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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 $\phi,\psi \colon X \rightarrow Y$ be two morphisms in $\mathsf{C}$. Is the following inequality true? $$[\operatorname{im}(\phi +\psi)\colon S] \leq [\operatorname{im}(\phi)\colon S]+[\operatorname{im}(\psi)\colon S]$$

I know that $\operatorname{im}(\phi), \operatorname{im}(\psi)$ and $\operatorname{im}(\phi+\psi)$ are subobjects of $\operatorname{im}(\phi)+ \operatorname{im}(\psi)$. This might be helpful but I don't know how.

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The following certainly works for modules, and should therefore also work in $\mathsf{C}$ by the Freyd–Mitchell embedding theorem: \begin{align*} \newcommand{\im}{\mathrm{im}} [\im(ϕ + ψ) : S] &≤ [\im(ϕ) + \im(ψ) : S] \\ &≤ [\im(ϕ) ⊕ \im(ψ) : S] \\ &= [\im(ϕ) : S] + [\im(ψ) : S] \,. \end{align*}

Let us check which properties of finite length objects we rely on.

We have the following general result:

  • If $0 \to X \to Y \to Z \to 0$ is a short exact sequence in $\mathsf{C}$, then $Y$ is of finite length if and only if both $X$ and $Z$ are of finite length, and then $$ [Y : S] = [X : S] + [Z : S] \,. $$

We need the following special cases of this general result:

  • For any two objects $X$ and $Y$ of finite lengths their direct sum $X ⊕ Y$ is again of finite length, with $$ [X ⊕ Y : S] = [X : S] + [Y : S] \,. $$

  • If $X$ is a subobject of an object $Y$ of finite length, then $X$ is again of finite length, with $$ [X : S] ≤ [Y : S] \,. $$

  • If $Z$ is a quotient object of an object $Y$ of finite length, then $Z$ is again of finite length, with $$ [Z : S] ≤ [Y : S] \,. $$

Finally, we need the following observations regarding $\im(ϕ) + \im(ψ)$:

  • We have an epimorphism $\im(ϕ) ⊕ \im(ψ) \to \im(ϕ) + \im(ψ)$, which is induced from the two canonical morphisms $\im(ϕ) \to \im(ϕ) + \im(ψ)$ and $\im(ψ) \to \im(ϕ) + \im(ψ)$.

  • We have an inclusion (of subobjects of $Y$) $\im(ϕ + ψ) \to \im(ϕ) + \im(ψ)$.

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  • $\begingroup$ Why do you need an embedding theorem? I think that the proof at the beginning should work in any abelian category (where objects have finite length). $\endgroup$
    – Aphelli
    Dec 26, 2022 at 14:57
  • $\begingroup$ I agree with @Aphelli. In fact, it’s not clear how you would use the Freyd-Mitchell embedding theorem, as the embedding it provides might not preserve the properties of being simple or of having finite length. But anyway, the proof you give uses nothing specific about module categories, so the embedding theorem is only a distraction. $\endgroup$
    – Jeremy Rickard
    Dec 26, 2022 at 15:13
  • $\begingroup$ @Aphelli I’m not saying that we need the embedding theorem. But I also haven’t checked that everything works out in an arbitrary abelian category, and I personally don’t like making claims which I haven’t properly verified. I am therefore trying to provide a meta-argument which tells us that everything is going to behave as expected once we decide to actually check the details. $\endgroup$
    – Jendrik Stelzner
    Dec 26, 2022 at 15:17
  • $\begingroup$ @JendrikStelzner Thank you! Do you have a reference for the general result (concerning the relationship between short exact sequences and the mutliplicity) you quote? $\endgroup$
    – Margaret
    Dec 27, 2022 at 19:13
  • $\begingroup$ @Margaret I sadly don’t, but the proof should go as follows: we have $Z = Y/X$. Take a composition series $0 = Z_0 ⊆ Z_1 ⊆ \dotsb ⊆ Z_n = Z$ of $Z$. This corresponds to a sequence $X = Y_0 ⊆ Y_1 ⊆ \dotsb ⊆ Y_n = Y$ of subobjects of $Y$, and $Y_i/Y_{i-1} ≅ Z_i/Z_{i-1}$ by the second isomorphism theorem. Also take a composition series $0 = X_0 ⊆ X_1 ⊆ \dotsb ⊆ X_m = X$ of $X$. Combine these series into a composition series $0 = X_0 ⊆ \dotsb ⊆ X_m = X = Y_0 ⊆ \dotsb ⊆ Y_n = Y$ of $Y$. We now have composition series for $X, Y, Z$ from which we can see $ℓ(Y) = ℓ(X)+ℓ(Z)$ and $[Y:S] = [X:S]+[Z:S]$. $\endgroup$
    – Jendrik Stelzner
    Dec 27, 2022 at 20:59

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