From ae599572920c1fa1ae3531572d9671d1583d6953 Mon Sep 17 00:00:00 2001 From: Danil Date: Sat, 10 Oct 2026 13:44:04 +0300 Subject: [PATCH] problem 1.2 b' --- boxes.typ | 40 ++ main.typ | 11 + problem11a.typ | 103 ++++++ problem11b.pdf | 936 +++++++++++++++++++++++++++++++++++++++++++++++ problem11b.typ | 169 +++++++++ problem12a.typ | 32 ++ problem12b.pdf | 643 ++++++++++++++++++++++++++++++++ problem12b.typ | 71 ++++ utils/boxes.typ | 38 ++ utils/macros.typ | 28 ++ 10 files changed, 2071 insertions(+) create mode 100644 boxes.typ create mode 100644 main.typ create mode 100644 problem11a.typ create mode 100644 problem11b.pdf create mode 100644 problem11b.typ create mode 100644 problem12a.typ create mode 100644 problem12b.pdf create mode 100644 problem12b.typ create mode 100644 utils/boxes.typ create mode 100644 utils/macros.typ diff --git a/boxes.typ b/boxes.typ new file mode 100644 index 0000000..4351bda --- /dev/null +++ b/boxes.typ @@ -0,0 +1,40 @@ +#let infobox( + body, + title: none, + stroke-color: black, + fill-color: luma(245), + border-width: 1pt, + radius: 4pt, + padding: 8pt, +) = { + block( + width: 100%, + stroke: (paint: stroke-color, thickness: border-width), + fill: fill-color, + radius: radius, + inset: padding, + )[ + #if title != none [ + *#title*\ + ] + #body + ] +} + +#let problembox(body, title: none) = infobox( + body, + title: title, + fill-color: aqua.transparentize(80%) +) + +#let answerbox(body, title: none) = infobox( + body, + title: title, + fill-color: red.transparentize(80%) +) + +#let addinfobox(body, title: none) = infobox( + body, + title: title, + fill-color: rgb("#FFB3BF").transparentize(80%) +) diff --git a/main.typ b/main.typ new file mode 100644 index 0000000..21b9d75 --- /dev/null +++ b/main.typ @@ -0,0 +1,11 @@ +#set text(lang: "ru", font: "New Computer Modern") +#set par(justify: true) + + +#include "problem11a.typ" +#pagebreak() +#include "problem11b.typ" +#pagebreak() +#include "problem12a.typ" +#pagebreak() +#include "problem12b.typ" diff --git a/problem11a.typ b/problem11a.typ new file mode 100644 index 0000000..3b48919 --- /dev/null +++ b/problem11a.typ @@ -0,0 +1,103 @@ +#import "utils/boxes.typ": * +#import "utils/macros.typ": * +#import "@preview/fletcher:0.5.8" as fletcher: diagram, node, edge + += 1.1 (Фильтрованные копределы в алгебраических категориях). + +#problembox[ + + а) Пусть дана фильтрованная poset-категория $cal(I)$ (т.\,е. для $i >= j$ имеем стрелку $phi_(i, j): i -> j$) и функтор $a: cal(I) -> bold("Set")$. Определим на $product.co_(i in I) a(i)$ следующее отношение эквивалентности: + $ a(i) in.rev t_i ~ t_j in a(j) <==> exists k in cal(I), i >= k, j >= k, "т. ч." a(phi_(i, k))(t_i) = a(phi_(j, k))(t_j) $ + + Проверить, что это отношение эквивалентности и что $(product.co_(i in I) a(i)) \/ ~ $ есть $colim_i a$. + +] + +== Решение + +=== Отношение эквивалентности + ++ *Рефлексивность*. Для $t_i$ можно выбрать $k = i$, т.