This post was translated from Korean by LLM (Kimi). The translation may contain errors or awkward sentences. The Korean original is the source of truth.
The category \(\Ch(\mathcal{A})\)
In [Category Theory] §Abelian Categories we mentioned that the category of chain complexes, \(\Ch(\mathcal{A})\), exists and is an abelian category. In this section we examine this claim more concretely. First, morphisms in \(\Ch(\mathcal{A})\) are given as follows.
Definition 1 For two chain complexes \(C_\bullet, D_\bullet\), a chain map \(f_\bullet:C_\bullet\rightarrow D_\bullet\) between them is a collection of maps \(f_n:C_n\rightarrow D_n\) making the following diagram
commute.
That is, saying \(f_\bullet\) is a chain map means that for all \(n\),
\[f_{n-1}\circ d_n^C=d_n^D\circ f_n\]holds. When there is no risk of confusion, we write both \(d^C\) and \(d^D\) simply as \(d\). Then defining the category \(\Ch(\mathcal{A})\) by
- the collection of chain complexes in \(\mathcal{A}\), \(\obj(\Ch(\mathcal{A}))\),
- the collection of chain maps from \(C_\bullet\) to \(D_\bullet\), \(\Mor_{\Ch(\mathcal{A})}(C_\bullet,D_\bullet)\)
it is obvious that \(\Ch(\mathcal{A})\) becomes a category. Here \(\id_{C_\bullet}\) is the chain map \(\id_\bullet:C_\bullet\rightarrow C_\bullet\) defined by \(\id_n:C_n\rightarrow C_n\) being \(\id_{C_n}\) for all \(n\), and composition of chain maps is given by
\[g_\bullet\circ f_\bullet=(g_n\circ f_n)_{n\in\mathbb{Z}}.\]Moreover, the category \(\Ch(\mathcal{A})\) is an \(\Ab\)-category. Given two chain maps \(f_\bullet,g_\bullet:C_\bullet\rightarrow D_\bullet\), defining their sum by
\[f_\bullet+g_\bullet=(f_n+g_n)_{n\in\mathbb{Z}}\]makes \(\Mor_{\Ch(\mathcal{A})}(C_\bullet,D_\bullet)\) an abelian group, and given two chain complexes \(C_\bullet,D_\bullet\), their product \(C_\bullet\oplus D_\bullet\) is1 the following chain complex
\[C_\bullet\oplus D_\bullet:\quad \cdots \longrightarrow C_{n+1}\oplus D_{n+1}\overset{d_{n+1}^C\oplus d_{n+1}^D}{\longrightarrow}C_n\oplus D_n\overset{d_{n}^C\oplus d_{n}^D}{\longrightarrow}C_{n-1}\oplus D_{n-1}\longrightarrow\cdots\]The zero object is the following chain complex
\[\cdots\rightarrow 0\rightarrow 0\rightarrow 0\rightarrow\cdots\]To show that \(\Ch(\mathcal{A})\) is an abelian category, it suffices to define the kernel and cokernel of an arbitrary chain map \(f_\bullet:C_\bullet \rightarrow D_\bullet\). First, for each \(n\), since \(f_n:C_n \rightarrow D_n\) is a morphism in the abelian category \(\mathcal{A}\), we have \(i_n:\ker f_n\rightarrow C_n\) and \(p_n:D_n\rightarrow \coker f_n\). Consider the following diagram
Then
\[f_{n-1}\circ d_n\circ i_n=d_n\circ f_n\circ i_n=0\]so from the universal property of \(\ker f_{n-1}\), a morphism \(\ker f_n\rightarrow \ker f_{n-1}\) is naturally induced. Similarly \(\coker f_n\rightarrow \coker f_{n-1}\) is also defined, and these data respectively form two chain complexes
\[\ker(f_\bullet):\qquad \cdots\longrightarrow \ker f_{n+1}\longrightarrow \ker f_n\longrightarrow \ker f_{n-1}\longrightarrow\cdots\]and
\[\coker(f_\bullet):\qquad \cdots\longrightarrow \coker f_{n+1}\longrightarrow \coker f_n\longrightarrow \coker f_{n-1}\longrightarrow\cdots\]To show that \(\ker(f_\bullet)\) defined this way is actually the kernel of \(f_\bullet\) in \(\Ch(\mathcal{A})\), we need to verify the following universal property
To do this, we first construct \(j_n:A_n\rightarrow \ker f_n\) through the following diagram for each \(n\)
and then show that \(j_\bullet\) is a chain map. That is, it suffices to prove that the upper square in the following diagram
commutes. Since kernels are always monomorphisms, for this it is enough to show that
\[i_{n-1}\circ(j_{n-1}\circ d_n^A)=i_{n-1}\circ(d_n\circ j_n)\]holds, which is obvious by definition. Thus \(\ker(f_\bullet)\) is indeed the kernel of \(f_\bullet\). Similarly one can verify that \(\coker(f_\bullet)\) is the cokernel of \(f_\bullet\) in \(\Ch(\mathcal{A})\).
