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I would like to understand the accounts of P. Gabriel (link text), pag 365, when he shows that the composition of this category is well defined.

Definition: Given a Serre subcategory $\mathcal{C}$ of $\mathcal{A}$, the quotient category $\mathcal{A}/\mathcal{C}$ of $\mathcal{A}$ with respect to $\mathcal{C}$ is defined as follows. The objects in $\mathcal{A}/\mathcal{C}$ are the objects in $\mathcal{A}$. Given two objects $A,B$ in $\mathcal{A}$, there is for each pair of subobjects $A' \subset A$ and $B'\subset B$ and induced map $Hom_{\mathcal{A}}(A,B) \to Hom_{\mathcal{A}}(A',B/B')$. The pairs $(A',B')$ such that both $A/A'$ and $B'$ lie in $\mathcal{C}$ form a directed set, and one obtains a direct system of abelian groups $Hom_{\mathcal{A}}(A',B/B')$. We define $$ Hom_{\mathcal{A}/\mathcal{C}} (A,B)= colim_{(A',B')} Hom_{\mathcal{A}}(A',B/B') $$ Let $\bar{f} \in Hom_{\mathcal{A}/\mathcal{C}} (M,N)$ and $\bar{g} \in Hom_{\mathcal{A}/\mathcal{C}} (N,P)$. Else, exist $f \in Hom_{\mathcal{A}}(M',N/N')$ and $g \in Hom_{\mathcal{A}}(N'',P/P')$. Thus, let $M''=f^{-1}((N''+N')/N')$. Why $M/M'' \in \mathcal{C}$? (first question).

The author then takes the following object: $g(N'' \cap N')$ lie in $\mathcal{C}$. Why? (second quastion).

Let $P''=P'+ g(N'' \cap N')$, too $P'' \in \mathcal{C}$, why? (third question)

Now, let the morphism $f' : M'' \to (N''+N')/N'$ induced $f$ and the morphism $g': N''/(N''\cap N') \to P/P''$ induced $g$.

Finally, it takes the following morphism: $M'' \xrightarrow{f'} (N''+N')/N' \cong N''/(N''\cap N') \xrightarrow{g'} P/P''$. He claims that this morphism is the representative of composite function $\bar{g}\bar{f}$ independently of morphisms $f$ and $g$. I do not understand this statement. (fourth question).

Comments: Let $f: A \to B$ and $A' \subset A$, $f(A')$ is the image of the $ A'\hookrightarrow A \xrightarrow{f} B$. If $B'\subset B$, $f^{-1}(B')$ is the kernel of the $ A \xrightarrow{f} B \twoheadrightarrow B/B'$.

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up vote 4 down vote accepted

First, you actually want $((N'' + N')/N')$ every time you have $(N'' + N')/N$. If this was the source of your confusion, sweet. If not, here goes. (Also, the double underscores in the NUMDAM url got messed up by MO's processing; here's a working link.)

  1. There's a short exact sequence $0 \to M'/M'' \to M/M'' \to M/M' \to 0$, and $M'/M'' \cong N/(N'' + N')$, since $M''$ is the kernel of the composition $M' \to N/N' \to N/(N'' + N')$; finally, $N/(N'' + N')$ is a quotient of $N/N''$, which is in $\mathcal{C}$, so it's in $\mathcal{C}$ as well.

  2. $\mathcal{C}$ is closed under taking subobjects and quotient objects. $N'$ is in $\mathcal{C}$, $N' \cap N''$ is a subobject, and $g(N' \cap N'')$ is a quotient of that.

  3. A sum of two subobjects of another object is a quotient of their direct sum. Here, $P'$ and $g(N'' \cap N')$ are in $\mathcal{C}$, so their direct sum is as well, and so this internal sum is as well.

  4. The point of this was to compose $\overline{f}$ and $\overline{g}$ in the quotient category, but to do this, we had to lift them to maps $f$ and $g$ in the original category. Gabriel's saying that we get the same composition in the quotient category even if we had picked different lifts of $\overline{f}$ and $\overline{g}$. Let's say we picked $\tilde{f}:\widetilde{M'} \to N/\widetilde{N'}$ and $\tilde{g}:\widetilde{N''} \to P/\widetilde{P'}$ instead. Since these have to become $\overline{f}$ and $\overline{g}$ in the direct limit, then, for example, $\widetilde{f}$ and $f$ must be the same as maps from some subobject of $M' \cap \widetilde{M'}$ to some quotient of $N/(N' + \widetilde{N'})$, and likewise for $g$ and $\widetilde{g}$.

    So we reduce to the case where $\widetilde{f}$ is defined 'further along in the direct system' than $f$. That is, $\widetilde{f}$ is a map $\widetilde{M'} \to N/\widetilde{N'}$ where $\widetilde{M'}$ and $\widetilde{N'}$ are subobjects of $M'$ and $N'$ respectively, with the obvious restrictions on their quotients, and we have to show that $gf$ and $g\widetilde{f}$ become the same map in the quotient category (and then do the same but vary $g$). I'm not going to do this whole proof for you, but if you're confused about things like this, it's probably worth doing yourself. Good luck!

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First, thank you for the answer. With your help, I wrote the details but is missing justify an assertion of P. Gabriel. When he wrote that the morphism $g'$ is induced for g. I wonder how would be a category of modules, for example. But I can not justify in a abelian category. Would use it? $$ N'' \xrightarrow{g} P/P' \twoheadrightarrow P/P' $$ –  Marcelo Silva May 19 '13 at 14:01
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$g$ is a map $N'' \to P/P'$; taking a further quotient induces $N'' \to P/P''$; and then by definition, $N' \cap N''$ is a subobject of the kernel of this map, so it passes to $N''/(N'\cap N'')\to P/P''$. There's a good reason why arguments about abelian categories look like arguments about modules: the Freyd-Mitchell embedding theorem, which says that any small abelian category embeds into some category of modules over a ring. So any statement you can phrase in terms of an arbitrary small set of objects in an abelian category, and prove when those objects are modules, is true in general! –  Paul VanKoughnett May 19 '13 at 23:08
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