From 05a1eb7e92c738d2a2205d12dc23bfb81d9c908b Mon Sep 17 00:00:00 2001 From: Yuhang Eric Wei Date: Thu, 28 Nov 2024 16:29:59 +0800 Subject: [PATCH] Fix typos Typos and Possible Improvements #15 --- ha/5-cc.typ | 10 +++++----- ha/6-df.typ | 8 ++++---- ha/7-balance.typ | 6 +++--- 3 files changed, 12 insertions(+), 12 deletions(-) diff --git a/ha/5-cc.typ b/ha/5-cc.typ index 73854e5..eb1c03d 100644 --- a/ha/5-cc.typ +++ b/ha/5-cc.typ @@ -577,7 +577,7 @@ By finding a resolution of a potentially "complicated" object $M$, we can work w $ d_1(P_1) = M_0 = Ker epsilon_0. $ - Thus the chain is exact at $P_0$. The procedure above can be then iterated for any $n >= 1$ and the resultant chain is infinitely long. + Thus the chain is exact at $P_0$. The procedure above can be then iterated for any $n >= 1$ and the resultant chain can be made arbitrarily long. // Set $d_0 = epsilon_0$. // Suppose we now have monomorphism $i_(n-1) : M_(n-1) -> P_(n-1)$ and $d_(n-1) : P_(n-1) -> P_(n-2)$ // By induction, givenlet $epsilon_n: P_n -> M_(n-1)$ be an epimorphism, where $P_n$ is projective, and let $M_n = ker epsilon_n$. Thus we have @@ -653,10 +653,10 @@ By finding a resolution of a potentially "complicated" object $M$, we can work w arr((0, 1), (1, 1), [$f_n oo d_(n+1)$]), ), ) - where since $P_(n+1)$ is an projective object, + where since $P_(n+1)$ is a projective object, the morphism $f_(n+1) : P_(n+1) -> Q_(n+1)$ exists such that the diagram commutes, i.e. $d'_(n+1) oo f_(n+1) = f_n oo d_(n+1)$. - For the uniqueness, let $h: P_cx -> Q_cx$ be another chain map lifting $f'$. We want to construct homotopy $s$ with terms $s_n: P_n -> Q_(n+1)$ such that + For uniqueness, let $h: P_cx -> Q_cx$ be another chain map lifting $f'$. We want to construct homotopy $s$ with terms $s_n: P_n -> Q_(n+1)$ such that $ h_n - f_n = d'_(n+1) s_n + s_(n-1) d_n $ for all $n >= -1$. @@ -723,7 +723,7 @@ By finding a resolution of a potentially "complicated" object $M$, we can work w ), ) - where $P_(n+1)$ is projective and $d'_(n+2) : Q_(n+2) -> Z_(n+1) (Q)$ is an epimorphism. + because $P_(n+1)$ is projective and $d'_(n+2) : Q_(n+2) -> Z_(n+1) (Q)$ is an epimorphism. Now $ d'_(n+1) (h_(n+1) - f_(n+1) - s_n d_(n+1)) &= d'_(n+1) (h_(n+1) - f_(n+1)) - d'_(n+1) s_n d_(n+1) \ @@ -801,7 +801,7 @@ $ $tilde(epsilon^(prime prime)) : P_0^(prime prime) arrow.r A$. The direct sum of $tilde(epsilon^(prime prime))$ and $i_A epsilon^prime colon P_0^prime arrow.r A$ gives a map $epsilon colon P_0 arrow.r A$. Then the diagram below - commutes： + commutes: // #align(center,image("../imgs/2023-11-04-13-59-23.png",width:80%)) // https://t.yw.je/#N4Igdg9gJgpgziAXAbVABwnAlgFyxMJZARgBpiBdUkANwEMAbAVxiRAGkYAnAChjWwMCAcgCUIAL6l0mXPkIoyAJiq1GLNp179BBcVJnY8BImQDMq+s1aIO3PgKxCwwsZOkgMR+USXlL6jYgAArCAPoADO6GciYofirUVhq2wZHRnrLGCsh+FkmBbKHhUQaZ3nHIZv4F1mwAgsIZXrE51Ylqdbb1zVk+KNX5nSkgjU1lLdlEETXDQaUek-3IACyzyfO9FTkzHRtsCzFTKGt7hbaH5a3TpEP7F1vXJ7cBXSCXS5VkEa8jH31fUgrX6bCYAnJ+H61P6PY65IEgg6w5bVKFzJFg7ZEarA6Gg1QwKAAc3gRFAADMuBAALZIACc1BwECQETKlJpLMZzMQZjZVNpPK5SAAbHi2DonCIMuyBaKQEykMRWR4ZYqyPLucQ+RzEOqFYgVtqBWsNUgAOxi2wS5zS-nmoW6pRGxXVU2IJ0qu3uh0AVmdiB9DoAHJaQNaRONPTqQ27iIaowLiIG3cqKV6ZrH-XruR60zq-LHhVmLW7eQmkK79fG88bff7k-riEH-SX9c3y4g5fqzVmGW6ix2Y-qlKyKBIgA #align( diff --git a/ha/6-df.typ b/ha/6-df.typ index 1ab5c10..3565213 100644 --- a/ha/6-df.typ +++ b/ha/6-df.typ @@ -94,7 +94,7 @@ 0 quad& n >= 2\, ) $ - form a homological $delta$-functor (or a cohomological + forms a homological $delta$-functor (or a cohomological $delta$-functor with $T^0 eq T_1$ and $T^1 eq T_0$).] #proof[Apply the @snake[Snake Lemma] to the commutative diagram // https://t.yw.je/#N4Igdg9gJgpgziAXAbVABwnAlgFyxMJZABgBpiBdUkANwEMAbAVxiRGJAF9T1Nd9CKAIzkqtRizYBBLjxAZseAkQBMo6vWatEIAEKzeigUQDM68VrYBhA-L5LByACznNknR26H+ylGSFibtrstgo+jiIBGhLBMl52Rr7IalEW7nqh9sYoZqlB1pmJji55MWyeYjBQAObwRKAAZgBOEAC2SCIgOBBIAGzRljpots1tSGpdPYgA7APpw-Gj7Yhmk0gAHHPBC3JLSGRriEKLLcud3eMnY4gTFytXy6t3Tg9IAKzUd72viP2H0z9Zod1j9NocAJycCicIA @@ -165,7 +165,7 @@ == Derived Functors -The main object of this section is to show that in an abelian category with