Support proof environments

content/Top-embeds-in-LRS.md

@@ -11,23 +11,27 @@ For much of this development, we will be dealing with the case of $\LRS_k$ where
 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 $(X, \O_X) \to (X, \underline{k})$ 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. <span class="qed">$\square$</span>
+::: 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.
+:::
 
 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 $V(f) \coloneqq \{x\in U : f \in \m_{X,x}\} = \{x \in U : f(x) = 0\}$, 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 $x \in X$ 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 $S$ 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
+::: 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 $(X, \underline{k})$, 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.
 
 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. <span class="qed">$\square$</span>
+From here, it is straightforward to show that this does in fact define an adjunction.
+:::
 
 _Remark._ In the special case where $k$ is a finite field, we have
 $$\textstyle X_0 \cap V(f) = \bigcap_{a \in k^\times} (X_0 \cap D(f-a)),$$
@@ -37,7 +41,9 @@ which is already open in the subspace topology. Therefore, in this case, $X_0$ i
 For any non-trivial commutative ring $R$, fix a quotient field $k$. Then the functor $K_R : \Top \to \LRS_R$ of equipping a topological space with the constant sheaf $\underline{k}$ is fully faithful; has a right adjoint; and preserves all inhabited limits.
 :::
 
-_Proof._ The functor $K_R$ is the composition of $K : \Top \to \LRS_k$ and the forgetful functor $\LRS_k \to \LRS_R$. Since $\LRS_k$ is equivalent to the slice category of $\LRS_R$ over the subterminal object $\Spec k$, the forgetful functor is fully faithful; has right adjoint ${-} \times \Spec k$; and preserves all inhabited limits. Therefore, from the previously established results on $K$, the result follows. <span class="qed">$\square$</span>
+::: Proof
+The functor $K_R$ is the composition of $K : \Top \to \LRS_k$ and the forgetful functor $\LRS_k \to \LRS_R$. Since $\LRS_k$ is equivalent to the slice category of $\LRS_R$ over the subterminal object $\Spec k$, the forgetful functor is fully faithful; has right adjoint ${-} \times \Spec k$; and preserves all inhabited limits. Therefore, from the previously established results on $K$, the result follows.
+:::
 
 ::: Corollary 4
 Let $R$ be any non-trivial commutative ring. Then:<br>
@@ -48,4 +54,6 @@ Let $R$ be any non-trivial commutative ring. Then:<br>
 (e) $\LRS_R$ does not have effective cocongruences.
 :::
 
-_Proof._ We already know that $\Top$ does not satisfy any of these properties. In order to conclude that $\LRS_R$ does not satisfy any of them either, we fix a quotient field of $R$ as above and consider the functor $K_R$. In each case, this is an easy application of a contrapositive of a result from [here](/content/subcategories) to the functor $K_R$. Namely, (a) follows from Lemma 5; (b) from Lemma 4; (c) from Lemma 7; (d) from the dual of Lemma 2 with the observation that $K_R$ preserves epimorphisms since it has a right adjoint; and (e) from the dual of Lemma 8. <span class="qed">$\square$</span>
+::: Proof
+We already know that $\Top$ does not satisfy any of these properties. In order to conclude that $\LRS_R$ does not satisfy any of them either, we fix a quotient field of $R$ as above and consider the functor $K_R$. In each case, this is an easy application of a contrapositive of a result from [here](/content/subcategories) to the functor $K_R$. Namely, (a) follows from Lemma 5; (b) from Lemma 4; (c) from Lemma 7; (d) from the dual of Lemma 2 with the observation that $K_R$ preserves epimorphisms since it has a right adjoint; and (e) from the dual of Lemma 8.
+:::

content/cocongruences_of_groups.md

@@ -27,7 +27,8 @@ is also a monomorphism.
 The category $\Grp$ has good pushouts of monomorphisms.
 :::
 
-_Proof._ Consider a diagram as above. We regard every monomorphism in it as an inclusion. Choose a system of representatives $S \subseteq B$ for the right $A$-cosets in $B$, meaning that the multiplication map $\cdot : A \times S \to B$ is bijective. Likewise, choose $T \subseteq C$ such that the multiplication map $\cdot : A \times T \to C$ is bijective. We may assume that $1 \in S$ and $1 \in T$.
+::: Proof
+Consider a diagram as above. We regard every monomorphism in it as an inclusion. Choose a system of representatives $S \subseteq B$ for the right $A$-cosets in $B$, meaning that the multiplication map $\cdot : A \times S \to B$ is bijective. Likewise, choose $T \subseteq C$ such that the multiplication map $\cdot : A \times T \to C$ is bijective. We may assume that $1 \in S$ and $1 \in T$.
 
