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    Appendix: homological algebra

    Contents

    Exact sequences

    Exactness of $A\to B\to C$

    Exactness at the middle term means image equals kernel.Suggest a slogan

    A sequence of homomorphisms \(A \mapsvia{f_1} B \mapsvia{f_2} C\) is exact at \(B\) if and only if \(\im f_1 = \ker f_2\). In that case \(f_2 \circ f_1 = 0\); the converse fails, since \(f_2\circ f_1 = 0\) only gives \(\im f_1\subseteq\ker f_2\).

    For maps of abelian groups (\(R\)-modules):

    • \(0 \to A \xrightarrow{f} B\) is exact if and only if \(f\) is injective.
    • \(B\xrightarrow{g} C \to 0\) is exact if and only if \(g\) is surjective.
    • \(0\to A \xrightarrow{f} B \to 0\) is exact if and only if \(f\) is an isomorphism.
    • If \(A \xrightarrow{a} B \xrightarrow{b} C \xrightarrow{c} D \xrightarrow{d} E\) is exact, \(a\) is surjective, and \(d\) is injective, then \(C = 0\).
    • If \(0\to A \xrightarrow{f} B \xrightarrow{g} C \xrightarrow{h} D\to 0\) is exact, then \(g\) is an isomorphism if and only if \(A = D = 0\).
    • If \(0\to A\xrightarrow{f} B \xrightarrow{g} C \to 0\) is exact, then \(g\) induces \(C \cong B/\im f\).
    • If \(0\to A\to B \to C \to 0\) is exact and \(C\) is free, then the sequence splits.

    (Proof of the fifth item)

    Suppose \(g\) is an isomorphism. Exactness at \(B\) gives \(\im f = \ker g = 0\), and \(f\) is injective, so \(A=0\). Exactness at \(C\) gives \(\ker h = \im g = C\), so \(h=0\), and \(h\) is surjective, so \(D=0\). Conversely, if \(A=D=0\), then exactness at \(B\) gives \(\ker g = 0\) and exactness at \(C\) gives \(\im g = \ker(C\to 0) = C\).

    (A short exact sequence that splits with $C$ not free)

    \(0\to\ZZ/2\to\ZZ/2\oplus\ZZ/2\to\ZZ/2\to 0\), with the inclusion of the first factor and the projection onto the second, splits, and \(\ZZ/2\) is not free.

    Split short exact sequence

    Let \[ 1 \to A \mapsvia{f} B \mapsvia{g} C \to 1 \] be a short exact sequence of groups. The sequence splits if there exists a homomorphism \(s\colon C\to B\) with \(g\circ s = \id_C\).

    For a short exact sequence \(0\to A\mapsvia{f} B\mapsvia{g} C\to 0\) of abelian groups, or more generally of modules over a ring, the following are equivalent: the sequence splits; there exists a homomorphism \(r\colon B\to A\) with \(r\circ f = \id_A\); there exists an isomorphism \(B\cong A\oplus C\) under which \(f\) is the inclusion of \(A\) and \(g\) is the projection onto \(C\). For groups, if \(s\) splits the sequence, then \(B \cong A\rtimes_\varphi C\), where \(\varphi\colon C\to\Aut(A)\) is given by \(f(\varphi(c)(a)) = s(c)\,f(a)\,s(c)^{-1}\).

    [@DF04].

    (Exact sequences)

    • \(0 \to \ZZ \xrightarrow{\times 2} \ZZ \xrightarrow{\operatorname{mod} 2} \ZZ/2 \to 0\).
    • For a normal subgroup \(N\) of a group \(G\), \(1 \to N \xrightarrow{\iota} G \xrightarrow{p} G/N \to 1\) with \(\iota\) the inclusion and \(p\) the quotient map.
    • For \(n\geq 3\), \(1 \to \ZZ/n \xrightarrow{\iota} D_{2n} \xrightarrow{s} \ZZ/2 \to 1\), with \(\iota\) the inclusion of the rotations and \(s\) sending rotations to \(0\) and reflections to \(1\).
    • For ideals \(I,J\) of a commutative ring \(R\), \(0 \to I \intersect J \xrightarrow{x\mapsto(x,x)} I \oplus J \xrightarrow{(x,y) \mapsto x-y} I + J \to 0\).
    • For ideals \(I,J\) of a commutative ring \(R\), \(0 \to R/(I \intersect J) \xrightarrow{x\mapsto(x,x)} R/I \oplus R/J \xrightarrow{(x,y) \mapsto x-y} R/(I + J) \to 0\).
    • For a bounded Lipschitz domain \(\Omega\subseteq\RR^3\) homeomorphic to a ball, the Sobolev de Rham sequence \(0\to\RR \to H^1(\Omega) \xrightarrow{\nabla} H(\operatorname{curl},\Omega) \xrightarrow{\nabla \cross} H(\operatorname{div},\Omega) \xrightarrow{\nabla \cdot} L^2(\Omega) \to 0\) is exact, with \(\RR\) the constant functions.

