From eff40ed9a9cbeb5a37b22830a171097d109ee444 Mon Sep 17 00:00:00 2001 From: Daniel Schepler Date: Sat, 6 Jun 2026 16:49:21 -0400 Subject: [PATCH 1/4] Add several negative properties of LRS_R deduced from those of Top --- content/LRS-not-cartesian-closed.md | 63 +++++++++++++++++++ databases/catdat/data/categories/LRS_R.yaml | 16 +++++ .../cartesian closed.yaml | 9 +++ 3 files changed, 88 insertions(+) create mode 100644 content/LRS-not-cartesian-closed.md diff --git a/content/LRS-not-cartesian-closed.md b/content/LRS-not-cartesian-closed.md new file mode 100644 index 00000000..deac6003 --- /dev/null +++ b/content/LRS-not-cartesian-closed.md @@ -0,0 +1,63 @@ +--- +title: $\LRS_R$ is not cartesian closed +description: A proof that $\LRS_R$ is not cartesian closed for any ring $R$ +author: Daniel Schepler +--- + +For most of this development, we will be dealing with the case of $\LRS_k$ where $k$ is a field. We begin by describing $\Top$ as a reflective subcategory of $\LRS_k$. + +::: Lemma 1 +The forgetful functor $U : \LRS_k \to \Top$ has a right adjoint $K : \Top \to \LRS_k$ of equipping a topological space $X$ with the constant sheaf of $k$. Furthermore, the functor $K$ is fully faithful, thus making $\Top$ into a reflective subcategory of $\LRS_k$. +::: + +_Proof._ In this adjunction, the counit $UK \to \id$ is just the identity. To describe the unit $\id \to KU$, we need to define a morphism $(X, \O_X) \to (X, k)$ for any locally ringed space $(X, \O_X)$. This morphism will be the identity on topological spaces, and the pullback operation $k \to \O_X$ will be the unique morphism of sheaves induced by the given structure of $\O_X$ as a sheaf of $k$-algebras. It is now straightforward to check this indeed defines an adjunction; and since the counit is an isomorphism, that implies that $K$ is fully faithful.$\square$ + +We now show that this reflective subcategory is in fact also a coreflective subcategory. + +::: Lemma 2 +For each object $X$ of $\LRS_k$, let $X_0$ be the set of points of $X$ such that the induced morphism from $k$ to the residue field $\kappa(x)$ is an isomorphism. We give $X_0$ the following strengthening of the subspace topology: it will be the topology where a neighborhood subbasis at $x \in X_0$ is the collection of sets of the form $X_0 \cap Z(f)$ where $f \in O_X(U)$ for some neighborhood $U$ of $x$ in $X$, and $x \in Z(f)$. Then $S : \LRS_k \to \Top$, $X \mapsto X_0$, is a right adjoint of $K$. +::: + +_Proof._ First, to see that $S$ defines a functor, suppose we have a morphism $f : (X, \O_X) \to (Y, \O_Y)$. Then for $x \in X_0$, we have a sequence $k \to \kappa(f(x)) \to \kappa(x)$ where the composition is an identity. Thus, $\kappa(f(x)) \to \kappa(x)$ is a surjective morphism of fields, and therefore an isomorphism. It follows that $k \to \kappa(f(x))$ is also an isomorphism of fields, so $f(x) \in Y_0$. To see that the restriction map $X_0 \to Y_0$ is continuous, suppose $g \in O_Y(V)$ is such that $f(x) \in Z(g)$. Then $x \in Z(f^\sharp g)$, and +$$X_0 \cap f^{-1}(Y_0 \cap Z(g)) = X_0 \cap Z(f^\sharp g)$$ +where $f^\sharp g \in \O_X(f^{-1}(V))$. In other words, we have shown that the inverse image in $X_0$ of any subbasic neighborhood of $f(x)$ is a neighborhood of $x$. + +Now if we apply the functor $S$ to a space of the form $(X, k)$, then since by definition any section of the constant sheaf $k$ is locally constant, we see that we recover exactly $X$ with its original topology. Thus, we can define the unit $\id \to SK$ of the adjunction to be the