 It is well known (see, for example, Serre's book _Trees_, Ch. I, §1, Thm. 1) that every element of the amalgamated free product $B \sqcup_A C$ has a unique representation of the form
 $$w = a \cdot x_1 \cdots x_n,$$
@@ -41,13 +42,15 @@ is injective. Indeed, if $b_1,b_2 \in B$ satisfy $A' b_1 = A' b_2$, then $b_1 b_
 
 Therefore, we may extend $S$ to a system of representatives $S' \subseteq B'$ for the right $A'$-cosets in $B'$. Likewise, we may extend $T$ to a system of representatives $T' \subseteq C'$ for the right $A'$-cosets in $C'$.
 
-With respect to these systems, an element $w \in B \sqcup_A C$ written in normal form as above remains in normal form after being mapped to $B' \sqcup_{A'} C'$. This shows that the induced map is injective. <span class="qed">$\square$</span>
+With respect to these systems, an element $w \in B \sqcup_A C$ written in normal form as above remains in normal form after being mapped to $B' \sqcup_{A'} C'$. This shows that the induced map is injective.
+:::
 
 ::: Proposition 2
 Let $\C$ be a balanced category with good pushouts of monomorphisms and equalizers of monomorphisms. Then every cocongruence in $\C$ is effective.
 :::
 
-_Proof._ Let $X \in \C$ be an object, and let $i_1,i_2 : X \rightrightarrows Y$ be a cocongruence. Since it is coreflexive, there exists a morphism $r : Y \to X$ satisfying
+::: Proof
+Let $X \in \C$ be an object, and let $i_1,i_2 : X \rightrightarrows Y$ be a cocongruence. Since it is coreflexive, there exists a morphism $r : Y \to X$ satisfying
 
 $$r \circ i_1 = \id_X, \quad r \circ i_2 = \id_X.$$
 
@@ -96,7 +99,8 @@ Since $\C$ has good pushouts of monomorphisms, it suffices to verify that both s
 
 $$a = r \circ i_1 \circ a = r \circ i_2 \circ b = b.$$
 
-Thus, $a$ is simply a morphism equalizing $i_1$ and $i_2$, so it factors uniquely through $e$. <span class="qed">$\square$</span>
+Thus, $a$ is simply a morphism equalizing $i_1$ and $i_2$, so it factors uniquely through $e$.
+:::
 
 ::: Corollary 3
 Every cocongruence in the category $\Grp$ is effective.

content/cogenerators_in_product_categories.md

@@ -9,5 +9,6 @@ description: How to construct a cogenerator in a product category
 For a family of categories $(\C_i)_{i \in I}$, each having a cogenerator $Q_i$ which is weakly terminal, the object $(Q_i)_{i \in I}$ is a cogenerator in the product category $\prod_{i \in I} \C_i$.
 :::
 
-_Proof._
-Let $(f_i: A_i \to B_i)_{i \in I}$ and $(g_i: A_i \to B_i)_{i \in I}$ be two parallel morphisms in the product category which are coequalized by any morphism $(B_i \to Q_i)_{i \in I}$. Let $i_0 \in I$. We claim that $f_{i_0},g_{i_0} : A_{i_0} \rightrightarrows B_{i_0}$ are coequalized by all morphisms $B_{i_0} \to Q_{i_0}$, and hence are equal: Indeed, for all $i \neq i_0$ we may choose some morphism $B_i \to Q_i$ since $Q_i$ is weakly terminal. Thus, the morphism $B_{i_0} \to Q_{i_0}$ extends to a morphism $(B_i \to Q_i)_{i \in I}$ in the product category. By assumption, it coequalizes $(f_i)_{i \in I}$ and $(g_i)_{i \in I}$. By looking at the $i_0$-component, we are done. <span class="qed">$\square$</span>
+::: Proof
+Let $(f_i: A_i \to B_i)_{i \in I}$ and $(g_i: A_i \to B_i)_{i \in I}$ be two parallel morphisms in the product category which are coequalized by any morphism $(B_i \to Q_i)_{i \in I}$. Let $i_0 \in I$. We claim that $f_{i_0},g_{i_0} : A_{i_0} \rightrightarrows B_{i_0}$ are coequalized by all morphisms $B_{i_0} \to Q_{i_0}$, and hence are equal: Indeed, for all $i \neq i_0$ we may choose some morphism $B_i \to Q_i$ since $Q_i$ is weakly terminal. Thus, the morphism $B_{i_0} \to Q_{i_0}$ extends to a morphism $(B_i \to Q_i)_{i \in I}$ in the product category. By assumption, it coequalizes $(f_i)_{i \in I}$ and $(g_i)_{i \in I}$. By looking at the $i_0$-component, we are done.
+:::