enough projectives, left derived functors, defined as follows, are homological $delta$-functors. +The main objective of this section is to show that in an abelian category with enough projectives, left derived functors, defined as follows, are homological $delta$-functors. #definition[ Let $cA$ and $cB$ be two abelian categories and let $F : cA -> cB$ be a right exact functor. Assume that $cA$ has enough projectives. For any $A in cA$, pick a projective resolution $P_(cx) -> A$ by @enough-resolution. Then $L_i F$ given by $ L_i F(A) := H_i (F(P)) $ is called the *$i$-th left derived functor*. @@ -184,7 +184,7 @@ The main object of this section is to show that in an abelian category with enou + Calculate the $i$-th homology $H_i (F (P))$ of this chain complex. // Now let $f: A -> B$ be a morphism in $cA$. To find $L_i F (f)$, we can find projective resolutions $P_cx -> A$ and $Q_cx -> B$, and by the @comparison[Comparison Theorem], there exists a chain map $f_cx : P_cx -> Q_cx$ lifting $f$. Then $L_i F (f) := H_i (F(f_cx))$, obtained in a similar fashion as above. ] -In fact, our definition of the "functor" $L_i F$ is still incomplete as we have not defined how it maps the morphisms in $cA$. However, we first need to show that for any object $A in cA$, our definition of $L_i F (A)$ is independent of the choice of projective resolution $P_cx -> A$. The following implies the case when $i = 0$. +In fact, our definition of the "functor" $L_i F$ is still incomplete as we have not defined how it maps the morphisms in $cA$. However, we first need to show that for any object $A in cA$, our definition of $L_i F (A)$ is independent of the choice of projective resolution $P_cx -> A$. The following shows that this is true for $i = 0$. #lemma[ $L_0 F(A) iso F(A)$. ] @@ -229,7 +229,7 @@ In fact, our definition of the "functor" $L_i F$ is still incomplete as we have Now we complete the definition of $L_i F$ and prove that it is indeed a functor. #lemma[ - If $f : A' -> A$ a morphism in $cA$, then there is a natural map $ L_i F(f) : L_i F(A') -> L_i F(A). $ + If $f : A' -> A$ is a morphism in $cA$, then there is a natural map $ L_i F(f) : L_i F(A') -> L_i F(A). $ ] #proof[ Let $P'_cx -> A'$ and $P_cx -> A$ be projective resolutions. By the @comparison[Comparison Theorem], $f$ lifts to a chain map $tilde(f) : P'_cx -> P_cx$, which gives a map $tilde(f_ast) : H_i F(P') -> H_i F(P)$. As any other lift is chain homotopic to $tilde(f)$, the map $tilde(f_ast)$ is independent of the lift. diff --git a/ha/7-balance.typ b/ha/7-balance.typ index e4e088a..9bcd5b0 100644 --- a/ha/7-balance.typ +++ b/ha/7-balance.typ @@ -14,7 +14,7 @@ In particular, $Ext_cA^0 (A, B) = Hom(A) (A, B)$. ] -Notice that the contravariant functor $Hom(A)(-, B): cA^op -> Ab$ is also left exact by @hom-left-exact-2. Assume that $cA$ has enough projectives, so $cA^op$ has enough injectives. Let $P_cx -> A$ be an projective resolution in $cA$, which can be seen as an injective resolution in $cA^op$. We can thus define another right derived functor $Ext_cA^i (-, B)$, given by +Notice that the contravariant functor $Hom(A)(-, B): cA^op -> Ab$ is also left exact by @hom-left-exact-2. Assume that $cA$ has enough projectives, so $cA^op$ has enough injectives. Let $P_cx -> A$ be a projective resolution in $cA$, which can be seen as an injective resolution in $cA^op$. We can thus define another right derived functor $Ext_cA^i (-, B)$, given by $ Ext_cA^i (-, B)(A) := R^i Hom(A)(-, B) (A) = H^i (Hom(A)(P_cx, B)). $ @@ -55,9 +55,9 @@ This isomorphism is called the *balancing of $Ext$*. Before proving the balancin (4) $=>$ (2). Let $ses(A', A, A'')$ be a short exact sequence in $cA$, which induces the #lest $ - 0 -> Hom(A) (A', B) -> Hom(A) (A, B) -> Hom(A) (A'', B) -> Ext^1_cA (A', B) -> dots. + 0 -> Hom(A) (A'', B) -> Hom(A) (A, B) -> Hom(A) (A', B) -> Ext^1_cA (A'', B) -> dots. $ - Since $Ext^1_cA (A', B) = 0$ by assumption, $Hom(A) (-, B)$ is an exact functor. + Since $Ext^1_cA (A'', B) = 0$ by assumption, $Hom(A) (-, B)$ is an exact functor. ] #proposition[ The following are equivalent: