Machine-translated from Chinese.
基本概念
对链复形
$$
\cdots\rightarrow C_{n+1}\xrightarrow{\partial} C_n\xrightarrow{\partial} C_{n-1}\rightarrow\cdots,
$$
我们作用$\operatorname{Hom}(-,G)$函子, 得到上链复形
$$
\cdots\leftarrow C_{n+1}^\ast \xleftarrow{\delta} C_n^\ast \xleftarrow{\delta} C_{n-1}^\ast \leftarrow \cdots
$$
诱导了上同调群.
类似于同调, 我们有由如下短正合列诱导的相对上同调长正合列.
$$
0\leftarrow C^n(A,B;G)\xleftarrow{i^\ast } C^n(X,B;G)\xleftarrow{j^\ast } C^n(X,A;G)\leftarrow 0,
$$
用泛系数定理或对偶地推导, 也有挖去定理
$$
i^\ast :H^n(X,A;G)\cong H^n(X\setminus Z,A\setminus Z;G),\quad Z\Subset A\subset X.
$$
以及MV序列(以及相对版本)
$$
\cdots \leftarrow H^{n}(A\cap B;G)\leftarrow H^n(A;G)\oplus H^n(B;G)\leftarrow H^n(A\cup B;G)\leftarrow H^{n-1}(A\cap B;G)\leftarrow \cdots
$$
上同调泛系数定理
考虑$h:H^n(C;G)\rightarrow \operatorname{Hom}(H_n(C),G).$
$H^n(C;G)$中的元素为$\varphi\in \operatorname{Hom}(C_n,G),$
满足$\delta \varphi=0,$ 即$\varphi\partial=0.$
因此$\varphi$在$B_n=\operatorname{Im}\partial$上消失.
这样限制$\varphi_0=\varphi|_{Z_n}$就诱导了$\overline\varphi_0:Z_n/B_n=H_n(C)\rightarrow G.$
若$\varphi=\delta\psi=\psi\partial,$ 那么$\varphi_0=0.$
于是这就定义好了$h:\varphi\mapsto \overline\varphi.$
由于$B$是自由的, 下面的正合列分裂:
$$
0\rightarrow Z_n\rightarrow C_n\rightarrow B_{n-1}\rightarrow 0
$$
从而存在$p:C_n\rightarrow Z_n$为投影.
那么任意选取$\varphi_0:Z_n\rightarrow G,$
取$\varphi=\varphi_0p:C_n\rightarrow G,$ 就有$\varphi|_{Z_n}=\varphi_0.$
若可以诱导映射$\overline\varphi_0:H_n(C)=Z_n/B_n\rightarrow G,$
则$\varphi\partial=0=\delta \varphi,$ $\varphi\in \ker\delta.$
从而$h$为满射:
$$
p^\ast :\operatorname{Hom}(H_n(C),G)\rightarrow \ker\delta\rightarrow H^n(C;G), \quad h\circ p^\ast =\mathrm{id},
$$
因此, 我们就得到了可裂短正合列:
$$
0\rightarrow\ker h\rightarrow H^n(C;G)\rightarrow \operatorname{Hom}(H_n(C),G)\rightarrow 0
$$
考虑可裂短正合列$0\rightarrow Z_n\rightarrow C_n\rightarrow B_{n-1}\rightarrow 0$的对偶短正合列,
诱导长正合列:
$$
\cdots\leftarrow B_n^\ast \xleftarrow{i^\ast _n} Z_n^\ast \leftarrow H^n(C;G)\xleftarrow{\delta} B_{n-1}^\ast \xleftarrow{i^\ast _{n-1} } Z_{n-1}^\ast \leftarrow \cdots,
$$
可以拆为短正合列
$$
0\leftarrow \ker i^\ast _n\leftarrow H^n(C;G)\leftarrow \operatorname{coker}i^\ast _{n-1}\leftarrow 0
$$
可以看出$\ker i^\ast _n=\operatorname{Hom}(H_n(C),G),$
因为$\,\forall\,\varphi_0\in \ker i^\ast _n,$
$\varphi_0:Z_n\rightarrow C_n,$ $i^\ast _n\varphi_0=\varphi_0|_{B_n}=0.$
$\varphi_0\mapsto \overline\varphi_0:H_n(C)\rightarrow G$给出了相等关系.
