Academia.eduAcademia.edu

Figure 15 – uploaded by Dennis Sieg

See full PDFdownloadDownload figure
The morphism jz is unique with this property because of the universal proper- ties of (1) and (3). This shows (1). Since f is an I-strict monomorphism it follows from (I) that ¢€ is a strict monomorphism. Then € 0 y~, = 20 Sp is also a strict monomorphism and by Proposition 4 it follows that sr is a strict monomorphism, hence k is an [-strict monomorphism which shows (k, pr) € Ec. E3): Since C has products and cokernels, the pushout of any two morphisms does exist. The pair (f,g) is in Ec if and only if (g°?, f°?) is in Eco». Then E3) follows from (E2) by duality. El): Let 0 X Ly £7 —+0and0 > X' 57 4% Z' 0 be short exact 1,01 = 119 SR = 2 029 SR = [lg 0 €0 A,

Figure 15 The morphism jz is unique with this property because of the universal proper- ties of (1) and (3). This shows (1). Since f is an I-strict monomorphism it follows from (I) that ¢€ is a strict monomorphism. Then € 0 y~, = 20 Sp is also a strict monomorphism and by Proposition 4 it follows that sr is a strict monomorphism, hence k is an [-strict monomorphism which shows (k, pr) € Ec. E3): Since C has products and cokernels, the pushout of any two morphisms does exist. The pair (f,g) is in Ec if and only if (g°?, f°?) is in Eco». Then E3) follows from (E2) by duality. El): Let 0 X Ly £7 —+0and0 > X' 57 4% Z' 0 be short exact 1,01 = 119 SR = 2 029 SR = [lg 0 €0 A,

Related Figures (19)

be a pushout square. Proposition 6 Let C be a semi-abelian category and
be a pushout square. Proposition 6 Let C be a semi-abelian category and
be a pullback square. Proposition 7 Let C be a semi-abelian category and The dual argument shows:
be a pullback square. Proposition 7 Let C be a semi-abelian category and The dual argument shows:
is a pushout square and f is a strict monomorphism, then s7 is a monomor- phism.
is a pushout square and f is a strict monomorphism, then s7 is a monomor- phism.
where (s,Sy,87) is the pushout of f and ¢t and c is a cokernel of sy, the morphism s7 is a monomorphism, but not necessarily a strict monomorphism. Those strict epimorphisms and monomorphisms that are preserved under pullback, resp. pushout, deserve a special name: morphism t : X — T we have a commutative diagram
where (s,Sy,87) is the pushout of f and ¢t and c is a cokernel of sy, the morphism s7 is a monomorphism, but not necessarily a strict monomorphism. Those strict epimorphisms and monomorphisms that are preserved under pullback, resp. pushout, deserve a special name: morphism t : X — T we have a commutative diagram
the morphism pr is also a strict epimorphism.
the morphism pr is also a strict epimorphism.
the morphism sv is also a strict monomorphism. Using these notions a quasi-abelian category is a semi-abelian category in which every strict epimorphism and every strict monomorphism is [-strict
the morphism sv is also a strict monomorphism. Using these notions a quasi-abelian category is a semi-abelian category in which every strict epimorphism and every strict monomorphism is [-strict
is a pullback of g'og and t, if (P, pr, pz) is a pullback of g’ and t and (Q, gp, qv) is a pullback of g and pz. By Proposition 4 iii) the morphism pr o qp is a strict epimorphism, hence gq’ 0 g is an [-strict epimorphism. ii) This follows as above from the transitivity of the pullback by using Propo- sition 4 ii) instead of iii).
is a pullback of g'og and t, if (P, pr, pz) is a pullback of g’ and t and (Q, gp, qv) is a pullback of g and pz. By Proposition 4 iii) the morphism pr o qp is a strict epimorphism, hence gq’ 0 g is an [-strict epimorphism. ii) This follows as above from the transitivity of the pullback by using Propo- sition 4 ii) instead of iii).
Proof: (E0) is satisfied by Proposition 12. We show that Ec is closed under isomorphisms: Let (f,g) € Ec and let
Proof: (E0) is satisfied by Proposition 12. We show that Ec is closed under isomorphisms: Let (f,g) € Ec and let
in C with isomorphisms ¢ and w is a pullback square as well as a pushout square, hence f’ is an /-strict monomorphism and g’ an /-strict epimorphism by Remark 11 which shows (f’, 9’) € Ec. (E2): The pullback of any two morphisms does exist in C, since C has products and kernels. Let (f,g) € Ec, let be a commutative square in C' with isomorphisms 7x, iy and iz.
in C with isomorphisms ¢ and w is a pullback square as well as a pushout square, hence f’ is an /-strict monomorphism and g’ an /-strict epimorphism by Remark 11 which shows (f’, 9’) € Ec. (E2): The pullback of any two morphisms does exist in C, since C has products and kernels. Let (f,g) € Ec, let be a commutative square in C' with isomorphisms 7x, iy and iz.
hence by the universal property of (1) there is a unique morphism € : Qi — Qa with w2osr = coy, and dgof = €0¢1, so that the following cube is commutative:
hence by the universal property of (1) there is a unique morphism € : Qi — Qa with w2osr = coy, and dgof = €0¢1, so that the following cube is commutative:
Since k is a monomorphism, 77 is unique with this property, hence the diagram in (II.1) is a pullback square. We need the following Lemma: (II.1) The diagram
Since k is a monomorphism, 77 is unique with this property, hence the diagram in (II.1) is a pullback square. We need the following Lemma: (II.1) The diagram
Proof of (II.2): By Lemma 5 i) we have a commutative diagram
Proof of (II.2): By Lemma 5 i) we have a commutative diagram
Hence the universal property of the coproduct yields lx;.y = dor, making the following diagram commutative:
Hence the universal property of the coproduct yields lx;.y = dor, making the following diagram commutative:
the morphism pr is not a strict morphism. orphisms g : Y — Z and t: T — Z in (LC)*™ so that in the pullback square
the morphism pr is not a strict morphism. orphisms g : Y — Z and t: T — Z in (LC)*™ so that in the pullback square
580 California St., Suite 400
San Francisco, CA, 94104
© 2025 Academia. All rights reserved