Let \(\mathcal{A}\) be an abelian category. A subcategory \(\mathcal{B}\) of \(\mathcal{A}\) is an abelian subcategory of \(\mathcal{A}\) if \(\mathcal{B}\) is abelian and the inclusion \(\mathcal{B} \subseteq \mathcal{A}\) is an exact functor.
Exercise6.59.
Let \(\mathcal{B}\) be a full subcategory of the abelian category \(\mathcal{A}\text{.}\) Show that: a) \(\mathcal{B}\) is an additive category if and only if \(\mathcal{B}\) contains 0 and is closed under finite coproducts. b) \(\mathcal{B}\) is an abelian subcategory if and only if \(\mathcal{B}\) is additive and closed under kernels and cokernels.
Definition6.60.
Let \(T: \mathcal{A} \longrightarrow \mathcal{B}\) be an additive covariant functor between abelian categories. We say \(T\) is left exact if it takes every exact sequence
\begin{equation*}
0 \longrightarrow A \xrightarrow{f} B \xrightarrow{g} C \longrightarrow 0
\end{equation*}
A contravariant additive functor \(T: \mathcal{A} \longrightarrow \mathcal{B}\) between abelian categories is left exact if it takes every short exact sequence
\begin{equation*}
0 \longrightarrow A \xrightarrow{f} B \xrightarrow{g} C \longrightarrow 0
\end{equation*}
Let \(\mathcal{A}\) be an abelian category, and fix an object \(x\) in \(\mathcal{A}\text{.}\) The functors
\begin{equation*}
\begin{array}{lll}
\mathcal{A} \longrightarrow \boldsymbol{A b} & \text { and } & \mathcal{A} \longrightarrow \boldsymbol{A b} \\
y \longmapsto \operatorname{Hom}_{\mathcal{A}}(x, y) & & y \longmapsto \operatorname{Hom}_{\mathcal{A}}(y, x)
\end{array}
\end{equation*}
are left exact.
Proof.
We will show that \(\operatorname{Hom}_{\mathcal{A}}(x,-)\) is left exact. Notice that the contravariant functor \(\operatorname{hom}_{\mathcal{A}}(-, x)\) can be viewed as the covariant functor \(\operatorname{Hom}_{\mathcal{A}^{\mathrm{op}}}(x,-)\text{.}\) Since \(\mathcal{A}^{\mathrm{op}}\) is also an abelian category, it will then follow that \(\operatorname{Hom}_{\mathcal{A}^{\text {op }}}(x,-)\) is also left exact, or equivalently, that \(\operatorname{Hom}_{\mathcal{A}}(-, x)\) is left exact.
So let
\begin{equation*}
0 \longrightarrow A \xrightarrow{f} B \xrightarrow{g} C \longrightarrow 0
\end{equation*}
be an exact sequence in \(\operatorname{Ch}(\mathcal{A})\text{.}\) We want to show that
\begin{equation*}
0 \longrightarrow \operatorname{Hom}_{\mathcal{A}}(x, A) \stackrel{f_{*}}{\longrightarrow} \operatorname{Hom}_{\mathcal{A}}(x, B) \stackrel{g_{*}}{\longrightarrow} \operatorname{Hom}_{\mathcal{A}}(x, C)
\end{equation*}
is exact, and notice this last complex lives in the category of abelian groups.
We have three things to show:
Exactness at \(A\) is equivalent to \(f\) being a mono. By assumption, \(f\) is a mono, so \(f_{*}(h)=f h\) is injective.
Since \(g f=0\text{,}\) so is \(g_{*} f_{*}=(g f)_{*}\text{.}\)
We want to show that \(\operatorname{ker} g_{*}=\operatorname{im} f_{*}\text{,}\) and these are now maps of abelian groups. So we need to show that every \(h \in \operatorname{Hom}_{\mathcal{A}}(x, C)\) such that \(g h=0\) factors uniquely through \(f\text{,}\) meaning \(h=\operatorname{im} f_{*}\text{.}\) Our assumption that the original sequence is exact implies that \(f=\operatorname{im} f=\operatorname{ker} g\text{.}\) The universal of property of the kernel gives us that whenever \(g h=0, h\) must factor through \(\operatorname{ker} g=f=\operatorname{im} f\text{.}\)
Exercise6.62.