Finally, we must show that an arbitrary monomorphism \(f_\bullet\) equals \(\ker(\coker f_\bullet)\), and an arbitrary epimorphism \(f_\bullet\) equals \(\coker(\ker f_\bullet)\). This is done by first proving that \(f_\bullet\) being a monomorphism (resp. epimorphism) is equivalent to \(f_n\) being a monomorphism (resp. epimorphism) for all \(n\), then using the abelian category condition degree by degree and combining the results. This process is almost identical to the above, so we omit the detailed argument.
Homology
Fix a chain complex \(C_\bullet\) and define a new chain complex \(C_{\bullet-1}\) by
\[C_{\bullet-1}:\qquad\cdots\longrightarrow\underbrace{C_n}_\text{\small degree $n+1$}\overset{d_n}{\longrightarrow}\underbrace{C_{n-1}}_\text{\small degree $n$}\overset{d_{n-1}}{\longrightarrow}\underbrace{C_{n-2}}_\text{\small degree $n-1$}\longrightarrow\cdots\tag{1}\]That is, \(C_{\bullet-1}\) is the chain complex obtained by shifting \(C_\bullet\) one place. Now define the chain map \(d_\bullet:C_\bullet\rightarrow C_{\bullet-1}\) by the following diagram
Then \(\ker (d_\bullet)\) is well-defined, and considering the following diagram
from \(d_{n+1}\circ i_{n+1}=0\) and the fact that \(i_n\) is a monomorphism, we see that the differentials of \(\ker(d_\bullet)\) are all zero maps.
Similarly \(\coker(d_\bullet)\) is also well-defined, and the differentials between them are also zero. Unlike the \(n\)th component of \(\ker(d_\bullet)\) being \(\ker d_n\rightarrow C_n\), the \(n\)th component of \(\coker(d_\bullet)\) is \(C_{n-1}\rightarrow \coker(d_n)\), so there may be some confusion; thus we again shift degrees by one and consider the chain complex \(\coker(d_{\bullet+1})\).
Now thinking of
\[\im(d_{\bullet+1})=\ker(C_\bullet\rightarrow \coker(d_{\bullet+1}))\]and using the universal property of \(\ker(d_\bullet)\rightarrow C_\bullet\), we obtain a monomorphism
\[\im(d_{\bullet+1})\rightarrow\ker(d_\bullet)\]and through this a new chain complex
\[H_\bullet(C)=\ker(d_\bullet)/\im(d_{\bullet+1})\]Explicitly, \(H_n(C)\) is induced by the following diagram
However, \(\ker(d_n)\rightarrow H_n\) is an epimorphism, and \(\ker(d_n)\rightarrow\ker(d_{n-1})\) is \(0\), so \(H_n(C)\rightarrow H_{n-1}(C)\) is also a zero map. This shows that (just like \(\ker(d_\bullet)\) or \(\coker(d_\bullet)\)) \(H_\bullet(C)\) is not very meaningful as a chain complex; instead, it is the individual components \(H_n(C)\in\obj(\mathcal{A})\) of \(H_\bullet(C)\) that are interesting objects.