к. $i >= i$. Тогда $ t_i tilde t_i $ ++ *Симметричность*. Если $t_i tilde t_j$, то существует такое $k in cal(I)$, что $i >= k, j >= k$ и $a(phi_(i,k))(t_i) = a(phi_(j,k))(t_j)$. \ + Если же $t_j tilde t_i$, то существует такое $l in cal(I)$, что $j >= l, i >= l$ и $a(phi_(j,k))(t_j) = a(phi_(i,k))(t_i)$. Положим $l = k$ --- симметричность выполнена. ++ Пусть $t_i tilde t_j, t_j tilde t_k$, тогда существуют $l, m in cal(I)$ такие, что $i >= l, j >= l$ и $j >= m, k >= m$ и соответствующие образы равны. \ + Так как $cal(I)$ - фильтрованная poset-категория, то существует $n in cal(I)$ такой, что $l >= n, m >= n$. + Рассмотрим диаграмму + + #cdiagram({ + node((-2, -1), [$a(i)$]) + node((0, 1), [$a(l)$]) + node((2, -1), [$a(j)$]) + node((4, 1), [$a(m)$]) + node((6, -1), [$a(k)$]) + node((2, 3), [$a(n)$]) + edge((-2, -1), (0, 1), [$a(phi_(i,l))$], label-side: right, "->") + edge((2, -1), (0, 1), [$a(phi_(j,l))$], label-side: left, "->") + edge((2, -1), (4, 1), [$a(phi_(j,m))$], label-side: right, "->") + edge((6, -1), (4, 1), [$a(phi_(k,m))$], label-side: left, "->") + edge((0, 1), (2, 3), [$a(phi_(l,n))$], label-side: right, "->") + edge((4, 1), (2, 3), [$a(phi_(m,n))$], label-side: left, "->") + }) + + Имеем + $ a(phi_(i,l))(t_i) = a(phi_(j,l))(t_j) $ + Применяя $a(phi_(l,n))$, имеет + $ a(phi_(l,n) compose phi_(i,l)) (t_i) = a(phi_(l,n) compose phi_(j,l))(t_j) $ + По свойствам poset-категории существует единственные морфизмы $i arrow n$ и $j arrow n$: + $ a(phi_(i,n))(t_i) = a(phi_(j,n))(t_j) $ + Аналогично для $k arrow n$. Тогда + $ a(phi_(i,n))(t_i) = a(phi_(j,n))(t_j) = a(phi_(k,n))(t_k) $ + + Имеем + $ t_i tilde t_k $ + +=== Копредел + +Обозначим $S := product.co_i a(i) \/ tilde$, $iota_i: a(i) arrow.twohead S, t_i mapsto [t_i]$. +#cdiagram({ + node((0, 0), [$a(i)$]) + node((2, 2), [$S$]) + node((4, 0), [$a(j)$]) + edge((0, 0), (4, 0), [$a(phi_(i,j))$], label-side: left, "->") + edge((0, 0), (2, 2), [$iota_i$], label-side: right, "->") + edge((4, 0), (2, 2), [$iota_j$], label-side: left, "->") +}) + +Пусть $u := a(phi_(i,j))(t_i)$. Проверим $ [u] =^? [t_i]. $ +Эти два элемента эквивалентны тогда и только тогда, когда существует $k in cal(I)$ такой, что $i >= k, j >= k$. Положим $k = j$. По определению $phi_(i,j)$ имеем $i >= j$ и $j >= j$ (отображение $a(phi_(j,j)) = 1_(a(j))$), а по определению $u = a(phi_(i,j))(t_i)$. Получаем +$ u = a(phi_(i,j))(t_i) => u tilde t_i => [u] = [t_i] $ + +Тогда +$ iota_j compose a(phi_(i,j)) = iota_i $ + +В таком случае, ${iota_i: a(i) arrow.twohead S}_(i in cal(I))$ -- коконус. + +Пусть $X in bold("Set")$ и ${mu_i: a(i) arrow