In this short post we state the basic decomposition result that any morphism in an abelian category decomposes canonically into an epimorphism followed by a monomorphism, and derive some very useful consequences.

1. The decomposition

Fact 1.

In an abelian category, the canonical map $ \ker(\mathop{\rm coker}\nolimits f)\to\mathop{\rm coker}\nolimits(\ker f)$ is an isomorphism, and thus every morphism $ f:A\to B$ splits canonically as \begin{align*}

\begin{xy} \xymatrix{ A\ar[r]^e & \ker(\mathop{\rm coker}\nolimits f) \ar[r]^m & B \\ } \end{xy}
\end{align*} where $ e$ is an epimorphism and $ m$ is a monomorphism (associated with $ \ker(\mathop{\rm coker}\nolimits f)$ and the isomorphic $ \mathop{\rm coker}\nolimits(\ker f)$); moreover, $ \ker(\mathop{\rm coker}\nolimits f)$ is naturally isomorphic to the image $ \mathop{\rm im}\nolimits f$.

Note that we have canonical maps $ A\to\mathop{\rm coker}\nolimits(\ker f)$ and $ \ker(\mathop{\rm coker}\nolimits f)\to B$ which are respectively an epimorphism and a monomorphism (being a coequalizer and an equalizer), so the crucial bit is indeed the isomorphism $ \ker (\mathop{\rm coker}\nolimits f)\to \mathop{\rm coker}\nolimits(\ker f)$.

This lemma is extremely fundamental, and we might prove it some jolly day, but for now we'll be content to just give some useful corollaries illustrating the strength of this lemma.

2. Consequences

Corollary 1.

In an abelian category,

  1. if the canonical map $ e:A\to\ker(\mathop{\rm coker}\nolimits f)$ is a monomorphism, it is an isomorphism;
  2. dually, if the canonical map $ \ker(\mathop{\rm coker}\nolimits f)\to B$ is an epimorphism, it is an isomorphism.

Proof 1.

We have the diagram

\begin{align*}

\begin{xy} \xymatrix{ \ker f\ar@<2pt>[r]^{k}\ar@<-2pt>[r]_0& A\ar@<2pt>[r]^f\ar[d]^{c'} \ar@<-2pt>[r]_0\ar[dr]_g & B\ar@<2pt>[r]^c\ar@<-2pt>[r]_0 & \mathop{\rm coker}f \\ & \mathop{\rm coker}(\ker f)\ar[r]_h^{} &\ker(\mathop{\rm coker}f)\ar[u]_{k'} & } \end{xy}
\end{align*}

where $ g$ comes from $ cf=0$ and $ h$ comes from $ gk = 0$. Then $ gk=g0$ implies $ k=0$, and hence $ \mathop{\rm coker}\nolimits(\ker f)$ is the cokernel of two zero maps, which is isomorphic to $ A$; hence, $ c'$ is an isomorphism, and thus $ g=hc'$ is an isomorphism as well. The dual assertion then follows by, well, duality.

Corollary 2.

A morphism $ f$ in an abelian category that is both a monomorphism and an epimorphism is an isomorphism.

Proof 2.

Let $ f:A\to B$ decompose as $ A \xrightarrow{ \ e \ } \ker(\mathop{\rm coker}\nolimits f) \xrightarrow{ \ m \ } B$ for $ m$ a monomorphism and $ e$ an epimorphism. Then observe that $ m$ is also an epimorphism, for \begin{align*} am = bm \implies ame=bme \implies af=bf\implies a =b \end{align*} as $ f$ is an epimorphism, and similarly $ e$ is also a monomorphism, for \begin{align*} ea =eb \implies mea=meb \implies fa=fb\implies a=b \end{align*} as $ f$ is a monomorphism. It follows from the previous lemma that $ e$ and $ m$ are both isomorphisms, hence so is $ f$.

Corollary 3.

Let $ f:A\to B$ in an Abelian category decompose as $ A \xrightarrow{ \ e \ } \ker(\mathop{\rm coker}\nolimits f) \xrightarrow{ \ m \ } B$ for $ m$ a monomorphism and $ e$ an epimorphism.

  1. if $ f$ is an epimorphism, $ m$ is an isomorphism;
  2. dually, if $ f$ is a monomorphism, $ e$ is an isomorphism.

Proof 3.

We basically did that already; as we saw in the previous proof, if $ f$ is an epimorphism, $ m$ is an epimorphism; but it's also a monomorphism, hence an isomorphism by the previous lemma. Analogously for the dual.