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problem 1.2 b'

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2026-10-10 13:44:04 +03:00
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boxes.typ Normal file
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#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%)
)

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main.typ Normal file
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#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"

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problem11a.typ Normal file
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#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 $]

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#import "utils/boxes.typ": *
#import "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 $CRing$]
]

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#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
]
]

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71
problem12b.typ Normal file
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@@ -0,0 +1,71 @@
#import "utils/boxes.typ": *
#import "utils/macros.typ": *
#import "@preview/fletcher:0.5.8" as fletcher: diagram, node, edge
#problembox[
б') Пусть даны категории $cal(I), cal(J), cal(C)$. 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 $eta: lim beta -> lim (beta compose phi^op)$
]
]

38
utils/boxes.typ Normal file
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#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%)
)

28
utils/macros.typ Normal file
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@@ -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,
)
)