因此我们可将原先的短正合列写成:
$$
0\rightarrow \operatorname{coker}i_{n-1}^\ast \rightarrow H^n(C;G)\rightarrow \operatorname{Hom}(H_n(C),G)\rightarrow 0
$$
引理 1. 任给$F,F'$分别为abel群$H,H'$的自由分解, 那么$\alpha:H\rightarrow H'$可以被诱导到$F_i\rightarrow F_i'$上成为链映射. 进而相同的abel群的两个自由分解满足$H^n(F;G)\cong H^n(F';G).$
任意abel群$H$有自由分解
$$
0\rightarrow F_1\rightarrow F_0\rightarrow H\rightarrow 0,
$$
$F_0$由$H$生成元自由地生成, $F_1=\ker(F_0\rightarrow H).$
那么唯一非平凡的上同调为$H^1(F;G),$ 仅由$H,G$唯一决定.
记其为$\operatorname{Ext}(H,G).$ 即有正合列
$$
0\leftarrow \operatorname{Ext}(H,G)\leftarrow F_1^\ast \leftarrow F_0^\ast \leftarrow H^\ast \leftarrow 0.
$$
再来看自由分解
$$
0\rightarrow B_{n-1}\xrightarrow{i_{n-1} }Z_{n-1}\rightarrow H_{n-1}\rightarrow 0,
$$
对偶为
$$
0\leftarrow \operatorname{coker}i_{n-1}^\ast \leftarrow B_{n-1}^\ast \xleftarrow {i_{n-1}^\ast } Z_{n-1}^\ast \leftarrow \operatorname{Hom}(H_{n-1}(C),G)\leftarrow 0.
$$
于是由引理,
$\operatorname{coker}i_{n-1}^\ast \cong \operatorname{Ext}(H_{n-1}(C),G).$
这就给出了泛系数定理:
定理 2 (上同调泛系数定理). 自由链复形$C$的上同调$H^n(C;G)$由下面的分裂正合列决定:
$$
0\rightarrow \operatorname{Ext}(H_{n-1}(C),G)\rightarrow H^n(C;G)\xrightarrow{h}\operatorname{Hom}(H_n(C);G)\rightarrow 0.
$$
一些关于$\operatorname{Ext}$的性质有:
$\operatorname{Ext}(H\oplus H',G)\cong \operatorname{Ext}(H,G)\oplus \operatorname{Ext}(H',G).$
$\operatorname{Ext}(H,G)=0,$ 若$H$自由.
$\operatorname{Ext}(\mathbb{Z}_n,G)\approx G/nG.$
推论 3. 若$H_n,H_{n-1}$都是有限生成的, 挠子群为$T_n,T_{n-1},$ 那么 $H^n(C;\mathbb{Z})\cong (H_n/T_n)\oplus T_{n-1}.$
泛系数定理中的短正合列是自然的, 不过分裂本身并不是自然的. 由五引理,
我们有:
推论 4. 若链映射诱导同调群的同构, 则它也诱导上同调群的同构.
同调泛系数定理
类似地,
我们从可裂短正合列$0\rightarrow Z_n\rightarrow C_n\rightarrow B_{n-1}\rightarrow 0$出发,
诱导张量积短正合列
$$
0\rightarrow Z_n\otimes G\rightarrow C_n\otimes G\rightarrow B_{n-1}\otimes G\rightarrow 0,
$$
给出长正合列:
$$
\cdots\rightarrow B_n\otimes G\xrightarrow{i_n\otimes 1} Z_n\otimes G\rightarrow H_n(C;G)\rightarrow B_{n-1}\otimes G\xrightarrow{i_{n-1}\otimes 1} Z_{n-1}\otimes G\rightarrow \cdots,
$$
拆分为可裂短正合列:
$$
0\rightarrow \operatorname{coker}(i_n\otimes 1)\rightarrow H_n(C;G)\rightarrow \ker(i_{n-1}\otimes 1)\rightarrow 0.
