Composition of morphisms in Quotient category

Question

I am having trouble understanding the composition of morphisms in the quotient category of an abelian category, following Gabriel's thesis on abelian categories.

Let $\mathcal{A}$ be an abelian category with a thick subcategory $\mathcal{T}.\;$ Let $\bar{f}\in Mor_{\mathcal{A/T}}(A,B)\; $ and $\bar{g}\in Mor_{\mathcal{A/T}}(B,C)\;.\;$ These morphisms are the images of some $f \in Mor_{\mathcal{A}}(A',B/B')\;$ and $g \in Mor_{\mathcal{A}}(B'',C/C')\;$ respectively in the corresponding colimits. Also, we have $A/A',B',B/B'',C' \in \mathcal{T}.\;$

We get the monic map $(B'+ B'')/B' \rightarrow B/B'$, and as $B/B' \in \mathcal{T}$ we have $(B'+ B'')/B' \in \mathcal{T}.\;$ We also get the pullback induced by $f,\; A'':= f^{-1}((B'+ B'')/B' )$ and the map $f'':A''\rightarrow(B'+ B'')/B' )$

Also, from $B' \in \mathcal{T}$ we have $B'\cap B'' \in \mathcal{T}.\;$ Therefore, for $\;i:B \cap B'' \rightarrow B'',\; $ we get the epic map $B'\cap B'' \rightarrow Img(g \circ i)\;$ making $\;Img(g\circ i ) \in \mathcal{T}.\;$ And we also have a monic map $\;Img(g\circ i ) \rightarrow C/C'.\;$

After this we consider the object $C'':= Img(g\circ i)+ C'.$ I do not see how $Img(g\circ i)$ can be a sub-object of $C$ so that we are able to take the sum. Therefore it would be helpful if anyone can explain the rest of the construction.

Thanks in advance!

Answer 1

It seems that some apostrophes are missing from your post, but I think I get the idea (that $i$ should really be $B' \cap B'' \rightarrow B''$).

Anyway, $\mathrm{Im}(g \circ i)$ is only a subobject of $C/C'$, but that's not a serious issue since one can always cosider the subobject $C''' \subseteq C$ with $C'''/C'=\mathrm{Im}(g \circ i)$: it is the pullback of the inclusion $\mathrm{Im}(g \circ i) \hookrightarrow C/C'$ along the projection $C \rightarrow C/C'$. Then the pullback exact sequence $$0 \rightarrow C''' \rightarrow \mathrm{Im}(g \circ i) \oplus C' \rightarrow C/C' $$ shows that $C'''$ is still torsion (i.e. $\in \mathcal{T}$) since $\mathrm{Im}(g \circ i)$ and $C'$ are.

So in the end, $g$ induces a map $$(B''+B')/B' \simeq B''/(B' \cap B'') \rightarrow C/C'''$$ (that differs from $g$ only by something torsion on domain and codomain), and precomposition with $f''$ should give you a representative of the composition.