identity. As for the counit $KS \to \id$, for any locally ringed space $(X, \O_X)$ we need to define a morphism $(X_0, k) \to (X, \O_X)$. The map of topological spaces will be the inclusion map $i : X_0 \hookrightarrow X$, which is continuous since in particular for $U$ an open neighborhood of $x \in X_0$, $X_0 \cap U = X_0 \cap Z(0_U)$ where $0_U \in \O_X(U)$ is the zero element. The pullback map $\O_X \to i_* k$ will be defined to take $f \in \O_X(U)$ to the function $X_0 \cap U \to k$ where $x \in X_0 \cap U$ maps to the inverse image of $f_x + \m_{X, x} \in \kappa(x)$ under the isomorphism $k \to \kappa(x)$. An alternative description of this pullback is that $x \in X_0 \cap U$ maps to the unique $a\in k$ such that $x \in Z(f-a)$. Since $Z(f-a)$ is a neighborhood of $x$ in $X_0$ by definition, this shows that we get a locally constant function to $k$ as required. + +From here, it is straightforward to show that this does in fact define an adjuction.$\square$ + +_Remark._ In the special case where $k$ is a finite field, we have +$$X_0 \cap Z(f) = \bigcap_{a \in k^*} (X_0 \cap D(f-a)),$$ +which is already open in the subspace topology. Therefore, in this case, $X_0$ is given exactly the subspace topology. + +Now, a formal argument shows that in this situation, cartesian closedness passes to the subcategory. + +::: Lemma 3 +Suppose $\D$ is a coreflective subcategory of $\C$ such that $\D$ has binary products and the inclusion functor preserves these binary products. If $\C$ is cartesian closed, then so is $\D$. +::: + +_Proof._ Let $U : \D \to \C$ be the inclusion functor with right adjoint $R : \C \to \D$. Then for any objects $X, Y, Z$ of $\D$ we have +$$\begin{align*} +\Hom_\D(Z\times X, Y) & \simeq \Hom_\C(UZ \times UX, UY) \\ +& \simeq \Hom_\C(UZ, (UY)^{UX}) \\ +& \simeq \Hom_\D(Z, R((UY)^{UX})). +\end{align*}$$ +It is also easy to see this isomorphism is natural in $Z$. Thus, $R((UY)^{UX})$ forms an exponential $Y^X$ in $\D$.$\square$ + +::: Corollary 4 +If $k$ is a field, then $\LRS_k$ is not cartesian closed. +::: + +_Proof._ This follows immediately from Lemma 3 since we know that the reflective and coreflective subcategory $\Top$ of $\LRS_k$ is not cartesian closed.$\square$ + +::: Corollary 5 +If $\C$ is a cartesian closed category and $P$ is a subterminal object (i.e. an object such that the unique morphism $P \to 1$ is a monomorphism), then the slice category $\C / P$ is also cartesian closed. +::: + +_Proof._ The forgetful functor $\C / P \to \C$ is fully faithful; it has right adjoint ${-} \times P$; and it preserves binary products (in fact all inhabited limits).$\square$ + +::: Corollary 6 +For any non-trivial ring $R$, $\LRS_R$ is not cartesian closed. +::: + +_Proof._ Choose any maximal ideal $\m$ of $R$ with quotient field $k$. Then $\LRS_k$ is equivalent to the slice category $\LRS_R / \Spec k$ where $\Spec k$ is subterminal.$\square$ diff --git a/databases/catdat/data/categories/LRS_R.yaml b/databases/catdat/data/categories/LRS_R.yaml index 112b01fe..f5417843 100644 --- a/databases/catdat/data/categories/LRS_R.yaml +++ b/databases/catdat/data/categories/LRS_R.yaml @@ -53,6 +53,22 @@ unsatisfied_properties: Alternatively, using the usual adjunction between affine schemes and locally ringed spaces (EGA I (1971), Ch. 1, Prop. 1.6.3), a generating set in $\LRS_R$ would induce a generating set in the category of affine $R$-schemes, which contradicts the fact that $\CAlg(R)$ does not have a cogenerating set. + - property: cartesian closed + proof: This is proved