Definition 2 For an arbitrary chain complex
\[C_\bullet:\quad\cdots\overset{d_{n+2}}{\longrightarrow} C_{n+1}\overset{d_{n+1}}{\longrightarrow} C_n\overset{d_n}{\longrightarrow} C_{n-1}\overset{d_{n-1}}{\longrightarrow}\cdots\]the module of \(n\)-cycles of \(C_\bullet\) is given by \(Z_n(C)=\ker (d_n)\). Also, we call \(\im(d_{n+1})\) the module of \(n\)-boundaries of \(C_\bullet\), and write it as \(B_n(C)\). Then the \(n\)th homology of \(C_\bullet\) is defined by \(H_n(C)=Z_n(C)/B_n(C)\).
Similarly, for a cochain complex one defines \(n\)-cocycles, \(n\)-coboundaries, and \(n\)th cohomology.
\(H_\bullet\) itself is a functor from \(\Ch(\mathcal{A})\) to \(\Ch(\mathcal{A})\), but each component \(H_n\) defines a functor from \(\Ch(\mathcal{A})\) to \(\mathcal{A}\).
Proposition 3 For any \(n\), \(H_n\) is a functor from \(\Ch(\mathcal{A})\) to \(\mathcal{A}\).
Proof
For an arbitrary chain map \(f_\bullet:C_\bullet\rightarrow D_\bullet\), applying §Diagram Chasing, ⁋Lemma 4 shows that \(f_n\) induces \(Z_n(C)\rightarrow Z_n(D)\) and \(B_n(C)\rightarrow B_n(D)\). Therefore \(f_n\) induces
\[H_n(f):H_n(C)\rightarrow H_n(D)\]Checking that the data obtained this way form a functor is all straightforward.
Double complexes and translation
Finally we define double complexes. Intuitively, this can be thought of as an object of \(\Ch(\Ch(\mathcal{A}))\), i.e., a chain complex made of abelian categories \(\Ch(\mathcal{A})\). The chain complex \(C_{\bullet-1}\) of (1) above and the chain map below it are examples of this.
However, when doing calculations later, it is sometimes cleaner to have the signs of each row alternate.2 Adopting this sign convention, we define as follows.
Definition 4 A double complex consisting of objects in an abelian category \(\mathcal{A}\) is a collection of objects with two maps
\[d^h:C_{p,q}\rightarrow C_{p-1,q},\qquad d^v:C_{p,q}\rightarrow C_{p,q-1}\]satisfying
\[(d^h)^2=(d^v)^2=d^vd^h+d^hd^v=0.\]
The condition \(d^vd^h+d^hd^v=0\) means \(d^vd^h=-d^hd^v\), so double complexes are not objects of \(\Ch(\Ch(\mathcal{A}))\); however, defining from \(d^v\) the formula
\[f_{p,q}=(-1)^pd_{p,q}^v:C_{p,q}\rightarrow C_{p,q-1}\]allows us to treat them as objects of \(\Ch(\Ch(\mathcal{A}))\).
Definition 5 The total complex \(\Tot(C)_\bullet\) of a double complex \(C_{p,q}\) is defined in each degree \(n\) by
\[(\Tot(C))_n = \bigoplus_{p+q=n} C_{p,q}\]and its differential is given by
\[d = d^h + (-1)^p d^v : (\Tot(C))_n \rightarrow (\Tot(C))_{n-1}\](Equivalently one may write \(d^v + (-1)^q d^h\).)
