Let \(\mathcal{A}\) be an abelian category. A chain complex or simply complex \((C, \partial)\text{,}\) which we sometimes write just a \(C\text{,}\) is a sequence of objects and arrows
such that \(\partial_{n-1} \partial_{n}=0\) for all \(n\text{.}\) A map of complexes \(f: C \longrightarrow D\) between two chain complexes is a sequence of arrows \(f_{n}\) such that the diagram
commutes. The category of (chain) complexes over \(\mathcal{A}\text{,}\) denoted \(\operatorname{Ch}(\mathcal{A})\text{,}\) is the category that has objects all chain complexes in \(\mathcal{A}\) and arrows all the chain complex maps.
Lemma6.42.
If \(\mathcal{A}\) is an abelian category, then \(\operatorname{Ch}(\mathcal{A})\) is also an abelian category.
Proof.
Proof sketch. First, note that \(\operatorname{Ch}(\mathcal{A})\) is a preadditive category: given two maps of complexes \(f\) and \(g, f+g\) is obtained degreewise, by taking
The facts that \(\operatorname{Hom}_{\operatorname{Ch}(\mathcal{A})}(C, D)\) is an abelian group and that composition is bilinear follow from the analogous facts in \(\mathcal{A}\) (exercise). The zero object is the zero complex, which has the zero object in \(\mathcal{A}\) in each degree. Given two complexes \(C\) and \(D\text{,}\) their product is taken degreewise:
and each of the projection maps in each degree assemble to make a map of complexes. So \(\operatorname{Ch}(\mathcal{A})\) is an additive category.
Let \(f: C \rightarrow D\) be a map of complexes. We need to show that both the both the kernel and cokernel of \(f\) exist. The universal property of \(\operatorname{ker} \partial_{n}\) gives us a unique arrow \(\delta_{n+1}\) such that
commutes. The commutativity of \(\operatorname{ker} f_{n+1} \longrightarrow C_{n+1}\) together with \(\partial_{n} \partial_{n+1}=0\) and the
fact that \(\operatorname{ker} f_{n-1}\) is a mono imply that \(\delta_{n} \delta_{n+1}=0\text{.}\) Finally, we conclude that
is a complex in \(\operatorname{Ch}(\mathcal{A})\text{,}\) and the canonical maps \(\operatorname{ker} f_{n} \rightarrow C_{n}\) assemble into a map of complexes. One can check that the universal property of the kernels \(\operatorname{ker} f_{n}\) forces this complex we just constructed to be \(\operatorname{ker} f\text{.}\) In particular, \(\operatorname{Ch}(\mathcal{A})\) has all kernels. Similarly, we construct cokernels in \(\operatorname{Ch}(\mathcal{A})\text{,}\) building on the fact that \(\mathcal{A}\) has all cokernels.
Finally, it remains to show that every mono is the kernel of its cokernel and every epi is the cokernel of its kernel. This boils down to the fact that \(f\) is a mono if and only if all the \(f_{n}\) are monos, and dually that \(f\) is an epi if and only if all the \(f_{n}\) are epis. The conclusion will then follow from our construction of kernels and cokernels and the fact that \(\mathcal{A}\) is abelian. Our claim follows from Exercise 83 and the fact that \(f=0\) if and only if all \(f_{n}=0\text{.}\)
Definition6.43.
Let \(\mathcal{A}\) be an abelian category. For each \(C\) in \(\operatorname{Ch}(\mathcal{A})\text{,}\) we define its cycles \(Z_{n}(C)\) and boundaries \(B_{n}(C)\) by
\begin{equation*}
Z_{n}(C):=\text { source ker } \partial_{n} \quad \text { and } \quad B_{n}(C):=\text { source im } \partial_{n+1}
\end{equation*}
Remark6.44.