$$
对于自由分解
$$
0\rightarrow B_n\xrightarrow{i_n}Z_n\rightarrow H_n(C)\rightarrow 0,
$$
诱导正合列
$$
0\rightarrow\ker (i_n\otimes 1)\rightarrow B_n\otimes G\xrightarrow{i_n\otimes 1}Z_n\otimes G\rightarrow H_n(C)\otimes G\rightarrow 0.
$$
因此$\operatorname{coker}(i_n\otimes 1)\cong H_n(C)\otimes G.$
类似上同调版本, 我们有引理:
引理 5. 相同abel群的两个自由分解满足$H_n(F\otimes G)\cong H_n(F'\otimes G).$
对于自由分解$0\rightarrow F_1\rightarrow F_0\rightarrow H\rightarrow 0,$
我们有正合列
$$
0\rightarrow \operatorname{Tor}(H,G)\rightarrow F_1\otimes G\rightarrow F_0\otimes G\rightarrow H\otimes G\rightarrow 0.
$$
$\operatorname{Tor}(H,G)$由$H,G$唯一决定. 比对两种自由分解,
我们有$\ker(i_n\otimes 1)\cong \operatorname{Tor}(H,G).$ 这样,
我们就得到了同调版本的泛系数定理:
定理 6 (同调泛系数定理). 自由链复形的任意系数同调群$H_n(C;G)$由下面的分裂正合列决定:
$$
0\rightarrow H_n(C)\otimes G\rightarrow H_n(C;G)\rightarrow \operatorname{Tor}(H_{n-1},G)\rightarrow 0.
$$
可以考虑相对同调的版本. 类似地, 短正合列是自然的, 但分裂不自然.
$\operatorname{Tor}$函子满足的一些性质是:
$\operatorname{Tor}(A,B)\cong \operatorname{Tor}(B,A).$
$\operatorname{Tor}(\bigoplus_iA_i,B)\cong \bigoplus_i\operatorname{Tor}(A_i,B).$
$\operatorname{Tor}(A,B)=0,$ 若$A,B$至少有一个是自由的,
或更一般的无挠的.
$\operatorname{Tor}(A,B)=\operatorname{Tor}(T(A),B),$
$T(A)$为$A$的挠子群.
$\operatorname{Tor}(\mathbb{Z}_n,A)\cong \ker(A\xrightarrow{n}A).$
短正合列$0\rightarrow B\rightarrow C\rightarrow D\rightarrow 0$自然地诱导正合列
$$
0\rightarrow \operatorname{Tor}(A,B)\rightarrow \operatorname{Tor}(A,C)\rightarrow \operatorname{Tor}(A,D)\rightarrow A\otimes B\rightarrow A\otimes C\rightarrow A\otimes D\rightarrow 0.
$$
特别地,
$\operatorname{Tor}(\mathbb{Z}_m,\mathbb{Z}_n)=\mathbb{Z}_{[m,n]}\cong \mathbb{Z}_m\otimes \mathbb{Z}_n.$
由$\operatorname{Tor}$函子性质, 我们有:
推论 7. 若$H_n(X;\mathbb{Z})$有限生成, 则$H_n(X;\mathbb{Q})\cong H_n(X;\mathbb{Z})\otimes \mathbb{Q}.$ 若$H_{n-1}(X;\mathbb{Z})$也是有限生成的, 那么对于质数$p$, $H_n(X;\mathbb{Z}_p)$中有$H_n(X;\mathbb{Z})$中$\mathbb{Z},\mathbb{Z}_{p^k}$项的个数加上$H_{n-1}(X;\mathbb{Z})$中$\mathbb{Z}_{p^k}$项的个数和个$\mathbb{Z}_p.$
推论 8. $\widetilde H_\ast (X;\mathbb{Z})=0$当且仅当$\widetilde H_\ast (X;\mathbb{Q})=\widetilde H_\ast (X;\mathbb{Z}_p)=0,$ $\,\forall\,$质数$p.$ $f$诱导$\mathbb{Z}$系数同调群的同构当且仅当它也诱导$\mathbb{Q},\mathbb{Z}_p$系数同调群的同构, $\,\forall\,$质数$p$.