    A sequence \(\cdots\to A_{k-1}\xrightarrow{f_{k-1}} A_k\xrightarrow{f_k} A_{k+1}\to\cdots\) with \(f_k\circ f_{k-1} = 0\) for all \(k\) is a chain complex, and it is exact if and only if its homology \(\ker f_k/\im f_{k-1}\) vanishes in every degree.

    A long exact sequence \(\cdots\to A_{k-1}\xrightarrow{f_{k-1}} A_k\xrightarrow{f_k} A_{k+1}\to\cdots\) splits into short exact sequences: with \(C_k \coloneqq \ker f_k = \im f_{k-1} \cong \coker f_{k-2}\), each \[ 0\to C_k\to A_k\xrightarrow{f_k} C_{k+1}\to 0 \] is exact.

    The five lemma

    The five lemma

    Exact rows and isomorphisms on the neighboring terms force the middle map to be an isomorphism.Suggest a slogan
    Suppose the rows in the commutative diagram

    are exact. If \(\alpha\) is surjective, \(\beta\) and \(\delta\) are isomorphisms, and \(\varepsilon\) is injective, then \(\gamma\) is an isomorphism [@Hat02].

    For injectivity, let \(c\in C\) satisfy \(\gamma(c)=0\). Then \[ \delta(h(c))=h'(\gamma(c))=0. \] Since \(\delta\) is injective, \(h(c)=0\). Exactness of the top row gives \(c=g(b)\) for some \(b\in B\). Now \[ 0=\gamma(g(b))=g'(\beta(b)), \] so exactness of the bottom row gives \(a'\in A'\) with \(\beta(b)=f'(a')\). Choose \(a\in A\) with \(\alpha(a)=a'\) using surjectivity of \(\alpha\). Commutativity gives \[ \beta(b)=f'(\alpha(a))=\beta(f(a)). \] Since \(\beta\) is injective, \(b=f(a)\), hence \(c=g(f(a))=0\). Thus \(\gamma\) is injective.

    For surjectivity, let \(c'\in C'\). Since \(\delta\) is surjective, choose \(d\in D\) with \(\delta(d)=h'(c')\). Then \[ \varepsilon(i(d))=i'(\delta(d))=i'(h'(c'))=0. \] Since \(\varepsilon\) is injective, \(i(d)=0\). Exactness of the top row gives \(c\in C\) with \(h(c)=d\). Hence \[ h'(c'-\gamma(c))=h'(c')-\delta(h(c))=0. \] By exactness of the bottom row, \(c'-\gamma(c)=g'(b')\) for some \(b'\in B'\). Choose \(b\in B\) with \(\beta(b)=b'\) using surjectivity of \(\beta\). Then \[ c'=\gamma(c)+g'(\beta(b))=\gamma(c)+\gamma(g(b))=\gamma(c+g(b)). \] Thus \(\gamma\) is surjective, so it is an isomorphism.

    Free resolutions

    For \(m\geq 1\), \[ 0 \to \ZZ \xrightarrow{\times m} \ZZ \xrightarrow{\operatorname{mod} m} \ZZ/m \to 0 \] is a free resolution of \(\ZZ/m\).

    For an abelian group \(G\) with generating set \(S\), let \(F\) be the free abelian group on \(S\) and \(f\colon F\to G\) the induced surjection. Subgroups of free abelian groups are free, so \[ 0 \to \ker(f) \to F \xrightarrow{f} G \to 0 \] is a free resolution of \(G\) of length \(1\).

    Properties of tensor products

    For abelian groups (\(R\)-modules over a commutative ring \(R\)) \(A\), \(B\), \(A_i\), \(B_j\), and integers \(m,n\geq 1\) with \(d=\gcd(m,n)\):

    • \(A\tensor B \cong B\tensor A\).
    • \(A \tensor_R R^n \cong A^n\).
    • \(\qty{\bigoplus_i A_i} \tensor \qty{\bigoplus_j B_j} \cong \bigoplus_i\bigoplus_j(A_i \tensor B_j)\).
    • \(\ZZ/m \tensor \ZZ/n \cong \ZZ/d\).
    • \(\ZZ/n \tensor A \cong A/nA\).