here. + check_redundancy: false + + - property: cartesian filtered colimits + proof: As a corollary of the results here, if we choose a quotient field $k$ of $R$, then the functor $\Top \to \LRS_R$ of equipping a topological space with the constant sheaf of $k$ is fully faithful, and preserves all colimits and all inhabited limits. Therefore, if $\LRS_R$ had cartesian filtered colimits, then $\Top$ would also, giving a contradiction. + + - property: regular + proof: 'As a corollary of the results here, if we choose a quotient field $k$ of $R$, then the functor $\Top \to \LRS_R$ of equipping a topological space with the constant sheaf of $k$ is fully faithful, and preserves all colimits and all inhabited limits. Therefore, if $\LRS_R$ were regular, then for every regular epimorphism $X \to Y$ and morphism $Z \to Y$, the pullback $f : Z \times_Y X \to X$ would satisfy that the canonical morphism from the quotient of the kernel pair of $f$ to $X$ is an isomorphism. This would be inherited by the subcategory $\Top$, and since $\Top$ is finitely complete and has coequalizers, that would imply $\Top$ is also regular, giving a contradiction.' + + - property: cofiltered-limit-stable epimorphisms + proof: 'As a corollary of the results here, if we choose a quotient field $k$ of $R$, then the functor $\Top \to \LRS_R$ of equipping a topological space with the constant sheaf of $k$ is fully faithful, and preserves all colimits and all inhabited limits. From here, the proof is similar to the one from $\Top$: we apply the contrapositive of the dual of this lemma to the functor $\Set \to \LRS_k$ which equips a set with the indiscrete topology and the constant sheaf of $k$.' + + - property: effective cocongruences + proof: As a corollary of the results here, if we choose a quotient field $k$ of $R$, then the functor $\Top \to \LRS_R$ of equipping a topological space with the constant sheaf of $k$ is fully faithful, and preserves all colimits and all inhabited limits. Therefore, if $\LRS_R$ had effective cocongruences, then for every cocongruence $E$ on $X$, the canonical morphism from the cokernel pair of the equalizer of $E$ to $E$ would be an isomorphism. This would be inherited by the subcategory $\Top$, giving a contradiction. + special_objects: initial object: description: empty space diff --git a/databases/catdat/data/category-implications/cartesian closed.yaml b/databases/catdat/data/category-implications/cartesian closed.yaml index a952b113..e4dab838 100644 --- a/databases/catdat/data/category-implications/cartesian closed.yaml +++ b/databases/catdat/data/category-implications/cartesian closed.yaml @@ -107,3 +107,12 @@ - locally cartesian closed proof: Each slice is thin, semi-strongly connected, and has a terminal object. Thus, it corresponds to a linear order with a largest element $1$. Every such category is cartesian closed, where the exponential $a \Rightarrow b$ (Heyting implication) is $1$ when $a \leq b$ and otherwise $b$. is_equivalence: false + +- id: cartesian_closed_thin_implies_lcc + assumptions: + - cartesian closed + - thin + conclusions: + - locally cartesian closed + proof: First, note that the hypotheses imply that the category is finitely complete. Now every object is subterminal, so the result follows from Corollary 5 here. + is_equivalence: false From d78b892682cce389314486707bf055d20dac806a Mon Sep 17 00:00:00 2001 From: Script Raccoon Date: Sun, 7 Jun 2026 00:25:20 +0200 Subject: [PATCH 2/4] remove latex from title and description; add heading --- content/LRS-not-cartesian-closed.md | 6 ++++-- 1 file changed, 4 insertions(+), 2 deletions(-) diff --git a/content/LRS-not-cartesian-closed.md b/content/LRS-not-cartesian-closed.md index deac6003..20b45f23 100644 --- a/content/LRS-not-cartesian-closed.md +++ b/content/LRS-not-cartesian-closed.md @@ -1,9 +1,11 @@ --- -title: $\LRS_R$ is not cartesian closed -description: A proof that $\LRS_R$ is not cartesian closed for any ring $R$ +title: The category of locally ringed spaces is not cartesian closed +description: A proof that the category of locally ringed spaces over a non-trivial ring R is not cartesian closed author: Daniel Schepler --- +## The category of locally ringed spaces is not cartesian closed + For most of this development, we will be dealing with the case of $\LRS_k$ where $k$ is a field. We begin by describing $\Top$ as a reflective subcategory of $\LRS_k$. ::: Lemma 1 From c65bede169e73f5b30aca7727d4bc054ca25c81b Mon Sep 17 00:00:00 2001 From: Script Raccoon Date: Sun, 7 Jun 2026 09:56:26 +0200 Subject: [PATCH 3/4] improve formatting and wording in LRS proof --- .cspell.json | 1 + content/LRS-not-cartesian-closed.md | 50 +++++++++++++++++------------ 2 files changed, 30 insertions(+), 21 deletions(-) diff --git a/.cspell.json b/.cspell.json index 158edf39..b4091af6 100644 --- a/.cspell.json +++ b/.cspell.json @@ -45,6 +45,7 @@ "Čech", "characterisation", "clopen", + "closedness", "Clowder", "coaccessible", "cocartesian", diff --git a/content/LRS-not-cartesian-closed.md b/content/LRS-not-cartesian-closed.md index 20b45f23..a20728b2 100644 --- a/content/LRS-not-cartesian-closed.md +++ b/content/LRS-not-cartesian-closed.md @@ -9,27 +9,29 @@ author: Daniel Schepler For most of this development, we will be dealing with the case of $\LRS_k$ where $k$ is a field. We begin by describing $\Top$ as a reflective subcategory of $\LRS_k$. ::: Lemma 1 -The forgetful functor $U : \LRS_k \to \Top$ has a right adjoint $K : \Top \to \LRS_k$ of equipping a topological space $X$ with the constant sheaf of $k$. Furthermore, the functor $K$ is fully faithful, thus making $\Top$ into a reflective subcategory of $\LRS_k$. +The forgetful functor $U : \LRS_k \to \Top$ has a right adjoint $K : \Top \to \LRS_k$ of equipping a topological space $X$ with the constant sheaf $\underline{k}$. Furthermore, the functor $K$ is fully faithful, thus making $\Top$ into a reflective subcategory of $\LRS_k$. ::: -_Proof._ In this adjunction, the counit $UK \to \id$ is just the identity. To describe the unit $\id \to KU$, we need to define a morphism $(X, \O_X) \to (X, k)$ for any locally ringed space $(X, \O_X)$. This morphism will be the identity on topological spaces, and the pullback operation $k \to \O_X$ will be the unique morphism of sheaves induced by the given structure of $\O_X$ as a sheaf of $k$-algebras. It is now straightforward to check this indeed defines an adjunction; and since the counit is an isomorphism, that implies that $K$ is fully faithful.$\square$ +_Proof._ In this adjunction, the counit $UK \to \id$ is just the identity. To describe the unit $\id \to KU$, we need to define a morphism $(X, \O_X) \to (X, \underline{k})$ for any locally ringed space $(X, \O_X)$ over $k$. This morphism will be the identity on topological spaces, and the pullback operation $\underline{k} \to \O_X$ will be the unique morphism of sheaves induced by the given structure of $\O_X$ as a sheaf of $k$-algebras. It is now straightforward to check this indeed defines an adjunction; and since the counit is an isomorphism, that implies that $K$ is fully faithful. $\square$ -We now show that this reflective subcategory is in fact also a coreflective subcategory. +We now show that this reflective subcategory is in fact also a coreflective subcategory. Recall that for $f \in \O_X(U)$ we have its vanishing set $V(f) \coloneqq \{x \in U : f(x) = 0\}$, where $f(x) \in \kappa(x)$ is the image of $f_x \in \O_{X,x}$ in the residue field. ::: Lemma 2 -For each object $X$ of $\LRS_k$, let $X_0$ be the set of points of $X$ such that the induced morphism from $k$ to the residue field $\kappa(x)$ is an isomorphism. We give $X_0$ the following strengthening of the subspace topology: it will be the topology where a neighborhood subbasis at $x \in X_0$ is the collection of sets of the form $X_0 \cap Z(f)$ where $f \in O_X(U)$ for some neighborhood $U$ of $x$ in $X$, and $x \in Z(f)$. Then $S : \LRS_k \to \Top$, $X \mapsto X_0$, is a right adjoint of $K$. +For each object $X$ of $\LRS_k$, let $X_0$ be the set of points $x \in X$ such that the induced morphism from $k$ to the residue field $\kappa(x)$ is an isomorphism. We give $X_0$ the following strengthening of the subspace topology: it will be the topology where a neighborhood subbasis at $x \in X_0$ is the collection of sets of the form $X_0 \cap V(f)$ where $f \in \O_X(U)$ for some neighborhood $U$ of $x$ in $X$, and $x \in V(f)$. Then $X \mapsto X_0$ defines a functor $S : \LRS_k \to \Top$ that is right adjoint to $K$. ::: -_Proof._ First, to see that $S$ defines a functor, suppose we have a morphism $f : (X, \O_X) \to (Y, \O_Y)$. Then for $x \in X_0$, we have a sequence $k \to \kappa(f(x)) \to \kappa(x)$ where the composition is an identity. Thus, $\kappa(f(x)) \to \kappa(x)$ is a surjective morphism of fields, and therefore an isomorphism. It follows that $k \to \kappa(f(x))$ is also an isomorphism of fields, so $f(x) \in Y_0$. To see that the restriction map $X_0 \to Y_0$ is continuous, suppose $g \in O_Y(V)$ is such that $f(x) \in Z(g)$. Then $x \in Z(f^\sharp g)$, and -$$X_0 \cap f^{-1}(Y_0 \cap Z(g)) = X_0 \cap Z(f^\sharp g)$$ +_Proof._ First, to see that $S$ is a functor, suppose we have a morphism $f : (X, \O_X) \to (Y, \O_Y)$. Then for $x \in X_0$, we have a sequence $k \to \kappa(f(x)) \to \kappa(x)$ where the composition is an isomorphism. Thus, $\kappa(f(x)) \to \kappa(x)$ is a surjective morphism of fields, and therefore an isomorphism. It follows that $k \to \kappa(f(x))$ is also an isomorphism of fields, so $f(x) \in Y_0$. To see that the restriction map $X_0 \to Y_0$ is continuous, suppose $g \in \O_Y(V)$ is such that $f(x) \in V(g)$. Then $x \in V(f^\sharp g)$, and +$$X_0 \cap f^{-1}(Y_0 \cap V(g)) = X_0 \cap V(f^\sharp g)$$ where $f^\sharp g \in \O_X(f^{-1}(V))$. In other words, we have shown that the inverse image in $X_0$ of any subbasic neighborhood of $f(x)$ is a neighborhood of $x$. -Now if we apply the functor $S$ to a space of the form $(X, k)$, then since by definition any section of the constant sheaf $k$ is locally constant, we see that we recover exactly $X$ with its original topology. Thus, we can define the unit $\id \to SK$ of the adjunction to be the identity. As for the counit $KS \to \id$, for any locally ringed space $(X, \O_X)$ we need to define a morphism $(X_0, k) \to (X, \O_X)$. The map of topological spaces will be the inclusion map $i : X_0 \hookrightarrow X$, which is continuous since in particular for $U$ an open neighborhood of $x \in X_0$, $X_0 \cap U = X_0 \cap Z(0_U)$ where $0_U \in \O_X(U)$ is the zero element. The pullback map $\O_X \to i_* k$ will be defined to take $f \in \O_X(U)$ to the function $X_0 \cap U \to k$ where $x \in X_0 \cap U$ maps to the inverse image of $f_x + \m_{X, x} \in \kappa(x)$ under the isomorphism $k \to \kappa(x)$. An alternative description of this pullback is that $x \in X_0 \cap U$ maps to the unique $a\in k$ such that $x \in Z(f-a)$. Since $Z(f-a)$ is a neighborhood of $x$ in $X_0$ by definition, this shows that we get a locally constant function to $k$ as required. +Now if we apply the functor $S$ to a space of the form $(X, \underline{k})$, then since by definition any section of the constant sheaf $\underline{k}$ is locally constant, we see that we recover exactly $X$ with its original topology. Thus, we can define the unit $\id \to SK$ of the adjunction to be the identity. -From here, it is straightforward to show that this does in fact define an adjuction.$\square$ +As for the counit $KS \to \id$, for any locally ringed space $(X, \O_X)$ over $k$ we need to define a morphism $(X_0, \underline{k}) \to (X, \O_X)$. The map of topological spaces will be the inclusion map $i : X_0 \hookrightarrow X$, which is continuous since in particular for $U$ an open neighborhood of $x \in X_0$ we have $X_0 \cap U = X_0 \cap V(0_U)$, where $0_U \in \O_X(U)$ is the zero element. The pullback map $\O_X \to i_* \underline{k}$ takes $f \in \O_X(U)$ to the function $X_0 \cap U \to k$ where $x \in X_0 \cap U$ maps to the inverse image of $f(x) \in \kappa(x)$ under the isomorphism $k \to \kappa(x)$. An alternative description of this pullback is that $x \in X_0 \cap U$ maps to the unique $a\in k$ such that $x \in V(f-a)$. Since $X_0 \cap V(f-a)$ is a neighborhood of $x$ in $X_0$ by definition, this shows that we get a locally constant function to $k$ as required. + +From here, it is straightforward to show that this does in fact define an adjunction. $\square$ _Remark._ In the special case where $k$ is a finite field, we have -$$X_0 \cap Z(f) = \bigcap_{a \in k^*} (X_0 \cap D(f-a)),$$ +$$\textstyle X_0 \cap V(f) = \bigcap_{a \in k^\times} (X_0 \cap D(f-a)),$$ which is already open in the subspace topology. Therefore, in this case, $X_0$ is given exactly the subspace topology. Now, a formal argument shows that in this situation, cartesian closedness passes to the subcategory. @@ -38,28 +40,34 @@ Now, a formal argument shows that in this situation, cartesian closedness passes Suppose $\D$ is a coreflective subcategory of $\C$ such that $\D$ has binary products and the inclusion functor preserves these binary products. If $\C$ is cartesian closed, then so is $\D$. ::: -_Proof._ Let $U : \D \to \C$ be the inclusion functor with right adjoint $R : \C \to \D$. Then for any objects $X, Y, Z$ of $\D$ we have -$$\begin{align*} -\Hom_\D(Z\times X, Y) & \simeq \Hom_\C(UZ \times UX, UY) \\ -& \simeq \Hom_\C(UZ, (UY)^{UX}) \\ -& \simeq \Hom_\D(Z, R((UY)^{UX})). -\end{align*}$$ -It is also easy to see this isomorphism is natural in $Z$. Thus, $R((UY)^{UX})$ forms an exponential $Y^X$ in $\D$.$\square$ +_Proof._ Let $U : \D \to \C$ be the inclusion functor with right adjoint $R : \C \to \D$. Then for any objects $X, Y, Z$ of $\D$ we have natural isomorphisms + +$$ +\begin{align*} +\Hom_\D(Z\times X, Y) & \cong \Hom_\C(UZ \times UX, UY) \\ +& \cong \Hom_\C(UZ, [UX,UY]) \\ +& \cong \Hom_\D\bigl(Z, R([UX,UY])\bigr). +\end{align*} +$$ + +Thus, $R([UX,UY])$ forms an exponential $[X,Y]$ in $\D$. $\square$ ::: Corollary 4 If $k$ is a field, then $\LRS_k$ is not cartesian closed. ::: -_Proof._ This follows immediately from Lemma 3 since we know that the reflective and coreflective subcategory $\Top$ of $\LRS_k$ is not cartesian closed.