Let \(I\) be any small category. Show that if \(\mathcal{A}\) is an abelian category, then so is the category \(\mathcal{A}^{I}\) of functors \(I \longrightarrow \mathcal{A}\text{.}\)
SubsectionThe Yoneda Lemma: Revisited
We are now ready for the abelian category version of the Yoneda Lemma; this turns out to be a very useful result.
Theorem6.63.Yoneda Embedding for Abelian Categories.
Let \(\mathcal{A}\) be an abelian category. Recall that \(\boldsymbol{A b}^{\mathcal{A}^{o p}}\) denotes the category of contravariant functors \(\mathcal{A} \rightarrow \boldsymbol{A} \boldsymbol{b}\text{.}\) The covariant functor
\begin{equation*}
x \longrightarrow y \longrightarrow z
\end{equation*}
must also be exact.
Proof.
First, our functor is injective on objects because our axioms for a category include the assumption that the Hom-sets are all disjoint. Moreover, by the usual version of the Yoneda Lemma the assignment
is exact. Then \(g_{*} f_{*}=0\text{,}\) so \(g f=g_{*} f_{*}\left(1_{x}\right)=0\text{.}\)
It remains to show that \(\operatorname{ker} g=\operatorname{im} f\text{.}\) Let \(\psi\) be the canonical arrow \(\operatorname{im} f \longrightarrow \operatorname{ker} g\text{.}\) The exactness of
together with the fact that \(g_{*}(\operatorname{ker} g)=0\) imply that \(\operatorname{ker} g\) factors through \(f\text{.}\) By Remark 7.27, ker \(g\) must also factor through \(\operatorname{im} f\text{,}\) say by \(\varphi\text{.}\) The universal property of the kernels ker \(g\) and \(\operatorname{im} f\) will give us that \(\psi\) and \(\varphi\) are inverse isos.
When \(\mathcal{A}=R\)-Mod, the proof can be simplified: the exactness of
\begin{equation*}
\operatorname{Hom}_{R}(R, A) \stackrel{f_{*}}{\longrightarrow} \operatorname{Hom}_{\mathcal{A}}(R, B) \stackrel{g_{*}}{\longrightarrow} \operatorname{Hom}_{\mathcal{A}}(R, C)
\end{equation*}
together with the natural isomorphism between \(\operatorname{Hom}_{R}(R,-)\) and the identity functor give us that
\begin{equation*}
A \stackrel{f}{\longrightarrow} B \stackrel{g}{\longrightarrow} C
\end{equation*}
is exact.
Here is a fun and very useful application of Theorem 7.43.
Corollary6.64.
Let \((L, R)\) be an adjoint pair of additive functors \(\mathcal{A} \underset{R}{\stackrel{L}{\longleftrightarrow}} \mathcal{B}\) between abelian categories. Then \(L\) is right exact, and \(R\) is left exact.
Proof.
Consider a short exact sequence
\begin{equation*}
0 \longrightarrow x \longrightarrow y \longrightarrow z \longrightarrow 0
\end{equation*}
in \(\mathcal{B}\text{,}\) and let \(w\) be an object in \(\mathcal{A}\text{.}\) The adjointness of the pair \((L, R)\) gives us a commutative diagram
where the vertical maps are bijections of sets. For every \(w\) in \(\mathcal{A}, \operatorname{Hom}_{\mathcal{B}}(L w,-)\) is left exact, by Theorem 7.42, so the bottom row of the diagram above is exact. We claim this implies that the top row must also be exact. Our vertical maps are a priori only bijection on sets, but it is easy to see that these natural bijections restrict to a bijection between the images of each pair of corresponding maps. Moreover, for any objects \(A\) and \(B\text{,}\) the natural bijection \(\operatorname{Hom}_{\mathcal{A}}(A, R B) \cong \operatorname{Hom}_{\mathcal{A}}(L A, B)\) must always send 0 to 0, since
commutes. It is then routine to check that our bijections also restrict to bijections between the kernels of each pair of corresponding maps. The exactness of the bottom row then induces exactness of the top row. By Theorem 7.43, Hom reflects exactness, and we conclude that
\begin{equation*}
0 \longrightarrow R x \longrightarrow R y \longrightarrow R z
\end{equation*}
must also be exact. Thus \(R\) is a left exact functor.