For convenience of discussion we have written the differential somewhat concisely, but unpacking this reveals that actual computation is somewhat more complicated than the above formula. For example, suppose a double complex \(C_{p,q}\) is given, and for convenience assume this complex is nonzero only where \(p,q\geq 0\). Then in the total complex \(\Tot(C)_\bullet\) of this double complex, to compute the differential \(d:\Tot(C)_2\rightarrow \Tot(C)_1\) one must proceed as follows. First, an element of \(\Tot(C)_2\) has the form \((x_{0,2}, x_{1,1}, x_{2,0})\),3 and the differential of this element becomes the following element containing all the information
\[(d^v(x_{0,2})+d^h(x_{1,1}), -d^v(x_{1,1})+d^h(x_{2,0}))\in\Tot(C)_1=C_{0,1}\oplus C_{1,0}\]from
\[d^h(x_{0,2})+(-1)^0d^v(x_{0,2})=d^v(x_{0,2}), \quad d^h(x_{1,1})+(-1)^1d^v(x_{1,1}),\quad d^h(x_{2,0})+(-1)^2d^v(x_{2,0})=d^h(x_{2,0})\]That is, \(\Tot(C)\) packages the double complex \(C_{p,q}\) into a single complex, and in doing so simultaneously considers all components with total degree \(p+q=n\).
To show that this is a complex, we must verify \(d^2=0\), which follows from the formula
\[(d^h + (-1)^p d^v)^2 = (d^h)^2 + (-1)^p d^vd^h + (-1)^{p-1} d^hd^v + (d^v)^2\]Here \((d^h)^2=0\), \((d^v)^2=0\), and from the double complex condition \(d^vd^h + d^hd^v=0\) we have \(d^vd^h = -d^hd^v\), so the remaining two terms also cancel.
Meanwhile, since we used the above sign convention in defining double complexes, the \(p\)-th translation \(C[p]\) of \(C\) is also the chain complex given by
\[C[p]_n=C_{n+p}\]and its differential must be defined as \((-1)^pd\). For instance \(C[-1]\) means the same chain complex as \(C_{\bullet-1}\) defined above, but with the signs of all differentials flipped.
In any case, regardless of these newly created signs,
\[H_n(C[p])=H_{n+p}(C)\]is obvious, and moreover if \(f:C\rightarrow D\) is a chain map then defining \(f[p]\) by
\[f[p]_n=f_{n+p}\]we can make \(f[p]:C[p]\rightarrow D[p]\). That is, translation is a functor from \(\Ch(\mathcal{C})\) to itself.
Truncation
Finally, let an arbitrary chain complex \(C_\bullet\) be given. For any integer \(n\), the chain complex \((\tau_{\geq n}C)_\bullet\) defined by the formula
\[(\tau_{\geq n}C)_i=\begin{cases}0&\text{if $i < n$}\\ Z_n&\text{if $i=n$}\\ C_i&\text{if $i > n$}\end{cases}\]with the same differential as \(C_\bullet\) is called the intelligent truncation. Then
\[H_i(\tau_{\geq n}C)=\begin{cases}0&\text{if $i < n$}\\ H_i(C)&\text{if $i\geq n$}\end{cases}\]Similarly \((\tau_{< n}C)_\bullet\) can also be defined. On the other hand, the brutal truncation is defined by the formula
\[(\sigma_{\geq n}C)_i=\begin{cases}0&\text{if $i \leq n$}\\ C_i&\text{otherwise}\end{cases}\]The definition itself may seem more natural for \((\sigma_{\geq n}C)_\bullet\), but looking at the \(n\)th homology one can verify that the intelligent truncation gives a better operation.
References
[Wei] C.A. Weibel. An Introduction to Homological Algebra. Cambridge Studies in Advanced Mathematics. Cambridge University Press, 1995.
[Hu] S.T. Hu, Introduction to homological algebra. University Microfilms, 1979.
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Here \(\oplus\) is understood as the coproduct in \(\mathcal{A}\). Since one can prove that finite products and coproducts coincide in an abelian category, in fact \(A\oplus B\cong A\times B\). ↩
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For example, for the Dolbeault operator \(d=\partial+\bar{\partial}\), the condition \(d^2=0\) becomes \(0=\partial^2+(\partial\bar{\partial}+\bar{\partial}\partial)+\bar{\partial}^2\), and considering bidegree we know that \(\partial\bar{\partial}+\bar{\partial}\partial=0\). ↩
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Since we assumed for convenience that \(C_{p,q}\neq 0\) only where \(p,q\geq 0\), terms such as \(x_{-1,3}\) can be ignored. ↩
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