X}_(i in cal(I))$ -- коконус. +Хотим: единственное $theta: S arrow X$ такое, что $ theta compose iota_i = mu_i $ + +#cdiagram({ + node((0, 0), [$a(i)$]) + node((2, 2), [$S$]) + node((4, 0), [$a(j)$]) + node((2, 4), [$X$]) + edge((0, 0), (4, 0), [$a(phi_(i,j))$], label-side: left, "->") + edge((0, 0), (2, 2), [$iota_i$], label-side: right, "->") + edge((4, 0), (2, 2), [$iota_j$], label-side: left, "->") + edge((4, 0), (2, 4), [$mu_j$], label-side: left, "->", bend: 30deg) + edge((0, 0), (2, 4), [$mu_i$], label-side: right, "->", bend: -30deg) + edge((2, 2), (2, 4), [$theta$], label-side: right, "-->") +}) + +Положим $theta([t_i]) := mu_i (t_i)$. +Проверим корректность. \ +Пусть $t_i tilde t_j => [t_i] = [t_j]$. +$ mu_i (t_i) =^? mu_j (t_j) $ + +Существует $k in cal(I)$ такое, что $i >= k, j >= k$ и соответствующие образы равны. Тогда, так как $mu$ - коконус, верно +$ mu_k compose a(phi_(i,k))(t_i) = mu_i (t_i) $ +$ mu_k compose a(phi_(j,k))(t_j) = mu_j (t_j) $ + +Отсюда $mu_i (t_i) = mu_j (t_j)$. + +Тогда +#answerbox[$ product.co_i a(i) \/ tilde space = colim_i a $] \ No newline at end of file diff --git a/problem11b.pdf b/problem11b.pdf new file mode 100644 index 0000000..efbac47 --- /dev/null +++ b/problem11b.pdf @@ -0,0 +1,936 @@ +%PDF-1.7 +% + +1 0 obj +<> +endobj +2 0 obj +</Dest 139 0 R>> +endobj +3 0 obj +</Dest 140 0 R>> +endobj +4 0 obj +</Dest 141 0 R>> +endobj +5 0 obj +</Dest 142 0 R>> +endobj +6 0 obj +<> +endobj +7 0 obj +<>/K[136 0 R]/ParentTree<>/ParentTreeNextKey 3>> +endobj +8 0 obj +[11 0 R 11 0 R 12 0 R 13 0 R 14 0 R 14 0 R 14 0 R 14 0 R 14 0 R 14 0 R 14 0 R 15 0 R 15 0 R 16 0 R 16 0 R 16 0 R 16 0 R 16 0 R 16 0 R 16 0 R 16 0 R 16 0 R 16 0 R 16 0 R 16 0 R 17 0 R 18 0 R 19 0 R 20 0 R 20 0 R 20 0 R 20 0 R 20 0 R 20 0 R 20 0 R 20 0 R 20 0 R 20 0 R 20 0 R 20 0 R 21 0 R 22 0 R 23 0 R 24 0 R 24 0 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"utils/macros.typ": * +#import "@preview/fletcher:0.5.8" as fletcher: diagram, node, edge + +#problembox[ + б) Пусть на множествах из предыдущего пункта имеется структура кольца (ассоциативного, коммутативного с $1$), а $a(phi_(i, j))$ --- гомоморфизмы колец. Докажите, что на $(product.co_(i in cal(I)) a(i)) \/ ~ $ есть естественная структура кольца (ассоциативного, коммутативного с $1$) такая, что $(product.co_(i in cal(I)) a(i)) \/ ~ $ --- копредел в категории $CRing$ по диаграмме $a: cal(I) -> CRing$. +] + +== Решение + +Обозначим $S := product.co_i a(i) \/ tilde$, $iota_i: a(i) arrow S, t_i mapsto [t_i]$. +Также обозначим $phi^*_(i,j) := a(phi_(i,j))$ + +=== Определение операций + +Определим операцию сложения на $S$ следующим образом: +$ [t_i] + [t_j] := [phi^*_(i,k)(t_i) + phi^*_(j,k)(t_j)], space.quad [t_i], [t_j] in S $ + +Так как $cal(I)$ --- фильтрованная