Finally, by Remark 7.25, \(\mathcal{A}^{\mathrm{op}}\) and \(\mathcal{B}^{\mathrm{op}}\) are both abelian categories. Consider the opposite functors \(L^{\mathrm{op}}\) and \(R^{\mathrm{op}}\text{.}\) Notice that \(L^{\mathrm{op}}\) is the right adjoint to \(R^{\mathrm{op}}\text{,}\) so \(L^{\mathrm{op}}\) must be left exact. Therefore, \(L\) must be right exact.
This is possibly the first time we have encountered a proof that truly used duality in an essential and interesting way. In the case where \(\mathcal{A}=\mathcal{B}=R\)-Mod, the fact that \(R\) is left exact can be obtained using only methods from \(R\)-Mod; but the statement about \(L\) used the fact that \(\mathcal{A}^{\mathrm{op}}\) is an abelian category, while the opposite category of \(R\)-Mod is not another category of modules.
The Yoneda embedding from Theorem 7.43 is the first piece of the proof of a very important result.
Theorem6.65.Freyd-Mitchell embedding theorem.
Let \(\mathcal{A}\) be a small abelian category. There exists a ring \(R\text{,}\) possibly not commutative, and an exact, fully faithful embedding \(\mathcal{A} \longrightarrow R\)-Mod.
The full details of the proof are rather complicated, and can be found in [Fre03]. Here is a very rough map of the proof. By Theorem 7.43, we already have a fully faithful embedding of \(\mathcal{A}\) in \(\mathbf{A} \mathbf{b}^{\mathcal{A}^{\text {op }}}\text{,}\) so it is sufficient to show that there is a fully faithful embedding of \(\mathbf{A} \mathbf{b}^{\mathcal{A}^{\text {op }}}\) into some \(R\)-Mod. The idea is to quotient \(\mathbf{A b}^{\mathcal{A}^{\text {op }}}\) by an abelian subcategory \(L\) that contains all the kernels and cokernels of the arrows \(\operatorname{Hom}_{\mathcal{A}}(-, y) \rightarrow \operatorname{Hom}_{\mathcal{A}}(-, z)\) for all epis \(y \longrightarrow z\text{,}\) in such a way that the composite of the embedding in Theorem 7.43 with this quotient remains an embedding. Then one shows that this quotient category has all coproducts and also what is called a projective generator. Roughly speaking, this is a projective object \(P\) such that for every object \(M\) there exists an arrow \(P \rightarrow M\text{.}\) Then one shows that this implies that this category is equivalent to a full abelian subcategory of \(R\)-Mod for some \(R\text{.}\)
SubsectionThe Diagram Chasers: Revisited
Most of the theorems we have proved about \(R\)-Mod extend to any abelian category. Some of those theorems can in fact be deduced from the fact that they are true over \(R\)-Mod.
Theorem6.66.Snake Lemma.
Consider an abelian category \(\mathcal{A}\) and a commutative diagram
If the rows of the diagram are exact, then there exists an exact sequence
\begin{equation*}
\text { ker } f \longrightarrow \operatorname{ker} g \longrightarrow \operatorname{ker} h \stackrel{\partial}{\longrightarrow} \operatorname{coker} f \longrightarrow \operatorname{coker} g \longrightarrow \text { coker } h \text {. }
\end{equation*}
Theorem6.67.Long exact sequence in homology.
Given a short exact sequence in \(\operatorname{Ch}(R)\)
\begin{equation*}
0 \longrightarrow A \xrightarrow{f} B \xrightarrow{g} C \longrightarrow 0
\end{equation*}
there are connecting arrows \(\partial: \mathrm{H}_{n}(C) \longrightarrow \mathrm{H}_{n-1}(A)\) such that
Given an abelian category \(\mathcal{A}\text{,}\) consider the following commutative diagram in \(\mathcal{A}\) with exact rows:
If \(b\) and \(d\) are epi and \(e\) is a mono, then \(c\) is an epi. If \(b\) and \(d\) are mono and a is epi, then \(c\) is mono.
One can prove these by invoking the Freyd-Mitchell theorem and checking that one can go back and forth with our statements between some small subcategory of \(\mathcal{A}\) containing our diagram and all the necessary kernels, cokernels, etc, and some \(R\)-Mod where that category embeds. Alternatively, one can use what are called members, as in [ML98, VIII.4.5]. The theory of members is an attempt to fix the main difficulty when dealing with abelian categories: that the objects and arrows are not just sets and functions, so we can’t just talk about members of the objects and their images by each arrow.
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