proof can be generalized with the same approach we used in the proof of \ref{completenesscomma}. \end{remark} \begin{theorem} - Let $\mathcal{H}$ be an Heyting Algebra - i.e. an ordered bicartesian category which is also cartesian closed - then $Cov(\mathcal{H})$ is also bicartesian closed. Morover the projection $\Pi_2 : Cov(\mathcal{H}) \to \mathcal{H}$ is bicartesian closed. + Let $\mathcal{H}$ be an Heyting Algebra - i.e. an ordered bicartesian category which is also cartesian closed - then $Cov(\mathcal{H})$ is also bicartesian closed. Moreover the projection $\Pi_2 : Cov(\mathcal{H}) \to \mathcal{H}$ is bicartesian closed. \end{theorem} \begin{proof}\label{coverheyting} By \ref{cover of bicartesian} we know that $Cov(\mathcal{H})$ is bicartesian and that $\Pi_2$ is cartesian closed. So we need only to prove the existence of function spaces. \\ - Let $(X,a,f)$ and $(Y,b,g)$ be objects of $Cov(\mathcal{C})$. We expect the exponential $(Y,b,g)^{(X,a,f)}$ to be something similar to $(Y^X, a \to b, \xi)$ where $Y^X = Set(X,Y)$ is the function space in $Set$, $a \to b$ is the heyting implication in $\mathcal{H}$ and $\xi$ is a function somehow defined using that beeing $\mathcal{H}$ an ordered category if $a \to b$ in nonempty the in terminal in $Set$. This intuition is almost correct, but there is a problem: in order to make $(Y^X, a \to b, \xi)$ an object of $\mathcal{H}$ we need $\xi: Y^X \to \mathcal{H}(1_\mathcal{H}, a \to b)$ to be a $Set$-morphism which is not possible if $ 1_\mathcal{H} \nleq_{\mathcal{H}} a \to b $ and $Y^X \neq \emptyset$. Thus we claim $(Y,b,g)^{(X,a,f)}$ to be $( (Y^X)_{a \leq b}, a \to b , \xi)$ where $(Y^X)_{a \leq b}$ is $Set(X,Y)$ if $a \leq_{\mathcal{H}} b$ - and so $\mathcal{H}(1_\mathcal{H}, a \to b) \neq \emptyset$ - and to be empty otherwise: + Let $(X,a,f)$ and $(Y,b,g)$ be objects of $Cov(\mathcal{C})$. We expect the exponential $(Y,b,g)^{(X,a,f)}$ to be something similar to $(Y^X, a \to b, \xi)$ where $Y^X = Set(X,Y)$ is the function space in $Set$, $a \to b$ is the heyting implication in $\mathcal{H}$ and $\xi$ is a function somehow defined using that being $\mathcal{H}$ an ordered category if $a \to b$ in non-empty the in terminal in $Set$. This intuition is almost correct, but there is a problem: in order to make $(Y^X, a \to b, \xi)$ an object of $\mathcal{H}$ we need $\xi: Y^X \to \mathcal{H}(1_\mathcal{H}, a \to b)$ to be a $Set$-morphism which is not possible if $ 1_\mathcal{H} \nleq_{\mathcal{H}} a \to b $ and $Y^X \neq \emptyset$. Thus we claim $(Y,b,g)^{(X,a,f)}$ to be $( (Y^X)_{a \leq b}, a \to b , \xi)$ where $(Y^X)_{a \leq b}$ is $Set(X,Y)$ if $a \leq_{\mathcal{H}} b$ - and so $\mathcal{H}(1_\mathcal{H}, a \to b) \neq \emptyset$ - and to be empty otherwise: \begin{equation*} (Y^X)_{a \leq b} := \{ h \in Y^X | \forall x \in X (f(x)\in \mathcal{H}(1_\mathcal{H},a) \implies g(h(x)) \in \mathcal{H}(1_\mathcal{H},b) )\} \end{equation*} @@ -556,7 +556,7 @@ \begin{equation*} (\hat{u}, c \leq a \to b) : (Z,c,h) \to ( (Y^X)_{a \leq b}, a \to b, \xi ) \end{equation*} - This couple is clearly an arrow of $Cov(\mathcal{C})$\footnote{the arrow condition is forced since if every definition is coherent than the commutativity