# Composition of morphisms in Quotient category

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.

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.