Let \(\mathcal{A}\) be an abelian category, and \(C \stackrel{f}{\rightarrow} D \stackrel{g}{\rightarrow} E\) be arrows in \(\mathcal{A}\) such that \(g f=0\text{.}\) By Remark 7.22, we can factor \(f\) as an epi followed by \(\operatorname{im} f\text{.}\)
Since \(g \circ \operatorname{im} f \circ \operatorname{epi}=g f=0\text{,}\) we must have \(g \circ \operatorname{im} f=0\text{,}\) so \(\operatorname{im} f\) factors uniquely through ker \(g\text{.}\) Most importantly, there is a canonical arrow \(\operatorname{im} f \longrightarrow \operatorname{ker} g\text{.}\)
Definition6.45.
Definition 7.32. Let \(\mathcal{A}\) be an abelian category. A sequence of arrows \(C \stackrel{f}{\rightarrow} D \stackrel{g}{\rightarrow} E\) in \(\mathcal{A}\) is exact if \(g f=0\) and \(\operatorname{ker} g=\operatorname{im} f\text{.}\)
Remark6.46.
Remark 7.33. In our definition of exact sequence, we really mean that the canonical arrow \(\operatorname{im} f \longrightarrow \operatorname{ker} g\) we described in Remark 7.31 is an isomorphism. But notice that is equivalent to saying that the arrow \(\operatorname{im} f\) is a kernel for \(g\text{,}\) and \(\operatorname{ker} g\) is an image for \(f\text{,}\) hence the equality we wrote above, which is a more compact way of saying this.
This immediately generalizes to define an exact sequence, and once again a short exact sequence is one of the form
\begin{equation*}
0 \longrightarrow A \longrightarrow B \longrightarrow C \longrightarrow 0
\end{equation*}
Exercise6.47.
Exercise 86. Show that \(0 \longrightarrow A \stackrel{f}{\rightarrow} B\) is exact if and only if \(f\) is a mono, and \(B \stackrel{g}{\rightarrow} C \longrightarrow 0\) is exact if and only if \(g\) is an epi. Moreover,
\begin{equation*}
0 \longrightarrow A \xrightarrow{f} B \stackrel{g}{\longrightarrow} C \longrightarrow 0
\end{equation*}
is a short exact sequence if and only if
\(f\) is a mono. \(\quad \bullet g\) is an epi. \(f=\operatorname{im} f=\operatorname{ker} g\text{.}\)\(\bullet \operatorname{coker} f=g\text{.}\)
Remark6.48.
Remark 7.34. Let \(\mathcal{A}\) be an abelian category and \((C, \partial)\) be a complex in \(\operatorname{Ch}(R)\text{.}\) Since \(\partial_{n} \partial_{n+1}=0\) for all \(n\text{,}\) we get a canonical arrow \(B_{n}(C) \longrightarrow Z_{n}(C)\) for each \(n\text{.}\)
Exercise6.49.
Exercise 87. Given an additive category \(\mathcal{A}, B_{n}\) and \(Z_{n}\) are additive functors \(\operatorname{Ch}(\mathcal{A}) \longrightarrow \mathcal{A}\text{.}\) In particular, an arrow \(C \stackrel{f}{\rightarrow} D\) induces arrows \(Z_{n}(C) \stackrel{Z_{n}(f)}{\longrightarrow} Z_{n}(D)\) and \(B_{n}(c) \stackrel{B_{n}(f)}{\longrightarrow} B_{n}(D)\text{.}\)
SubsectionHomology in Abelian Categories
Definition6.50.
Let \(\mathcal{A}\) be an abelian category and \((C, \partial)\) a complex in \(\operatorname{Ch}(R)\text{.}\) The \(n\)th homology of \(C\) is the object
where \(B_{n}(C) \longrightarrow Z_{n}(C)\) is the canonical arrow we described in Remark 7.31.
In fact, the \(n\)th homology is an additive functor \(\mathrm{H}_{n}: \mathrm{Ch}(\mathcal{A}) \longrightarrow \mathcal{A}\text{.}\) But to see that, we first need to make sense of what homology does to maps of complexes.
