# 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.

Thanks in advance!

## 1 Answer

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.