категория, такой $k$ существует. + +Проверим корректность. +Пусть $t_i in a(i), t_j in a(j)$. Пусть также $t'_i' in a(i'), t'_j' in a(j')$ такие, что $t_i tilde t'_i', t_j tilde t'_j'$. +$ u = phi^*_(i,k)(t_i) + phi^*_(j,k)(t_j) $ +$ u' = phi^*_(i',k')(t'_i') + phi^*_(j',k')(t'_j') $ + +Найдём такой $p in cal(I)$, что образы $u, u'$ в $a(p)$ совпадают. + +Для $t_i, t'_i'$ выберем $l in cal(I)$: +$ phi^*_(i,l) (t_i) =_(t_i tilde t'_i') phi^*_(i',l)(t'_i') $ +Для $t_j, t'_j'$ выберем $m in cal(I)$: +$ phi^*_(j,m) (t_j) =_(t_j tilde t'_j') phi^*_(j',m)(t'_j') $ + +По фильтрованности $cal(I)$ существует индекс $p in cal(I)$ такой, что +$ p <= k, space.quad p <= k', space.quad p <= l, space.quad p <= m $ + +$ phi^*_(k,p)(u) = phi^*_(k,p) lr((phi^*_(i,k)(t_i) + phi^*_(j,k)(t_j)), size: #150%) = $ +$ = phi^*_(k,p) lr((phi^*_(i,k)(t_i)), size: #150%) + phi^*_(k,p) lr((phi^*_(j,k)(t_j)), size: #150%) $ +$ phi^*_(k,p)(u) = phi^*_(i,p)(t_i) + phi^*_(j,p)(t_j) $ + +Аналогично получаем +$ phi^*_(k', p)(u') = phi^*_(i', p)(t'_i') + phi^*_(j',p)(t'_j') $ + +$ phi^*_(i,p)(t_i) = phi^*_(l,p) lr(( phi^*_(i,l)(t_i) ), size: #150%) = phi^*_(l,p) lr(( phi^*_(i',l)(t'_i') ), size: #150%) = phi^*_(i', p)(t'_i') $ + +Аналогично +$ phi^*_(j,p)(t_j) = phi^*_(j',p)(t'_j') $ + +Получается, что +$ u tilde u' => [u] = [u'] $ + +Операция корректна + +Нейтральным элементом является $[0]$, т.к. гомоморфизмы сохраняют $0$. Обратным к $[t_i]$ является $[-t_i]$. + +Проверим ассоциативность: +$ +[t_i] + ([t_j] + [t_k]) = lr(| + #inlinediag( + node((0, 0), $j$), + node((2, 0), $k$), + node((1, 1), $l$), + edge((0, 0), (1, 1), "->"), + edge((2, 0), (1, 1), "->"), + ) +|) = [t_i] + [phi^*_(j,l)(t_j) + phi^*_(k,l)(t_k)] = +$ +$ += lr(| + #inlinediag( + node((0,0), $i$), + node((2,0), $j$), + node((4,0), $k$), + node((3,1), $l$), + node((1,3), $m$), + edge((0,0), (1,3), "->"), + edge((2,0), (1,3), "->"), + edge((2,0), (3,1), "->"), + edge((2,0), (3,1), "->"), + edge((4,0), (3,1), "->"), + edge((3,1), (1,3), "->"), + ) +|) = [ phi^*_(i,m)(t_i) + phi^*_(j,m)(t_j) + phi^*_(k,m)(t_k) ] +$ + +С другой стороны +$ +([t_i] + [t_j]) + [t_k] = lr(| + #inlinediag( + node((0,0), $i$), + node((2,0), $j$), + node((1,1), $m$), + edge((0,0), (1,1), "->"), + edge((2,0), (1,1), "->") + ) +|) = [phi^*_(i,m)(t_i) + phi^*_(j,m)(t_j)] + [t_k] = +$ +$ + = lr(| + #inlinediag( + node((0,0), $i$), + node((1,0), $j$), + node((2,0), $k$), + node((1,1), $m$), + edge((0,0), (1,1), "->"), + edge((1,0), (1,1), "->"), + edge((2,0), (1,1), "->"), + ) +|) = [phi^*_(i,m)(t_i) + phi^*_(j,m)(t_j) + phi^*_(k,m)(t_k)] +$ + +Коммутативность следует из коммутативность колец. + +Операцию умножения определим тем же образом: +$ [t_i] dot [t_j] = [phi^*_(i,k)(t_i) dot phi^*_(j,k)(t_j)] $ + +Она корректна по тем же соображениям. + +Нейтральным элементом является $[1]$, т.