Multiplicity in a composition series and a certain cokernel

Question

Let $\mathsf{C}$ be an abelian category in which every object has finite length. Let $S$ be a simple object in $\mathsf{C}$. For any object $X \in \mathsf{C}$ denote by $[X\colon S]$ the number of times $S$ is a composition factor of $X$. This number is well-defined by the Jordan–Hölder theorem.

Let $f,g \colon M \rightarrow N$ be morphisms in $\mathsf{C}$. Denote by $\pi \colon N \rightarrow N/\operatorname{im}(f)$ the cokernel of the map $f$. Assume $[\operatorname{im}(f):S]=0$. I am trying to show that then the following two equalities hold: \begin{gather*} [\operatorname{im}(f+g):S]= [\operatorname{im}(\pi \circ (f+g)):S] \,, \\[1em] [\operatorname{im}(g):S]= [\operatorname{im}(\pi \circ g):S] \,. \end{gather*} Why are they true? I am already stuck with the question how the image (as defined in an abelian category) behaves under composition.

Answer 1

We have the short exact sequence $$ \newcommand{\im}{\operatorname{im}} 0 \longrightarrow \im(f) \longrightarrow N \xrightarrow{\enspace π \enspace} \mathrm{coker}(f) \longrightarrow 0 \,. $$ Let us assume for a moment that $\mathsf{C}$ is an exact, full subcategory of $\mathrm{Mod}_R$ for some ring $R$.

• For every subobject $N'$ of $N$ we get the induced short exact sequence $$ 0 \longrightarrow \im(f) ∩ N' \longrightarrow N' \longrightarrow π(N') \longrightarrow 0 \,. $$ I will use two properties of multiplicities from my last post.

  1. Since $\im(f) ∩ N'$ is a subobject of $\im(f)$ we have $[\im(f) ∩ N' : S] ≤ [\im(f) : S] = 0$ and thus $[\im(f) ∩ N' : S] = 0$.
  2. By the additivity of multiplicities for short exact sequences, we have $$ [N' : S] = [\im(f) ∩ N' : S] + [π(N') : S] = 0 + [π(N') : S] = [π(N') : S] \,. $$

• Suppose now that $N' = \im(h)$ for some morphism $h \colon P \to N$. Then $$ π(N') = π(\im(h)) = \im(π ∘ h) \,, $$ and so the equation $[N' : S] = [π(N') : S]$ becomes $$ [\im(h) : S] = [\im(π ∘ h) : S] \,. $$

For the required formulas we respectively choose $h = f + g$ and $h = g$.

For the case of a general abelian category $\mathsf{C}$ we have two choices. We could use the Freyd–Mitchell embedding theorem to reduce the general case to the module case. But we can also do the following:

1. Let $i \colon X \to Y$ be a subobject of $Y$ and let $f \colon Y \to Z$ be some morphism. Define $f(X)$ as the image of the composite $f ∘ i$; this image comes with an epimorphism $p \colon X \to f(X)$ and an inclusion morphism $j \colon f(X) \to Z$ such that $j ∘ p = f ∘ i$.

   We claim that $\ker(f) ∩ X$ is the kernel of $p$. Since $j$ is a monomorphism, both $p$ and $j ∘ p = f ∘ i$ have the same kernel; so we want to show that $\ker(f) ∩ X$ is the kernel of $f ∘ i$. For this, we consider the following diagram: $$ \require{AMScd} \begin{CD} \ker(f) ∩ X @>>> \ker(f) @>>> 0 \\ @VVV @VVV @VVV \\ X @>>{i}> Y @>>{f}> Z \end{CD} $$ Both squares in this diagram are pullbacks, whence the overall diagram $$ \require{AMScd} \begin{CD} \ker(f) ∩ X @>>> 0 \\ @VVV @VVV \\ X @>>{f ∘ i}> Z \end{CD} $$ is again a pullback. This means that $\ker(f) ∩ X \to X$ is indeed the kernel of $f ∘ i$.

   We have therefore the short exact sequence $$ 0 \longrightarrow \ker(f) ∩ X \longrightarrow X \xrightarrow{\enspace p \enspace} f(X) \longrightarrow 0 \,. $$

2. Given morphisms $f \colon X \to Y$ and $g \colon Y \to Z$, we want to show that $g( \im(f) ) = \im(g ∘ f)$. We have an epimorphism $p \colon X \to \im(f)$ and the inclusion morphism $i \colon \im(f) \to X$ such that $f = i ∘ p$. We have defined $g(\im(f))$ as the image of the composite $g ∘ i$, i.e., as the kernel of the cokernel of $g ∘ i$. Similarly, $\im(g ∘ f)$ is the kernel of the cokernel of $g ∘ f$. But both $g ∘ i$ and $g ∘ f$ have the same cokernel, since $g ∘ f = (g ∘ i) ∘ p$ with $p$ being an epimorphism.

Our argumentation now makes sense in any abelian category.