Let \(\mathcal{A}\) be an abelian category and \(C \stackrel{f}{\rightarrow} D\) a map of complexes in \(\operatorname{Ch}(\mathcal{A})\text{.}\) Fix an integer \(n\text{.}\) We get induced arrows \(B_{n}(f)\) and \(Z_{n}(f)\text{,}\) since \(B_{n}\) and \(Z_{n}\) are additive functors. This gives us a commutative diagram
where \(\alpha\) and \(\beta\) are the canonical arrows. To construct \(\mathrm{H}_{n}(f)\text{,}\) we claim that there is a unique arrow coker \(\alpha \longrightarrow\) coker \(\beta\) making the diagram commute. This is all explained in the commutative diagram
where coker \(\beta \circ\left(Z_{n}(f)\right) \circ \alpha=\operatorname{coker} \beta \circ \beta \circ B_{n}(f)=0\text{,}\) which gives us a unique factorization \(\mathrm{H}_{n}(f)\) through coker \(\alpha\text{.}\)
Exercise6.51.
Given any abelian category \(\mathcal{A}, \mathrm{H}_{n}\) is an additive functor \(\operatorname{Ch}(\mathcal{A}) \longrightarrow \mathcal{A}\text{.}\)
Similarly, we can define homotopies.
Definition6.52.
Let \(\mathcal{A}\) be an abelian category and \(f, g: F \longrightarrow G\) be maps complexes in \(\mathrm{Ch}(\mathcal{A})\text{.}\) A homotopy, sometimes referred to as a chain homotopy, between \(f\) and \(g\) is a sequence of arrows \(h_{n}: F_{n} \longrightarrow G_{n+1}\)
for all \(n\text{.}\) If there exists a homotopy between \(f\) and \(g\text{,}\) we say that \(f\) and \(g\) are homotopic, and write \(f \simeq g\text{.}\) If \(f\) is homotopic to the zero map, we say \(f\) is null-homotopic. If \(f: F \rightarrow G\) and \(g: G \rightarrow F\) are maps of complexes such that \(f g\) is homotopic to the identity arrow \(1_{G}\) and \(g f\) is homotopic to the identity arrow \(1_{F}\text{,}\) we say that \(f\) and \(g\) are homotopy equivalences and \(F\) and \(G\) are homotopy equivalent.
Exercise6.53.
Homotopy is an equivalence relation in \(\operatorname{Ch}(\mathcal{A})\text{.}\)
Exercise6.54.
Let \(\mathcal{A}\) be an abelian category. Homotopic maps of complexes in \(\operatorname{Ch}(\mathcal{A})\) induce the same map on homology.
Remark6.55.
Let \(F\) be an additive functor between abelian categories. Then \(F\) must send complexes to complexes, and it induces a functor \(\operatorname{Ch}(\mathcal{A}) \rightarrow \operatorname{Ch}(\mathcal{A})\text{,}\) which we also call \(F\text{.}\) Now if \(h\) is a homotopy between two maps of complexes, \(F\) preserves the identities \(\delta_{n+1} h_{n}+h_{n-1} \delta_{n}=f_{n}-g_{n}\) for all \(n\text{,}\) so \(F(h)\) is a homotopy between \(F(f)\) and \(F(g)\text{.}\)
Definition6.56.
Let \(\mathcal{A}\) be an abelian category. A map of complexes \(f: A \rightarrow B\) in \(\operatorname{Ch}(\mathcal{A})\) is a quasi-isomorphism if \(\mathrm{H}_{n}(f)\) is an isomorphism for all \(n\text{.}\)
Finally, we set up some notation we will use later.
Definition6.57.
We will denote the full subcategory of \(\operatorname{Ch}(\mathcal{A})\) of complexes \(C\) such that \(C_{n}=0\) for all \(n<k\) by \(\mathrm{Ch}_{\geqslant k}(\mathcal{A})\text{.}\)
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