文章最后更新于 2022-10-16 22:28:28
Basic concepts
chain complex
$$
\cdots\rightarrow C_{n+1}\xrightarrow{\partial} C_n\xrightarrow{\partial} C_{n-1}\rightarrow\cdots,
$$
We apply the $\operatorname{Hom}(-,G)$ functor to get the cochain complex
$$
\cdots\leftarrow C_{n+1}^\ast \xleftarrow{\delta} C_n^\ast \xleftarrow{\delta} C_{n-1}^\ast \leftarrow \cdots
$$
induced Cohomology group.
Similar to cohomology, we have relatively cohomological long exact sequences induced by the following short exact sequences.
$$
0\leftarrow C^n(A,B;G)\xleftarrow{i^\ast } C^n(X,B;G)\xleftarrow{j^\ast } C^n(X,A;G)\leftarrow 0,
$$
Use the universal coefficient theorem or dual derivation, and there are also digging theorems
$$
i^\ast :H^n(X,A;G)\cong H^n(X\setminus Z,A\setminus Z;G),\quad Z\Subset A\subset X.
$$
and the MV sequence (and relative version)
$$
\cdots \leftarrow H^{n}(A\cap B;G)\leftarrow H^n(A;G)\oplus H^n(B;G)\leftarrow H^n(A\cup B;G)\leftarrow H^{n-1}(A\cap B;G)\leftarrow \cdots
$$
Cohomology universal coefficient theorem
Consider $h:H^n(C;G)\rightarrow \operatorname{Hom}(H_n(C),G).$
The elements in $H^n(C;G)$ are $\varphi\in \operatorname{Hom}(C_n,G),$
Satisfies $\delta \varphi=0,$ that is $\varphi\partial=0.$
Therefore $\varphi$ disappears on $B_n=\operatorname{Im}\partial$.
This restriction $\varphi_0=\varphi|_{Z_n}$ induces $\overline\varphi_0:Z_n/B_n=H_n(C)\rightarrow G.$
If $\varphi=\delta\psi=\psi\partial,$ then $\varphi_0=0.$
So this is defined $h:\varphi\mapsto \overline\varphi.$
Since $B$ is free, the following exact column splits:
$$
0\rightarrow Z_n\rightarrow C_n\rightarrow B_{n-1}\rightarrow 0
$$
Therefore, $p:C_n\rightarrow Z_n$ exists as a projection.
Then choose $\varphi_0:Z_n\rightarrow G,$ arbitrarily
Take $\varphi=\varphi_0p:C_n\rightarrow G,$ and get $\varphi|_{Z_n}=\varphi_0.$
If mapping can be induced $\overline\varphi_0:H_n(C)=Z_n/B_n\rightarrow G,$
Then $\varphi\partial=0=\delta \varphi,$ $\varphi\in \ker\delta.$
Therefore $h$ is surjective:
$$
p^\ast :\operatorname{Hom}(H_n(C),G)\rightarrow \ker\delta\rightarrow H^n(C;G), \quad h\circ p^\ast =\mathrm{id},
$$
Therefore, we get the splittable short exact sequence:
$$
0\rightarrow\ker h\rightarrow H^n(C;G)\rightarrow \operatorname{Hom}(H_n(C),G)\rightarrow 0
$$
Consider the dual short exact sequence of the splittable short exact sequence $0\rightarrow Z_n\rightarrow C_n\rightarrow B_{n-1}\rightarrow 0$,
Induced long exact sequence:
$$
\cdots\leftarrow B_n^\ast \xleftarrow{i^\ast _n} Z_n^\ast \leftarrow H^n(C;G)\xleftarrow{\delta} B_{n-1}^\ast \xleftarrow{i^\ast _{n-1} } Z_{n-1}^\ast \leftarrow \cdots,
$$
Can be split into short exact sequences
$$
0\leftarrow \ker i^\ast _n\leftarrow H^n(C;G)\leftarrow \operatorname{coker}i^\ast _{n-1}\leftarrow 0
$$
It can be seen that $\ker i^\ast _n=\operatorname{Hom}(H_n(C),G),$
Because$\,\forall\,\varphi_0\in \ker i^\ast _n,$
$\varphi_0:Z_n\rightarrow C_n,$ $i^\ast _n\varphi_0=\varphi_0|_{B_n}=0.$
$\varphi_0\mapsto \overline\varphi_0:H_n(C)\rightarrow G$ gives the equality relationship.