    Properties of Hom

    • \(\Hom_R \qty{\bigoplus_i A_i, \prod_j B_j} \cong \prod_i \prod_j \Hom_R(A_i, B_j)\).
    • \(\Hom_R(\wait,\wait)\) is contravariant in the first variable and covariant in the second.
    • Over a field, \(\Hom(\wait, B)\) and \(\Hom(A,\wait)\) are exact.

    $\Hom_R(R,A)\cong A$

    An $R$-linear map out of $R$ is determined uniquely by the image of $1$.Suggest a slogan

    Let \(R\) be a ring with \(1\) and \(A\) a left \(R\)-module. The map \(\Hom_R(R, A)\to A\), \(f\mapsto f(1)\), is an isomorphism of abelian groups, with inverse sending \(a\in A\) to the homomorphism \(r\mapsto ra\); if \(R\) is commutative, it is an isomorphism of \(R\)-modules.

    What is $\mathop{\mathrm{Hom}}_R(R, A)$ for a left $R$-module $A$?

    Properties of Tor

    For abelian groups, with \(\mathbf{T}G\) the torsion subgroup of \(G\):

    • \(\tor_0^R(A, B) \cong A \tensor_R B\).
    • \(\tor\qty{\bigoplus_i A_i, \bigoplus_j B_j} \cong \bigoplus_i \bigoplus_j \tor(\mathbf{T}A_i, \mathbf{T}B_j)\).
    • \(\tor(A, B) \cong \tor(B, A)\).
    • \(\tor(\ZZ/n, G) \cong \ker (G\xrightarrow{\times n} G) = \theset{g\in G\mid ng = 0}\).

    Properties of Ext

    For abelian groups:

    • \(\ext_R^0(A, B) \cong \Hom_R(A, B)\).
    • \(\ext\qty{\bigoplus_i A_i, \prod_j B_j} \cong \prod_i \prod_j \ext(A_i, B_j)\).
    • \(\ext(F, G) = 0\) if \(F\) is free.
    • \(\ext(\ZZ/n, G) \cong G/nG\).

    Computing Tor

    If \(\cdots \to F_1\to F_0\to A\to 0\) is a free resolution of \(A\), then \[ \tor_n(A, B) \cong H_n\qty{\cdots \to F_n \tensor B \to F_{n-1}\tensor B \to \cdots \to F_0\tensor B \to 0}. \]

    $\Tor^{\ZZ}_*(\ZZ/n, A)$

    For $\ZZ/n$, $\Tor_0=A/nA$, $\Tor_1=A[n]$, and all higher $\Tor$ groups vanish.Suggest a slogan

    Let \(n\geq 1\), let \(C_n\coloneqq\ZZ/n\), let \(A\) be an abelian group, and let \(A[n]\coloneqq\ts{a\in A \st na = 0}\). The groups $\Tor^{\ZZ}_k(C_n, A)$ are \[ \Tor^{\ZZ}_0(C_n, A)\cong C_n\otimes_{\ZZ}A\cong A/nA,\qquad \Tor^{\ZZ}_1(C_n, A)\cong A[n],\qquad \Tor^{\ZZ}_k(C_n, A) = 0\quad (k\geq 2) \] [@Hat02].

    What is $\operatorname{Tor}^{\mathbf{Z}}_*(C_n, A)$?

    Computing Ext

    If \(\cdots \to F_1\to F_0\to A\to 0\) is a free resolution of \(A\), then \[ \ext^n(A, B) \cong H^n\qty{0 \to \Hom(F_0, B) \to \Hom(F_1, B) \to \cdots \to \Hom(F_n, B) \to \cdots}. \]

    $\Ext^*_{\ZZ}(\ZZ/n, A)$

    For $\ZZ/n$, $\Ext^0$ detects $n$-torsion, $\Ext^1$ is quotient by $n$, and higher $\Ext$ vanishes.Suggest a slogan

    Let \(n\geq 1\), let \(A\) be an abelian group, and let \(A[n]\coloneqq\ts{a\in A \st na = 0}\). The groups $\Ext^k_{\ZZ}(\ZZ/n, A)$ are \[ \Ext^0_{\ZZ}(\ZZ/n, A)\cong\Hom(\ZZ/n, A)\cong A[n],\qquad \Ext^1_{\ZZ}(\ZZ/n, A)\cong A/nA,\qquad \Ext^k_{\ZZ}(\ZZ/n, A) = 0\quad (k\geq 2) \] [@Hat02].

    What is $\operatorname{Ext}^*_{\mathbf{Z}}(\mathbf{Z}/n, A)$?