$\square$ +_Proof._ This follows immediately from Lemma 3 since we know that the reflective and coreflective subcategory $\Top$ of $\LRS_k$ is not cartesian closed. $\square$ + +Next, we generalize this result to arbitrary base rings. ::: Corollary 5 -If $\C$ is a cartesian closed category and $P$ is a subterminal object (i.e. an object such that the unique morphism $P \to 1$ is a monomorphism), then the slice category $\C / P$ is also cartesian closed. +If $\C$ is a cartesian closed category and $P$ is a [subterminal object](https://ncatlab.org/nlab/show/subterminal+object) of $\C$, then the slice category $\C / P$ is also cartesian closed. ::: -_Proof._ The forgetful functor $\C / P \to \C$ is fully faithful; it has right adjoint ${-} \times P$; and it preserves binary products (in fact all inhabited limits).$\square$ +_Proof._ The forgetful functor $\C / P \to \C$ is fully faithful; it has right adjoint ${-} \times P$; and it preserves binary products (in fact all inhabited limits). Hence, Lemma 3 applies. $\square$ ::: Corollary 6 -For any non-trivial ring $R$, $\LRS_R$ is not cartesian closed. +For any non-trivial commutative ring $R$, the category $\LRS_R$ is not cartesian closed. ::: -_Proof._ Choose any maximal ideal $\m$ of $R$ with quotient field $k$. Then $\LRS_k$ is equivalent to the slice category $\LRS_R / \Spec k$ where $\Spec k$ is subterminal.$\square$ +_Proof._ Let $k$ be a residue field of $R$. Then $\LRS_k$ is equivalent to the slice category $\LRS_R / \Spec k$ where $\Spec k$ is subterminal. $\square$ From c07afb98e2baa894349516d0ea5600aaf06419fb Mon Sep 17 00:00:00 2001 From: Script Raccoon Date: Sun, 7 Jun 2026 10:32:52 +0200 Subject: [PATCH 4/4] extract results about cartesian closed categories --- content/LRS-not-cartesian-closed.md | 34 +++---------------- content/cartesian-closed-results.md | 27 +++++++++++++++ .../cartesian closed.yaml | 2 +- 3 files changed, 33 insertions(+), 30 deletions(-) create mode 100644 content/cartesian-closed-results.md diff --git a/content/LRS-not-cartesian-closed.md b/content/LRS-not-cartesian-closed.md index a20728b2..96af5c59 100644 --- a/content/LRS-not-cartesian-closed.md +++ b/content/LRS-not-cartesian-closed.md @@ -34,40 +34,16 @@ _Remark._ In the special case where $k$ is a finite field, we have $$\textstyle X_0 \cap V(f) = \bigcap_{a \in k^\times} (X_0 \cap D(f-a)),$$ which is already open in the subspace topology. Therefore, in this case, $X_0$ is given exactly the subspace topology. -Now, a formal argument shows that in this situation, cartesian closedness passes to the subcategory. - -::: Lemma 3 -Suppose $\D$ is a coreflective subcategory of $\C$ such that $\D$ has binary products and the inclusion functor preserves these binary products. If $\C$ is cartesian closed, then so is $\D$. -::: - -_Proof._ Let $U : \D \to \C$ be the inclusion functor with right adjoint $R : \C \to \D$. Then for any objects $X, Y, Z$ of $\D$ we have natural isomorphisms - -$$ -\begin{align*} -\Hom_\D(Z\times X, Y) & \cong \Hom_\C(UZ \times UX, UY) \\ -& \cong \Hom_\C(UZ, [UX,UY]) \\ -& \cong \Hom_\D\bigl(Z, R([UX,UY])\bigr). -\end{align*} -$$ - -Thus, $R([UX,UY])$ forms an exponential $[X,Y]$ in $\D$. $\square$ - -::: Corollary 4 +::: Corollary 3 If $k$ is a field, then $\LRS_k$ is not cartesian closed. ::: -_Proof._ This follows immediately from Lemma 3 since we know that the reflective and coreflective subcategory $\Top$ of $\LRS_k$ is not cartesian closed. $\square$ - -Next, we generalize this result to arbitrary base rings. +_Proof._ We know that $\Top$ is not cartesian closed. By Lemma 2, $\Top$ is a coreflective subcategory of $\LRS_k$. Moreover, the inclusion preserves binary products (in fact, all limits) by Lemma 1. Therefore, the claim follows from Lemma 1 [here](/content/cartesian-closed-results). $\square$ -::: Corollary 5 -If $\C$ is a cartesian closed category and $P$ is a [subterminal object](https://ncatlab.org/nlab/show/subterminal+object) of $\C$, then the slice category $\C / P$ is also cartesian closed. -::: +Finally, we generalize this result to arbitrary base rings. -_Proof._ The forgetful functor $\C / P \to \C$ is fully faithful; it has right adjoint ${-} \times P$; and it preserves binary products (in fact all inhabited limits). Hence, Lemma 3 applies. $\square$ - -::: Corollary 6 +::: Corollary 4 For any non-trivial commutative ring $R$, the category $\LRS_R$ is not cartesian closed. ::: -_Proof._ Let $k$ be a residue field of $R$. Then $\LRS_k$ is equivalent to the slice category $\LRS_R / \Spec k$ where $\Spec k$ is subterminal. $\square$ +_Proof._ Let $k$ be a residue field of $R$. Then $\LRS_k$ is equivalent to the slice category $\LRS_R / \Spec k$ where $\Spec k$ is subterminal. Therefore, the claim follows from Corollary 3 above and Corollary 2 [here](/content/cartesian-closed-results). $\square$ diff --git a/content/cartesian-closed-results.md b/content/cartesian-closed-results.md new file mode 100644 index 00000000..0c9f8593 --- /dev/null +++ b/content/cartesian-closed-results.md @@ -0,0 +1,27 @@ +--- +title: Results about cartesian closed categories +description: We prove in particular when a coreflective subcategory of a cartesian closed category is again cartesian closed. +author: Daniel Schepler +--- + +## Results about cartesian closed categories + +::: Lemma 1 +Suppose $\D$ is a coreflective subcategory of $\C$ such that $\D$ has binary products and the inclusion functor preserves these binary products. If $\C$ is cartesian closed, then so is $\D$. +::: + +_Proof._ Let $U : \D \to \C$ be the inclusion functor with right adjoint $R : \C \to \D$. Then for any objects $X, Y, Z$ of $\D$ we have natural isomorphisms + +$$ +\begin{align*} +\Hom_\D(Z\times X, Y) & \cong \Hom_\C(UZ \times UX, UY) \\ +& \cong \Hom_\C(UZ, [UX,UY]) \\ +& \cong \Hom_\D\bigl(Z, R([UX,UY])\bigr). +\end{align*} +$$ + +::: Corollary 2 +If $\C$ is a cartesian closed category and $P$ is a [subterminal object](https://ncatlab.org/nlab/show/subterminal+object) of $\C$, then the slice category $\C / P$ is also cartesian closed. +::: + +_Proof._ The forgetful functor $\C / P \to \C$ is fully faithful; it has right adjoint ${-} \times P$; and it preserves binary products (in fact all inhabited limits). Hence, Lemma 1 applies. $\square$ diff --git a/databases/catdat/data/category-implications/cartesian closed.yaml b/databases/catdat/data/category-implications/cartesian closed.yaml index e4dab838..88d76953 100644 --- a/databases/catdat/data/category-implications/cartesian closed.yaml +++ b/databases/catdat/data/category-implications/cartesian closed.yaml @@ -114,5 +114,5 @@ - thin conclusions: - locally cartesian closed - proof: First, note that the hypotheses imply that the category is finitely complete. Now every object is subterminal, so the result follows from Corollary 5 here. + proof: In a thin category, every object is subterminal. Thus, the result follows from Corollary 2 here. is_equivalence: false