к. гомоморфизмы в $CRing$ сохраняют единицу. +Дистрибутивность следует также из свойств гомоморфизма. + +Таким образом, на $S$ есть структура коммутативного кольца с единицей. + +=== Проверка коконуса + +Пусть $t, t' in a(i)$. +$ iota_i (t + t') = [t + t'] $ +$ +iota_i (t) + iota_i (t') = [t] + [t'] = lr( + | phi^*_(i,i) = 1_(a(i)) +|) = [t + t'] +$ +$ iota_i (t + t') = iota_i (t) + iota_i (t') $ +$ iota_i (0_i) = [0] space.quad "все нули эквивалентны" $ + +$ iota_i (t dot t') = [t dot t'] $ +$ +iota_i (t) dot iota_i (t') = [t] dot [t'] = lr(| + phi_(i,i) = 1_(a(i)) +|) = [t dot t'] +$ +$ iota_i (t dot t') = iota_i (t) dot iota_i (t') $ + + +$ iota_i (1_(a_i)) = [1] space.quad "все единицы эквивалентны" $ + +Тогда $iota: a ==> S$ - коконус в $CRing$. + +=== Универсальное свойство + +Пусть $X in CRing$, $mu: a ==> X$ --- коконус. +Хотим: единственный гомоморфизм $theta: S arrow X$. Определим, как и раньше +$ theta([t_i]) := mu_i (t_i) $ + +Проверим свойства гомоморфизма. Пусть $t_i in a(i), t_j in a(j)$: +$ theta lr( ([phi^*_(i,k)(t_i) + phi^*_(j,k)(t_j)]), size: #150%) = mu_k (phi^*_(i,k)(t_i)) + mu_k (phi^*_(j,k)(t_j)) = $ +$ =_("По свойствам коконуса") mu_i (t_i) + mu_j (t_j) = theta([t_i]) + theta([t_j]) $ + +Аналогично +$ theta lr( ([phi^*_(i,k)(t_i) dot phi^*_(j,k)(t_j)]) , size: #150%) = mu_i (t_i) dot mu_j (t_j) = theta([t_i]) dot theta([t_j]) $ + +Очевидно сохранение единицы: +$ theta([1_i]) = mu_i (1_i) = 1_X, space.quad 1_i in a(i) $ + +Тогда $theta: S arrow X$ определена корректно. + +#answerbox[ + #align(center)[ + $product.co_i a(i) \/ tilde space = colim a$ в $CRing$] +] diff --git a/problem12a.typ b/problem12a.typ new file mode 100644 index 0000000..66cfee0 --- /dev/null +++ b/problem12a.typ @@ -0,0 +1,32 @@ +#import "utils/boxes.typ": * +#import "utils/macros.typ": * +#import "@preview/fletcher:0.5.8" as fletcher: diagram, node, edge + += 1.2 (Прямой и обратный образ предпучка). + +#addinfobox(title: [Определение предпучка])[ + Пусть дано топологическое пространство $X$. Рассмотрим топологию на нём как poset и определим poset-категорию $Open(X)$. Тогда предпучок множеств на $X$ есть функтор $cal(F): Open(X)^"op" -> Set$. Отображения $r_(U, V): cal(F)(U) -> cal(F)(V)$ для $V subset U$ называются отображениями ограничения. Пусть дано непрерывное отображение $f: X -> Y$ и предпучки $cal(F)$ на $X$ и $cal(G)$ на $Y$. +] + +#problembox[ + + #v(1mm) + а) Определим предпучок $f_* cal(F)$ на $Y$ как $f_* cal(F)(U) = cal(F)(f^(-1)(U))$ для открытых $U subset Y$. Отображения ограничения определим как $r_(U, V) := r_(f^(-1)(U), f^(-1)(V))$. Убедитесь, что это предпучок. Этот предпучок называется *прямым образом предпучка* $cal(F)$. +] + +== Решение + +Будем