of the arrow condition square follows from $\mathcal{H}(1_\mathcal{H}, a \to b)$ be either termial or empty}. We claim that, if the abstraction is well defined, then the following diagram is commutative: + This couple is clearly an arrow of $Cov(\mathcal{C})$\footnote{the arrow condition is forced since if every definition is coherent than the commutativity of the arrow condition square follows from $\mathcal{H}(1_\mathcal{H}, a \to b)$ be either terminal or empty}. We claim that, if the abstraction is well defined, then the following diagram is commutative: \begin{equation}\label{exp goal} % https://q.uiver.app/#q=WzAsMyxbMCwyLCIoKFleWClfe2EgXFxsZXEgYn0sIGEgXFx0byBiLCBcXHhpKSAiXSxbMCwwLCIoWCwgYSxmKSBcXHRpbWVzIChaLGMsIGgpIl0sWzIsMiwiKFkpIl0sWzEsMiwiKHUsIGMgXFxsYW5kIGEgXFxsZXEgYikiXSxbMCwyLCIoRXZfe1NldH0sRXZfXFxtYXRoY2Fse0h9KSIsMl0sWzEsMCwiKFxcaGF0e3V9LCBjIFxcbGVxIGEgXFx0byBiICApIiwyXV0= \begin{tikzcd} @@ -568,11 +568,11 @@ \arrow["{(Ev_{Set},Ev_\mathcal{H})}"', from=3-1, to=3-3] \end{tikzcd} \end{equation} - Now, commutativity at the level of $Set$ is ensured by universal property of $Y^X$ in $Set$ plus the coerence condition of $(Y^X)_{a \leq b}$ with respect to $a \to b$. Now, in $\mathcal{H}$ we have that $c \land a \leq b$ we have that $c\land a \leq b \iff c \leq a \to b$ so we have commutativity at the level of $\mathcal{H}$. It remains to check the well definition of the abstraction: let $Z \neq \emptyset$, then $h(m) \in \mathcal{H}(1_\mathcal{H}, c)$ for $m \in Z$, hence $1_\mathcal{H} \leq c$ and so, since $c \land a \leq b$, then $a \leq b$ and so $\hat{u} (m) \in (Y^X)_{a \leq b}$ wihch ensure well definition of abstraction arrows. Notice that uniqueness of abstraction follows directly form uniqueness of abstractions in $Set$ and $\mathcal{H}$. Thus $Cov(\mathcal{H})$ is a bicartesian cloed category.\\ + Now, commutativity at the level of $Set$ is ensured by universal property of $Y^X$ in $Set$ plus the coherence condition of $(Y^X)_{a \leq b}$ with respect to $a \to b$. Now, in $\mathcal{H}$ we have that $c \land a \leq b$ we have that $c\land a \leq b \iff c \leq a \to b$ so we have commutativity at the level of $\mathcal{H}$. It remains to check the well definition of the abstraction: let $Z \neq \emptyset$, then $h(m) \in \mathcal{H}(1_\mathcal{H}, c)$ for $m \in Z$, hence $1_\mathcal{H} \leq c$ and so, since $c \land a \leq b$, then $a \leq b$ and so $\hat{u} (m) \in (Y^X)_{a \leq b}$ which ensure well definition of abstraction arrows. Notice that uniqueness of abstraction follows directly form uniqueness of abstractions in $Set$ and $\mathcal{H}$. Thus $Cov(\mathcal{H})$ is a bicartesian closed category.\\ Concerning $\Pi_2$, the proof is identical to \ref{completenesscomma} since the $\mathcal{H}$ level of the exponential object is the exponential object in $\mathcal{H}$ and so $\Pi_2$ is a bicartesian closed functor and we are done. \end{proof} \begin{remark}[Structure of $Cov(\mathcal{C})(1, A \oplus B)$] \label{global section in cover rmk} - Let $(u,v): 1 \to (X \oplus Y, a \lor b, f \oplus g)$ and consider $u: 1_{Set} \to X \oplus Y$. Since the coproudct in $Set$ is precisely the disjoint union, then $u(\star)$ is either in $X$ or in $Y$. Assume $u(\star) = x_0 = \iota_X(x_0) \in X$ for some $x_0 \in X$, then $(f \oplus g) \circ u = \Gamma v \circ un $ implies $v(\star)= (\iota_1)_\mathcal{C} \circ f (x_0)$. Since this argument is simmetrical if $u(\star) \in Y$ we have the following decomposition: + Let $(u,v): 1 \to (X \oplus Y, a \lor b, f \oplus g)$ and consider $u: 1_{Set} \to X \oplus Y$. Since the coproduct in $Set$ is precisely the disjoint union, then $u(\star)$ is either in $X$ or in $Y$. Assume $u(\star) = x_0 = \iota_X(x_0) \in X$ for some $x_0 \in X$, then $(f \oplus g) \circ u = \Gamma v \circ un $ implies $v(\star)= (\iota_1)_\mathcal{C} \circ f (x_0)$. Since this argument is symmetrical if $u(\star) \in Y$ we have the following decomposition: \[ Cov(\mathcal{C})(1, A \oplus B) \cong Cov(\mathcal{C})(1, A ) \sqcup Cov(\mathcal{C})(1, B) \] @@ -598,10 +598,10 @@ \end{tikzcd}\] \end{definition} \begin{lemma}\label{cover of cc is cc} - Let $\mathcal{C}$ be a category in \textbf{biCC}, then its freyd cover $Cov(\mathcal{C})$ is again in \textbf{biCC} + Let $\mathcal{C}$ be a category in \textbf{biCC}, then its Freyd cover $Cov(\mathcal{C})$ is again in \textbf{biCC} \end{lemma} \begin{proof} - We want to generalize contstruction in \ref{coverheyting} for arbitrary bicartesian closed category. We mimic the construction in \cite{angiuli2022canonicity}. Let $\mathcal{C}$ be a bicartesian closed category and let $(X,A,f)$ and $(Y,B,g)$, we want to construct the exponential object $(Y,B,g)^{(X,A,g)}$. As in \ref{coverheyting} we expect the exponential to mimic $(Y^X,B^C,h)$ where $h: Y^X \to \Gamma(B^C)$; however we know that $(Y^X)$ cannot be the rigth $Set$-component of the exponential. morevoer the constuction of $h$ in this situation is slithely more complex. Let's begin by noticing that the global set functor is a product preserving functor and that, since $\mathcal{C}$ is CC, in $\mathcal{C}$ there is the exponential object $B^A$ with evaluation $ev_\mathcal{C}: B^A \times B \to A$. We can form the composition : + We want to generalize the construction in \ref{coverheyting} for arbitrary bicartesian closed category. We mimic the construction in \cite{angiuli2022canonicity}. Let $\mathcal{C}$ be a bicartesian closed category and let $(X,A,f)$ and $(Y,B,g)$, we want to construct the exponential object $(Y,B,g)^{(X,A,g)}$. As in \ref{coverheyting} we expect the exponential to mimic $(Y^X,B^C,h)$ where $h: Y^X \to \Gamma(B^C)$; however we know that $(Y^X)$ cannot be the right $Set$-component of the exponential. The construction of $h$ in this situation is slightly more complex. Let's begin by noticing that the global set functor is a product preserving functor and that, since $\mathcal{C}$ is CC, in $\mathcal{C}$ there is the exponential object $B^A$ with evaluation $ev_\mathcal{C}: B^A \times B \to A$. We can form the composition : \begin{equation} u : \Gamma(A) \times \Gamma (B^A) \xrightarrow{\cong} \Gamma ( A \times B^A) \xrightarrow{\Gamma(ev_\mathcal{C})} \Gamma(B) \end{equation} @@ -620,7 +620,7 @@ \arrow["{\hat{u}}"', from=3-1, to=4-2] \arrow["{(\Gamma(B))^f}"', from=4-2, to=3-3] \end{tikzcd}\] - Where $(\Gamma (D))^f : \Gamma(B)^{\Gamma(A)} \to \Gamma(D)^X$ is defined by $\mu \to \mu \circ f$ and $g^X: Y^X \to \Gamma(B)$ is defined by $\omega \to g \circ \omega$. We claim $(H,B^A, h := p_1)$ to be the exponenetial object $(X,A,f)^{(Y,B,g)}$. We will prove this by checking that, provided $E = (Z,E,k)$, we have a natural ismoporphism $Cov(\mathcal{C})(E \times A, B)\cong Cov(\mathcal{C})(E, B^A)$\footnote{here we make an abuse of notation setting $A:= (X,A,f),B:=(Y,B,g)$}. \\ + Where $(\Gamma (D))^f : \Gamma(B)^{\Gamma(A)} \to \Gamma(D)^X$ is defined by $\mu \to \mu \circ f$ and $g^X: Y^X \to \Gamma(B)$ is defined by $\omega \to g \circ \omega$. We claim $(H,B^A, h := p_1)$ to be the exponential object $(X,A,f)^{(Y,B,g)}$. We will prove this by checking that, provided $E = (Z,E,k)$, we have a natural isomorphism $Cov(\mathcal{C})(E \times A, B)\cong Cov(\mathcal{C})(E, B^A)$\footnote{here we make an abuse of notation setting $A:= (X,A,f),B:=(Y,B,g)$}. \\ \textbf{Construction of $\Phi: \mathbf{Cov(\mathcal{C})(E \times A, B) \to Cov(\mathcal{C})(E, B^A)}$}: let $(\psi_{Set},\psi_{\mathcal{C}})$ in $Cov(\mathcal{C})(E \times A, B)$ then, $\psi_{Set}: Z \times X \to Y$ and $\psi_\mathcal{C}: E \times A \to B$. Thus, using that $Set$ and $\mathcal{C}$ are cartesian closed, we can take the abstractions $\hat{\psi}_{Set}: Z \to Y^X$ and $\hat{\psi}_{C}: E \to B^A$. Now, consider the following diagram: % https://q.uiver.app/#q=WzAsNixbMSwxLCJIIl0sWzEsMywiXFxHYW1tYShCXkEpIl0sWzMsMywiXFxHYW1tYShEKV5YIl0sWzMsMSwiWV5YIl0sWzAsMCwiWiJdLFswLDIsIlxcR2FtbWEoRSkiXSxbMCwxLCJwXzEiLDJdLFsxLDIsIlxcR2FtbWEoRCleZiBcXGNpcmMgXFxoYXR7dX0iLDJdLFszLDIsImdeWCJdLFswLDMsInBfMiJdLFswLDIsIiIsMCx7InN0eWxlIjp7Im5hbWUiOiJjb3JuZXIifX1dLFs0LDMsIlxcaGF0e1xccHNpfV97U2V0fSIsMCx7ImN1cnZlIjotMn1dLFs0LDUsImsiLDJdLFs1LDEsIlxcR2FtbWEoXFxoYXR7XFxwc2l9X3tcXG1hdGhjYWx7Q319KSIsMl0sWzQsMCwiXFxleGlzdCAhIFxccGhpX3tTZXR9IiwxXV0= \[\begin{tikzcd} @@ -670,12 +670,12 @@ \[ Ev_{Set} \circ (\hat{u} \times id_{\Gamma A}) \circ (\psi_2 \times f ) = \Gamma (Ev_{Set}) \circ \theta \circ (\psi_2 \times f) \] - Now, expliciting the definition of $\theta$ we get : + Now, making explicit the definition of $\theta$ we get : \[ Ev_{Set} \circ ( (\Gamma B^f \circ \hat{u} \circ \psi_2) \times id_X) = \Gamma ( \psi_{\mathcal{C}} ) \circ (k \times f) \] - Now, the arrow condition on $(\psi_{Set}, \psi_{\mathcal{C}})$ proves the equation \ref{pullback condition}, therefore by the universal property of $H$ there exists a unuque $\phi_{Set}: Z \to H$ that makes the pullback diagram commutative. Notice that the universal property of pullbacks ensure us that $(\phi_{Set}, \hat{\psi}_{\mathcal{C}})$ is an element of $Cov(\mathcal{C})(E, B^A)$. Morover, this construction is unique, hence we have defined a function $\Phi_1: Cov(\mathcal{C})(E \times A, B) \to Cov(\mathcal{C})(E, B^A)$ sending $(\psi_{Set}, \psi_{\mathcal{C}})$ to $(\phi_{Set}, \hat{\psi}_{\mathcal{C}})$. \\ - \textbf{Contrution of $\mathbf{\Psi:Cov(\mathcal{C})(E, B^A) \to Cov(\mathcal{C})(E \times A, B)}$}: let $(\psi_{Set}, \psi_\mathcal{C})$ in $Cov(\mathcal{C})(E,B^A)$, then the idea is to performe and "inverse construction" with respect to $\Phi$. Let's begin with the $Set$-component. \\ + Now, the arrow condition on $(\psi_{Set}, \psi_{\mathcal{C}})$ proves the equation \ref{pullback condition}, therefore by the universal property of $H$ there exists a unique $\phi_{Set}: Z \to H$ that makes the pullback diagram commutative. Notice that the universal property of pullbacks ensure us that $(\phi_{Set}, \hat{\psi}_{\mathcal{C}})$ is an element of $Cov(\mathcal{C})(E, B^A)$. Moreover, this construction is unique, hence we have defined a function $\Phi_1: Cov(\mathcal{C})(E \times A, B) \to Cov(\mathcal{C})(E, B^A)$ sending $(\psi_{Set}, \psi_{\mathcal{C}})$ to $(\phi_{Set}, \hat{\psi}_{\mathcal{C}})$. \\ + \textbf{Construction of $\mathbf{\Psi:Cov(\mathcal{C})(E, B^A) \to Cov(\mathcal{C})(E \times A, B)}$}: let $(\psi_{Set}, \psi_\mathcal{C})$ in $Cov(\mathcal{C})(E,B^A)$, then the idea is to perform an "inverse construction" with respect to $\Phi$. Let's begin with the $Set$-component. \\ We have $\psi_{Set}: Z \to H$, thus we define $\phi_1 := p_1 \circ \psi_{Set}$ and $\phi_2 := p_2 \circ \psi_{Set}$. Now, we use the exponential structure of $Set$ to define \[ \bar{\psi}_{Set} := Ev_{Set} \circ (\phi_1 \times id_X) : Z \times X \to Y @@ -709,22 +709,22 @@ \begin{align*} g \bar{\psi}_{Set} = Ev_{Set} \circ (u \times id_\Gamma A ) \circ (\Gamma \psi_\mathcal{C} \circ k , f) \end{align*} - Now, recalling the definition of $\hat{u}$ and expliciting the canonical isomorphism $\theta$ we get: + Now, recalling the definition of $\hat{u}$ and making explicit the canonical isomorphism $\theta$ we get: \begin{align*} g \circ \bar{\psi}_{Set} & = \Gamma (Ev_{Set}) \circ \theta \circ (\Gamma \psi_\mathcal{C} k, f) \\ & = \Gamma \bar{\psi}_\mathcal{C} \circ (k ,f) \end{align*} - Thus, since the universal property of exponentials and pullbacks grant us the unicity of $(\hat{\psi}_{Set}, \hat{\psi}_\mathcal{C})$ then we have succesfully constructed a morhpism $\Psi:Cov(\mathcal{C})(E, B^A) \to Cov(\mathcal{C})(E \times A, B)$. \\ - \textbf{ $\mathbf{\Phi}$ and $\mathbf{\Psi}$ gives a natural isomorphism}: this follows easily by construction. In fact evaulation and abstractions are inverse to each other and, since we are working in $Set$ with the two "projections" of set component the universal property of pullbacks grants us that our functions are inverse each other. Moreover our construction uses only mapping derived from natural isomorphisms - e.g. natural isomorphism of exponential objects in $Set$ and $\mathcal{C}$ and natural ismophism of pullbacks in $Set$. Therefore our assignment is natural. \\ + Thus, since the universal property of exponentials and pullbacks grant us the unicity of $(\hat{\psi}_{Set}, \hat{\psi}_\mathcal{C})$ then we have successfully constructed a morphism $\Psi:Cov(\mathcal{C})(E, B^A) \to Cov(\mathcal{C})(E \times A, B)$. \\ + \textbf{ $\mathbf{\Phi}$ and $\mathbf{\Psi}$ gives a natural isomorphism}: this follows easily by construction. In fact evaluation and abstractions are inverse to each other and, since we are working in $Set$ with the two "projections" of set component the universal property of pullbacks grants us that our functions are inverse each other. Moreover our construction uses only mapping derived from natural isomorphisms - e.g. natural isomorphism of exponential objects in $Set$ and $\mathcal{C}$ and natural isomorphism of pullbacks in $Set$. Therefore our assignment is natural. \\ We proved that there is a natural isomorphism: \[ Cov(\mathcal{C})(E \times A, B)\cong Cov(\mathcal{C})(E, B^A) \] - And so the, since the functor $(-)^A$ is right adjoint to $(-) \times A$, then our candidate $(H,B^A,h)$ is indeed the exponential objet. Thus $Cov(\mathcal{C})$ is cartesian closed and we are done. + And so the, since the functor $(-)^A$ is right adjoint to $(-) \times A$, then our candidate $(H,B^A,h)$ is indeed the exponential object. Thus $Cov(\mathcal{C})$ is cartesian closed and we are done. (AGGIUNGERE CONTI) \end{proof} \begin{remark} - Notice that the exponential construction in \ref{cover of cc is cc} agrees with exponential construction in \ref{coverheyting} if $\mathcal{C}$ is an ordered catgory + Notice that the exponential construction in \ref{cover of cc is cc} agrees with exponential construction in \ref{coverheyting} if $\mathcal{C}$ is an ordered category \end{remark} \begin{proposition} The Freyd Cover of a bicartesian closed category extend to an endofunctor @@ -753,11 +753,11 @@ where $F\cdot f: X \to \Gamma (FA)$ is the set function defined sending $x \in X$ into the global setion of $FA$ given by $F( f(x)) : 1_\mathcal{B} \to FA$ \end{proposition} \begin{proof} - By lemma \ref{cover of cc is cc} $Cov(-)$ exentend to a class function from \textbf{biCC} to itself. Then notice that the action of $Cov(-)$ over arrows of \textbf{biCC} is well defined since the commutativity condition of the arrow is preserved under the action of $F$. Indeed by definition of $F \cdot f$ and $F \cdot g$ we have: + By lemma \ref{cover of cc is cc} $Cov(-)$ extend to a class function from \textbf{biCC} to itself. Then notice that the action of $Cov(-)$ over arrows of \textbf{biCC} is well defined since the commutativity condition of the arrow is preserved under the action of $F$. Indeed by definition of $F \cdot f$ and $F \cdot g$ we have: \[ (F \cdot g) \circ h = F(\Gamma k) \circ (F \cdot f) \] - Ando so for $F: A \to B$ we proved $Cov(F): Cov(\mathcal{A}) \to Cov(\mathcal{B})$ is a well defined functor. Moreover, clearly $Cov(Id_\mathcal{A})= Id_{Cov(\mathcal{A})}$ and by definition of $F \cdot f$ thaking $F: \mathcal{A} \to \mathcal{B}$ and $G: \mathcal{B} \to \mathcal{C}$ we have that $Cov(G \cdot F) = Cov(G) \cdot Cov(F)$. Thus $Cov(-): \textbf{biCC} \to \textbf{biCC}$ is a well defined endofunctor + And so for $F: A \to B$ we proved $Cov(F): Cov(\mathcal{A}) \to Cov(\mathcal{B})$ is a well defined functor. Moreover, clearly $Cov(Id_\mathcal{A})= Id_{Cov(\mathcal{A})}$ and by definition of $F \cdot f$ taking $F: \mathcal{A} \to \mathcal{B}$ and $G: \mathcal{B} \to \mathcal{C}$ we have that $Cov(G \cdot F) = Cov(G) \cdot Cov(F)$. Thus $Cov(-): \textbf{biCC} \to \textbf{biCC}$ is a well defined endofunctor \end{proof} \begin{corollary} @@ -765,20 +765,20 @@ \[ Pcov(-): \textbf{bLat} \to \textbf{bLat} \] - where $Pcov(-) := Pos \circ Cov (-) : \textbf{bLat} \to\textbf{bLat}$ and \textbf{bLat} is the category of bounded lattices - i.e. lattice with top and bottom objects - with lattice homomorpishms as arrows. + where $Pcov(-) := Pos \circ Cov (-) : \textbf{bLat} \to\textbf{bLat}$ and \textbf{bLat} is the category of bounded lattices - i.e. lattice with top and bottom objects - with lattice homomorphisms as arrows. \end{corollary} \begin{proof} - By lemma \ref{cover of bicartesian} the cover of a lattice has a bicartesian structure and by lemma \ref{cover of cc is cc} the cover of a lattice is also a cartesian closed category. Therefore we are done, since the posettal reflection of a bicartesian closed category is a bounded lattice (the posettal reflection preserves bicartesian closed struture of the category) + By lemma \ref{cover of bicartesian} the cover of a lattice has a bicartesian structure and by lemma \ref{cover of cc is cc} the cover of a lattice is also a cartesian closed category. Therefore we are done, since the posettal reflection of a bicartesian closed category is a bounded lattice (the posettal reflection preserves bicartesian closed structure of the category) \end{proof} \section{Application to Logic} - After defining the construction of Freyd Cover and proving some baisc property of this constuction we are ready to use the categorical property of $Cov(-)$ to show some logical poperty of intuitionistic logic. In this section we will prove categorically that the propositional logic satisfy the disjunction property using the Freyd Cover construction. + After defining the construction of Freyd Cover and proving some basic property of this construction we are ready to use the categorical property of $Cov(-)$ to show some logical property of intuitionistic logic. In this section we will prove categorically that the propositional logic satisfy the disjunction property using the Freyd Cover construction. \begin{theorem} Let the following disjunction $\phi \lor \psi$ be derivable in the intuitionistic propositional logic. Then either $\phi$ or $\psi$ is derivable. \end{theorem} \begin{proof} \, \\ \textbf{Idea}: We will apply our construction to the Lindembaum-Tarski algebra of the initial intuitionistic propositional logic $\mathcal{T}_0$. Notice that the projection $\Pi_2: Cov(\mathcal{A}_i(\mathcal{T}_0)) \to \mathcal{A}_i(\mathcal{T}_0)$ "preserves" the Lindembaum component of element of $Cov(\mathcal{A}_i(\mathcal{T}_0))$,therefore a clever model build upon $Cov(\mathcal{A}_i(\mathcal{T}_0))$ respects the categorical semantics of $\mathcal{T}_0$. We will use this property to construct, from the fact that $\phi \lor \psi$ holds, a global section of $\phi$ or $\psi$, proving the disjunction property. \\ - \textbf{Proof}: Let $\mathcal{T}_0$ be the initial intuitionistic propositional theory and consider the Lindembaum-Tarki algebra $\mathcal{A}_i(\mathcal{T}_0)$. Notice that $Pcov(\mathcal{A}_i(\mathcal{T}_0))$is an heyting algebra, so we can define a model + \textbf{Proof}: Let $\mathcal{T}_0$ be the initial intuitionistic propositional theory and consider the Lindembaum-Tarski algebra $\mathcal{A}_i(\mathcal{T}_0)$. Notice that $Pcov(\mathcal{A}_i(\mathcal{T}_0))$is an heyting algebra, so we can define a model \[ \nu : Frm(\mathcal{T}_0) \to Pcov(\mathcal{A}_i(\mathcal{T}_0)) \] @@ -790,15 +790,15 @@ \[ X_A := \{ x = 0 | 1 \leq_{\mathcal{A}_i(\mathcal{T}_0)} [A] \} \] - notice that $[(X_A,[A],e_A)]$ is a well defined element of $Pcov(\mathcal{A}_i(\mathcal{T}_0))$. In fact if $A$ is the top in $\mathcal{A}_i(\mathcal{T}_0)$ then the set global section of the class $[A]$ is terminal and so $e_A$ is uniquely defined. Otherwise $X_A$ is inital in $Set$ and thus again $e_A$ is uniquely defined. Now, this definition can be extended inductively by lifting the Lindembaum-Tarski evalutation to the Freyd Cover, namely by setting $\nu(\psi \land psi) = \nu (\psi) \times \nu (\psi)$, $\nu (\psi \lor \psi) = \nu(\phi) \oplus \nu(\psi)$ and $\nu(\phi \to \psi)= \nu(\psi)^{\nu(\phi)}$. Therefore for an arbitrary formula $\phi$ we have: + notice that $[(X_A,[A],e_A)]$ is a well defined element of $Pcov(\mathcal{A}_i(\mathcal{T}_0))$. In fact if $A$ is the top in $\mathcal{A}_i(\mathcal{T}_0)$ then the set global section of the class $[A]$ is terminal and so $e_A$ is uniquely defined. Otherwise $X_A$ is initial in $Set$ and thus again $e_A$ is uniquely defined. Now, this definition can be extended inductively by lifting the Lindembaum-Tarski evaluation to the Freyd Cover, namely by setting $\nu(\psi \land psi) = \nu (\psi) \times \nu (\psi)$, $\nu (\psi \lor \psi) = \nu(\phi) \oplus \nu(\psi)$ and $\nu(\phi \to \psi)= \nu(\psi)^{\nu(\phi)}$. Therefore for an arbitrary formula $\phi$ we have: \[ \nu(\phi) = [(X_\phi, [\phi], e_\phi)] \] - for some $X_\phi$ in $Set$ and $e_\phi : X_\phi \to \Gamma [\phi]$. Notice that in this model the evaluation of a propositional variable $A$ and $\neg A$ is neither the top or the bottom (this follows immediatly by construction). \\ + for some $X_\phi$ in $Set$ and $e_\phi : X_\phi \to \Gamma [\phi]$. Notice that in this model the evaluation of a propositional variable $A$ and $\neg A$ is neither the top or the bottom (this follows immediately by construction). \\ Let $\phi \lor \psi$ be a tautology in $\mathcal{T}_0$, therefore \[\nu(\phi \lor \psi) = 1_{Pcov(\mathcal{A}_i(\mathcal{T}_0))} = [(1_{Set}, 1_{\mathcal{A}_i(\mathcal{T}_0)},un)] \] Now, since $\nu (\phi \lor \psi)= \nu(\phi) \oplus \nu(\psi)$ there exists a global section $g: 1_{Pcov(\mathcal{A}_i(\mathcal{T}_0))} \to [(X_\phi , [\phi] , e_\phi)] \oplus [(X_\psi , [\psi], e_\psi)]$. - By remark \ref{global section in cover rmk} $g = (g_1,g_2): 1_{Pcov(\mathcal{A}_i(\mathcal{T}_0))} \to \nu(\phi)$ or $g = (g_1,g_2): 1_{Pcov(\mathcal{A}_i(\mathcal{T}_0))} \to \nu(\psi)$. Now, in the first case $e_\phi (g_1(\star)) \in \Gamma [\phi]$ hence $\phi$ is provable in the logic. In the second case $e_\psi (g_1(\star)) \in \Gamma [\psi]$ and so $\psi$ is provable in the logic. Thus the disjunctin property is proved and so we are done. + By remark \ref{global section in cover rmk} $g = (g_1,g_2): 1_{Pcov(\mathcal{A}_i(\mathcal{T}_0))} \to \nu(\phi)$ or $g = (g_1,g_2): 1_{Pcov(\mathcal{A}_i(\mathcal{T}_0))} \to \nu(\psi)$. Now, in the first case $e_\phi (g_1(\star)) \in \Gamma [\phi]$ hence $\phi$ is provable in the logic. In the second case $e_\psi (g_1(\star)) \in \Gamma [\psi]$ and so $\psi$ is provable in the logic. Thus the disjunction property is proved and so we are done.$e_\phi (g_1(\star)) \in \Gamma [\phi]$ \end{proof} \printbibliography \end{document} \ No newline at end of file