1
$\begingroup$

Let $\mathcal{A}$ be an abelian category and $\mathcal{T}$ be a thick subcategory (i.e., closed under taking subquotient and extensions) of $\mathcal{A}$. Then we construct the quotient category $\mathcal{A/T}$ with objects same as category $\mathcal{A}$, and as for morphisms $Mor_{\mathcal{A/T}}(A,B)$ we consider the set $$\mathcal{I}=\{(A',B'): A' \subseteq A, B' \subseteq B,\; \; A/A' \in \mathcal{T},\; B' \in \mathcal {T}\}$$ w.r.t to the order $$(A',B')\leq (A'',B'') \iff A'\supseteq A'',\; \; B'\subseteq B''.$$ Indeed, $\; \mathcal{I}\; $ is a directed set w.r.t. pre-order $\; \leq\; $. So we define a diagram $$D:\mathcal{I}\rightarrow Mor_{\mathcal{A}}\\ (A',B')\rightarrow Mor_{\mathcal{A}}(A',B/B'),$$ and then we go on to define$$Mor_{\mathcal{A/T}}(A,B)= Colim_{(A',B')\in \mathcal{I}} Mor_{\mathcal{A}}(A',B/B').$$

My first question is, how is the diagram $D$ well defined? Since the object $B/B'$ depends on the choice of the monic map $B'\rightarrow B$ and we choose the fixed pair $(A',B')$ for all the monic maps $A' \rightarrow A$ and $B'\rightarrow B$.

Indeed, if we take $(A'\rightarrow A, B'\rightarrow B)$ pairs for fixed monic maps instead of $(A',B')$, the problem goes away and we have unique map $Mor_{\mathcal{A}} (A',B/B')\rightarrow Mor_{\mathcal{A}}(A'',B/B''). $

Secondly, why is it enough to have $B''\supseteq B'$, i.e., a monic map $B' \rightarrow B''$? Don't we need the condition $B' \rightarrow B'' \rightarrow B= B' \rightarrow B$ to have a map $B/B'\rightarrow B/B''$?

$\endgroup$
1
$\begingroup$

1) $A' \subseteq A, \; B' \subseteq B$ are treated as subobjects, so I would say that your reformulation (fixing the whole monomorphisms rather than just domains) seems to be the intended meaning (and the one that makes sense).

2) Similarly, the comparison of objects ``$A'' \subseteq A'$ '' seems to be meant as comparison in the category of subobjects of $A$, i.e. the monomorphism should be compatible with those to $A$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.