Therefore we can write the original short exact sequence as:
$$
0\rightarrow \operatorname{coker}i_{n-1}^\ast \rightarrow H^n(C;G)\rightarrow \operatorname{Hom}(H_n(C),G)\rightarrow 0
$$
Lemma 1. Given that $F,F'$ is a free decomposition of the abel group $H,H'$, then $\alpha:H\rightarrow H'$ can be induced to $F_i\rightarrow F_i'$ to become a chain map. Then the two free decompositions of the same abel group satisfy $H^n(F;G)\cong H^n(F';G).$
Any abel group $H$ has free decomposition
$$
0\rightarrow F_1\rightarrow F_0\rightarrow H\rightarrow 0,
$$
$F_0$ is freely generated from the generator $H$, $F_1=\ker(F_0\rightarrow H).$
Then the only non-trivial cohomology is $H^1(F;G),$ uniquely determined by $H,G$.
Record it as $\operatorname{Ext}(H,G).$, that is, there is an exact sequence
$$
0\leftarrow \operatorname{Ext}(H,G)\leftarrow F_1^\ast \leftarrow F_0^\ast \leftarrow H^\ast \leftarrow 0.
$$
Let’s look at free decomposition
$$
0\rightarrow B_{n-1}\xrightarrow{i_{n-1} }Z_{n-1}\rightarrow H_{n-1}\rightarrow 0,
$$
The dual is
$$
0\leftarrow \operatorname{coker}i_{n-1}^\ast \leftarrow B_{n-1}^\ast \xleftarrow {i_{n-1}^\ast } Z_{n-1}^\ast \leftarrow \operatorname{Hom}(H_{n-1}(C),G)\leftarrow 0.
$$
So by the lemma,
$\operatorname{coker}i_{n-1}^\ast \cong \operatorname{Ext}(H_{n-1}(C),G).$
This gives the universal coefficient theorem:
Theorem 2 (Cohomology universal coefficient theorem). The cohomology $H^n(C;G)$ of the free chain complex $C$ is determined by the following splitting exact sequence:
$$
0\rightarrow \operatorname{Ext}(H_{n-1}(C),G)\rightarrow H^n(C;G)\xrightarrow{h}\operatorname{Hom}(H_n(C);G)\rightarrow 0.
$$
Some properties of $\operatorname{Ext}$ are:
$\operatorname{Ext}(H\oplus H',G)\cong \operatorname{Ext}(H,G)\oplus \operatorname{Ext}(H',G).$
$\operatorname{Ext}(H,G)=0,$ If $H$ is free.
$\operatorname{Ext}(\mathbb{Z}_n,G)\approx G/nG.$
Corollary 3. If $H_n,H_{n-1}$ are all finitely generated, and the torsion subgroup is $T_n,T_{n-1},$, then $H^n(C;\mathbb{Z})\cong (H_n/T_n)\oplus T_{n-1}.$
The short exact sequence in the universal coefficient theorem is natural, but the split itself is not natural. From the five lemmas,
We have:
Corollary 4. If a chain map induces isomorphism in homology groups, it also induces isomorphism in cohomology groups.
Homology universal coefficient theorem
Similarly,
We start from the splittable short exact sequence $0\rightarrow Z_n\rightarrow C_n\rightarrow B_{n-1}\rightarrow 0$,
Induced tensor product short exact sequence
$$
0\rightarrow Z_n\otimes G\rightarrow C_n\otimes G\rightarrow B_{n-1}\otimes G\rightarrow 0,
$$
Give a long exact sequence:
$$
\cdots\rightarrow B_n\otimes G\xrightarrow{i_n\otimes 1} Z_n\otimes G\rightarrow H_n(C;G)\rightarrow B_{n-1}\otimes G\xrightarrow{i_{n-1}\otimes 1} Z_{n-1}\otimes G\rightarrow \cdots,
$$
Split into splittable short exact sequences:
$$
0\rightarrow \operatorname{coker}(i_n\otimes 1)\rightarrow H_n(C;G)\rightarrow \ker(i_{n-1}\otimes 1)\rightarrow 0.
