# Multiplicity in a composition series and a certain cokernel

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.

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.