    Hom, Ext, and Tor tables

    For \(m,n\geq 1\) and \(d = \gcd(m, n)\), with the first argument indexing rows and the second columns:

    \(\Hom\) \(\ZZ/m\) \(\ZZ\) \(\QQ\)
    \(\ZZ/n\) \(\ZZ/d\) \(0\) \(0\)
    \(\ZZ\) \(\ZZ/m\) \(\ZZ\) \(\QQ\)
    \(\QQ\) \(0\) \(0\) \(\QQ\)
    \(\tor\) \(\ZZ/m\) \(\ZZ\) \(\QQ\)
    \(\ZZ/n\) \(\ZZ/d\) \(0\) \(0\)
    \(\ZZ\) \(0\) \(0\) \(0\)
    \(\QQ\) \(0\) \(0\) \(0\)
    \(\ext\) \(\ZZ/m\) \(\ZZ\) \(\QQ\)
    \(\ZZ/n\) \(\ZZ/d\) \(\ZZ/n\) \(0\)
    \(\ZZ\) \(0\) \(0\) \(0\)
    \(\QQ\) \(0\) \(\mathbb{A}_f/\QQ\) \(0\)

    Here \(\mathbb{A}_f\) is the ring of finite adeles of \(\QQ\), containing \(\QQ\) diagonally. Applying \(\Hom(\QQ,\wait)\) to \(0\to\ZZ\to\QQ\to\QQ/\ZZ\to 0\), with \(\Hom(\QQ,\ZZ)=0\) and \(\ext(\QQ,\QQ)=0\), gives the exact sequence \(0\to\QQ\to\Hom(\QQ,\QQ/\ZZ)\to\ext(\QQ,\ZZ)\to 0\), and \(\Hom(\QQ,\QQ/\ZZ)\cong\mathbb{A}_f\).

    Tor groups among $\ZZ/n$, $\ZZ$ and $\QQ$

    Over $\ZZ$, $\Tor_1(\ZZ/n, \ZZ/m)\cong\ZZ/\gcd(m,n)$, $\Tor_1(A, B) = 0$ when $A$ or $B$ is torsion-free, and $\Tor_k = 0$ for $k\geq 2$.Suggest a slogan

    Let \(m, n\geq 1\) and \(d\coloneqq\gcd(m,n)\). For an abelian group \(A\) let \(A[n]\coloneqq\ts{a\in A \st na = 0}\). Then \(\Tor^{\ZZ}_0(\ZZ/n, A)\cong A/nA\), \(\Tor^{\ZZ}_1(\ZZ/n, A)\cong A[n]\), \(\Tor^{\ZZ}_k(\ZZ, A) = 0\) for \(k\geq 1\), \(\Tor^{\ZZ}_1(A, B)\cong\Tor^{\ZZ}_1(B, A)\), and \(\Tor^{\ZZ}_1(A, B) = 0\) whenever \(A\) or \(B\) is torsion-free [@Hat02]. The groups \(\Tor^{\ZZ}_0(V, W) = V\otimes_{\ZZ}W\) are

    \(V \otimes_{\ZZ} W\) \(W=\ZZ/m\) \(W=\ZZ\) \(W=\QQ\)
    \(V=\ZZ/n\) \(\ZZ/d\) \(\ZZ/n\) \(0\)
    \(V=\ZZ\) \(\ZZ/m\) \(\ZZ\) \(\QQ\)
    \(V=\QQ\) \(0\) \(\QQ\) \(\QQ\)

    and the groups \(\Tor^{\ZZ}_1(V, W)\) are

    \(\Tor^{\ZZ}_1(V, W)\) \(W=\ZZ/m\) \(W=\ZZ\) \(W=\QQ\)
    \(V=\ZZ/n\) \(\ZZ/d\) \(0\) \(0\)
    \(V=\ZZ\) \(0\) \(0\) \(0\)
    \(V=\QQ\) \(0\) \(0\) \(0\)

    All higher groups \(\Tor^{\ZZ}_k\), \(k\geq 2\), vanish.

    What are the groups $\operatorname{Tor}^{\mathbf{Z}}_*$ among $\ZZ/n$, $\ZZ$ and $\QQ$?

    The following functors are zero: \(\ext(\ZZ, \wait)\), \(\tor(\wait, \ZZ)\), \(\tor(\ZZ, \wait)\), \(\tor(\wait, \QQ)\), and \(\tor(\QQ, \wait)\). The following are naturally isomorphic to the identity functor: \(\Hom(\ZZ, \wait)\), \(\wait \tensor_\ZZ \ZZ\), and \(\ZZ \tensor_\ZZ \wait\).

    What links to this