For any object X denote by R(X) the set of strong subobjects of X. Since A has finite limits, the poset R(X) has meets. Suppose u: U --> X and v: V --> X are two strong subobjects. Suppose T = U + V is the sum of U and V and t: T --> X is the map induced by u and v. Then the strong image t(T) of T in X is the join U V of U and V in R(X). It follows that R(X) has joins. Thus R(X) is a lattice, with 0X: 0 --> X as zero and 1X: X --> X as one. An object Z has exactly one strong subobject (i.e. 0Z = 1Z) iff it is initial.
If u: U --> X is a mono, we denote by f-1(u) the pullback of u along f. Then f-1: R(X) --> R(Y) is a mapping preserves meets such that f-1(0X) = 0Y and f-1(1X) = 1Y (i.e. f-1 is bounded ). Also f-1 has a left adjoint f+1: R(Y) --> R(X) sending each strong subobject v: V --> Y to the strong image of the composite f.v: V --> X . If V = Y then f+1(Y) is simply the strong image of f.
Proof. Let e: X1 --> x+1(X1) and m: x+1(X1) --> X be the epi-strong-mono factorization of x. Form the pullback
Proof. We shall use the notation and diagram in the proof of
(1.5.1). Since x and f are disjoint,
the pullback Y1 of x and f is 0
. The map e1: 0 = Y1 --> T is an isomorphism
as it is an epic strong mono. It follows that f and m are
Proof. Since f-1 is bounded and preserves meets, we only need to verify that it also preserves joins. Suppose u: U --> X and v: V --> X are two strong subobjects. Let x: U + V --> X be the map induced by u and v. Then x+1(U + V) is the join of U and V . Since sums are stable, the pullback Y1 of U + V along f in the following pullback diagram is f-1(U) + f-1(V).
Proposition 1.5.4. If f: Y --> X is a coflat mono, then f-1f+1 is the identity R(Y) --> R(Y).
Proof. Suppose u: U --> Y is a strong mono. Consider the mono fu: U --> X , whose pullback along f is u .
Proof. Form the pullback for a strong mono u: U --> S