обозначать $r^U_V := r_(U,V)$ --- ограничения для $V subset.eq U$. + +По непрерывности $f$, $f^(-1)(U) in tau_X$. Тогда $cal(F) (f^(-1)(U))$ определено корректно. + +Если $V subset.eq U$, то $f^(-1)(V) subset.eq f^(-1)(U)$: +$ x in f^(-1)(V) => f(x) in V => f(x) in U => x in f^(-1)(U) $ + +Тогда отображения $r^(f^(-1)(U))_(f^(-1)(V))$ определены корректно. + +#answerbox[ + #align(center)[ + $f_* cal(F): Open(Y)^op arrow Set$ --- предпучок + ] +] diff --git a/problem12b.pdf b/problem12b.pdf new file mode 100644 index 0000000..8d3cef4 --- /dev/null +++ b/problem12b.pdf @@ -0,0 +1,643 @@ +%PDF-1.7 +% + +1 0 obj +<> +endobj +2 0 obj +</Dest 79 0 R>> +endobj +3 0 obj +</Dest 80 0 R>> +endobj +4 0 obj +</Dest 81 0 R>> +endobj +5 0 obj +<> +endobj +6 0 obj +<>/K[76 0 R]/ParentTree<>/ParentTreeNextKey 2>> +endobj +7 0 obj +[9 0 R 10 0 R 10 0 R 10 0 R 10 0 R 10 0 R 11 0 R 12 0 R 12 0 R 12 0 R 12 0 R 12 0 R 13 0 R 14 0 R 14 0 R 14 0 R 14 0 R 14 0 R 15 0 R 16 0 R 16 0 R 16 0 R 16 0 R 16 0 R 16 0 R 17 0 R 17 0 R 18 0 R 18 0 R 18 0 R 18 0 R 18 0 R 18 0 R 18 0 R 18 0 R 18 0 R 18 0 R 18 0 R 19 0 R 20 0 R 20 0 R 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Пусть даны функторы $phi: cal(J) -> cal(I)$, $alpha: cal(I) -> cal(C)$, $beta: cal(I)^"op" -> cal(C)$. Определите естественные морфизмы $colim(alpha compose phi) -> colim(alpha)$ и $lim(beta) -> lim(beta compose phi^"op")$ в предположении, что все пределы существуют. (Это будет на след. лекции, но попробуйте догадаться самостоятельно). +] + +== Решение + +=== Для копределов + +Обозначим $iota: alpha ==> colim alpha$, $tau: (alpha compose phi) ==> colim (alpha compose phi)$ --- коконусы. + +Хотим: коконус $psi: (alpha compose phi) ==> colim alpha$. +Построим его с помощью переиндексации +$ psi_j := iota_(phi(j)) $ +Очевидно, это коконус. + +По универсальному свойству, $forall X in cal(C)$ и коконуса $mu: (alpha compose phi) ==> X$ существует единственное $theta: colim(alpha compose phi) -> X$ т.ч +$ theta compose tau_(j) = mu_(j) $ + +Пололжим $X = colim alpha, mu = psi$. +#cdiagram({ + node((-1, -1), [$alpha(phi(j))$]) + node((1, 1), [$op("colim") (alpha compose phi)$]) + node((3, -1), [$alpha (phi(j'))$]) + node((1, 3), [$op("colim") alpha$]) + edge((-1, -1), (3, -1), [$alpha(phi(f))$], label-side: left, "->") + edge((-1, -1), (1, 1), [$tau_j$], label-side: right, "->") + edge((3, -1), (1, 1), [$tau_j'$], label-side: left, "->") + edge((-1, -1), (1, 3), [$psi_j$], label-side: left, "->", bend: -30deg) + edge((3, -1), (1, 3), [$psi_j'$], label-side: left, "->", bend: 30deg) + edge((1, 1), (1, 3), [$theta$], label-side: left, "-->") +}) + +#answerbox[ + #align(center)[ + Определили $theta: colim (alpha compose phi) -> colim alpha$ + ] +] + +=== Для пределов + +Имеем пределы с