Embedding categories with exactness into their abelianization
Journal of the Australian Mathematical Society
We give a partial answer to the following questions: Given an additive category and a class of sequences, under what conditions is the universal functor to the abelian category faithful and what other sequences are taken to exact sequences? The "answers" to these questions appear as theorems 2.2 and 3.2 respectively and amount to a weakening of the condition that there be enough relative projectives. We then characterize those additive categories with exactness which have the property that all
... elative exact sequences are determined by a small set of functors into the category of abelian groups. of use, available at https://www.cambridge.org/core/terms. https://doi.org/10.1017/S1446788700018905 Downloaded from https://www.cambridge.org/core. IP address: 22.214.171.124, on 03 May 2019 at 08:21:00, subject to the Cambridge Core terms DEFINITION 5.2. If % is determined by a single functor into < § then (&,%) will be called "S-determined. DEFINITION 5.3. (Freyd (1973) §3). An abelian category is called abelian concrete if there is an exact faithful functor from it to < S. PROPOSITION 5.4. (si,%) is ^-determined if and only if %(si) is abelian concrete. PROOF. Suppose (si, %) is ^-determined. Let F: si -* ^ be the determining functor, and let F:%(si)^> ( & be the unique exact functor such that commutes. Then (£) = 0 iff (E)E% iff F(%) = 0 hence F is faithful. The converse is trivial. LEMMA If si is a well-powered abelian category and if is a Serre class then si/if is a well-powered abelian category. PROOF. Let the diagram represent the map gf ' in si/if. Then g is mono in si/if iff Ker g G if. So such a diagram represents a subobject of silif. Clearly g and gf' 1 represent the same subobject so we need consider only subobjects of the form P -?-* A with K e r g G^ and gE.si. Let P-*I g -*A represent the image of g. Then / g -* A is a subobject of A in si. Given with Ker h E if. Suppose J g -* A and /" -* A represent the same subobject of A in si. Then there is an isomorphism p: I g -»I k so that np = I. Now Ker k = Ker g E.if and Ker m = Ker h E if; and since Cok h and Cok m are both zero and hence in if, it follows that k and m are isomorphisms in si/if. The diagram of use, available at https://www.cambridge.org/core/terms. https://doi.