$$
For free decomposition
$$
0\rightarrow B_n\xrightarrow{i_n}Z_n\rightarrow H_n(C)\rightarrow 0,
$$
Induced Exact Column
$$
0\rightarrow\ker (i_n\otimes 1)\rightarrow B_n\otimes G\xrightarrow{i_n\otimes 1}Z_n\otimes G\rightarrow H_n(C)\otimes G\rightarrow 0.
$$
Therefore $\operatorname{coker}(i_n\otimes 1)\cong H_n(C)\otimes G.$
Similar to the cohomology version, we have the lemma:
Lemma 5. Two free decompositions of the same abel group satisfy $H_n(F\otimes G)\cong H_n(F'\otimes G).$
For free decomposition $0\rightarrow F_1\rightarrow F_0\rightarrow H\rightarrow 0,$
We have the exact column
$$
0\rightarrow \operatorname{Tor}(H,G)\rightarrow F_1\otimes G\rightarrow F_0\otimes G\rightarrow H\otimes G\rightarrow 0.
$$
$\operatorname{Tor}(H,G)$ is uniquely determined by $H,G$. Comparing the two free decompositions,
We have $\ker(i_n\otimes 1)\cong \operatorname{Tor}(H,G).$ like this,
We then get the homologous version of the universal coefficient theorem:
Theorem 6 (Homology universal coefficient theorem). The arbitrary coefficient homology group $H_n(C;G)$ of the free chain complex is determined by the following splitting exact series:
$$
0\rightarrow H_n(C)\otimes G\rightarrow H_n(C;G)\rightarrow \operatorname{Tor}(H_{n-1},G)\rightarrow 0.
$$
A relatively homologous version can be considered. Similarly, short exact sequences are natural, but splitting is unnatural.
Some properties satisfied by the $\operatorname{Tor}$ functor are:
$\operatorname{Tor}(A,B)\cong \operatorname{Tor}(B,A).$
$\operatorname{Tor}(\bigoplus_iA_i,B)\cong \bigoplus_i\operatorname{Tor}(A_i,B).$
$\operatorname{Tor}(A,B)=0,$ If at least one of $A,B$ is free,
Or more generally inflexible.
$\operatorname{Tor}(A,B)=\operatorname{Tor}(T(A),B),$
$T(A)$ is the deflection subgroup of $A$.
$\operatorname{Tor}(\mathbb{Z}_n,A)\cong \ker(A\xrightarrow{n}A).$
Short exact sequences $0\rightarrow B\rightarrow C\rightarrow D\rightarrow 0$ naturally induce exact sequences
$$
0\rightarrow \operatorname{Tor}(A,B)\rightarrow \operatorname{Tor}(A,C)\rightarrow \operatorname{Tor}(A,D)\rightarrow A\otimes B\rightarrow A\otimes C\rightarrow A\otimes D\rightarrow 0.
$$
In particular,
$\operatorname{Tor}(\mathbb{Z}_m,\mathbb{Z}_n)=\mathbb{Z}_{[m,n]}\cong \mathbb{Z}_m\otimes \mathbb{Z}_n.$
From the $\operatorname{Tor}$ functor property, we have:
Corollary 7. If $H_n(X;\mathbb{Z})$ is finitely generated, then $H_n(X;\mathbb{Q})\cong H_n(X;\mathbb{Z})\otimes \mathbb{Q}.$ If $H_{n-1}(X;\mathbb{Z})$ is also finitely generated, then for the prime number $p$, $H_n(X;\mathbb{Z}_p)$ has the number of $\mathbb{Z},\mathbb{Z}_{p^k}$ terms in $H_n(X;\mathbb{Z})$ plus the number of $\mathbb{Z}_{p^k}$ terms in $H_{n-1}(X;\mathbb{Z})$ and $\mathbb{Z}_p.$
Corollary 8. $\widetilde H_\ast (X;\mathbb{Z})=0$ if and only if $\widetilde H_\ast (X;\mathbb{Q})=\widetilde H_\ast (X;\mathbb{Z}_p)=0,$ $\,\forall\,$ prime number $p.$ $f$ induces the isomorphism of $\mathbb{Z}$ coefficient homology group if and only if it also induces the isomorphism of $\mathbb{Q},\mathbb{Z}_p$ coefficient homology group, $\,\forall\,$ prime number $p$.
The article was last updated on 2022-10-16 22:28:28