их каноническими конусами: +- $pi: lim beta ==> beta$ для функтора $beta: cal(I)^"op" -> cal(C)$, +- $q: lim (beta compose phi^"op") ==> (beta compose phi^"op")$ для функтора $beta compose phi^"op": cal(J)^"op" -> cal(C)$. + +Построим конус $p: lim beta ==> (beta compose phi^op)$ с помощью переиндексации: +$ p_j := pi_phi(j) $ + +Тогда по универсальному свойству предела $lim(beta compose phi^"op")$ существует единственный морфизм $eta: lim beta -> lim (beta compose phi^"op")$, делающий диаграмму коммутативной: + +#cdiagram({ + node((0, 4), [$beta(phi(j))$]) + node((4, 4), [$beta(phi(j'))$]) + node((2, 2), [$lim (beta compose phi^"op")$]) + node((2, 0), [$lim beta$]) + edge((4, 4), (0, 4), [$beta(phi(f))$], label-side: left, "->") + edge((2, 2), (0, 4), [$q_j$], label-side: right, "->") + edge((2, 2), (4, 4), [$q_(j')$], label-side: left, "->") + edge((2, 0), (2, 2), [$eta$], label-side: right, "-->") + edge((2, 0), (0, 4), [$p_j$], label-side: right, "->", bend: -30deg) + edge((2, 0), (4, 4), [$p_j'$], label-side: left, "->", bend: 30deg) +}) + +#answerbox[ + #align(center)[ + Определили $eta: lim beta -> lim (beta compose phi^op)$ + ] +] diff --git a/utils/boxes.typ b/utils/boxes.typ new file mode 100644 index 0000000..b597df9 --- /dev/null +++ b/utils/boxes.typ @@ -0,0 +1,38 @@ +#let infobox( + body, + title: none, + stroke-color: black, + fill-color: luma(245), + border-width: 1pt, + radius: 4pt, + padding: 8pt, +) = { + block( + width: 100%, + stroke: (paint: stroke-color, thickness: border-width), + fill: fill-color, + radius: radius, + inset: padding, + )[ + #if title != none [ + *#title*\ + ] + #body + ] +} + +#let problembox(body, title: none) = infobox( + body, + title: title, + fill-color: aqua.transparentize(80%) +) + +#let answerbox(body, title: none) = infobox( + body,title: title, fill-color: red.transparentize(80%) +) + +#let addinfobox(body, title: none) = infobox( + body, + title: title, + fill-color: rgb("#FFB3BF").transparentize(80%) +) diff --git a/utils/macros.typ b/utils/macros.typ new file mode 100644 index 0000000..1ece2ad --- /dev/null +++ b/utils/macros.typ @@ -0,0 +1,28 @@ +#import "@preview/fletcher:0.5.8" as fletcher + +#let CRing = math.bold("CRing") +#let Open = math.op("Open") +#let Set = math.bold("Set") + + +#let colim = math.limits(math.op("colim")) + +#let cdiagram(..args) = { + let named = args.named() + // Если spacing не задан явно при вызове, берем дефолтный + if "spacing" not in named { + named.insert("spacing", (10mm, 7mm)) + } + align(center, fletcher.diagram(..args.pos(), ..named)) +} + +// Макрос для инлайн-диаграмм внутри формул +#let inlinediag(..args) = box( + baseline: 50%, + fletcher.diagram( + spacing: (4mm, 3.5mm), + node-inset: 1pt, + edge-stroke: 0.6pt, + ..args, + ) +)