alternative lecture notes - Rational points and algebraic cycles

Example 7.5. C = Top, W is the subcategory of weak equivalences, FIB is the subcategory
of Serre fibrations (have right lifting property with respect to all A ↩→ A × I where A is a
CW-complex), and COF = ⊥ (FIB ∩W ).
Example 7.6. Fix a ring R. C is the category of ≥ 0 complexes of R-modules, W is the
subcategory of quasi-isomorphisms, FIB is the subcategory of maps that are onto for k > 0,
COF is the subcategory of monomorphisms with projective cokernel for k ≥ 0.
Example 7.7. Fix a ring R. C is the category of ≤ 0 complexes of R-modules, W is
the subcategory of quasi-isomorphisms, FIB is the subcategory of maps that are onto with
injective kernelfor k ≤ 0, COF is the subcategory of maps that are monomorphisms for
k < 0.
Example 7.8. C is the category of simplicial sets, W is the subcategory of weak equivalences,
FIB is the subcategory of Kan fibrations (maps of simplicial sets that have the right lifting
property with respect to all horns Λ n k ↩→ ∆n ; then being fibrant is the same as being a Kan
simplicial set), COF is the subcategory of monomorphisms levelwise.
Example 7.9. C is the category of simplicial sets with G-action, W is the subcategory
of weak equivalences, FIB is the subcategory of Kan fibrations, COF is the subcategory of
monomorphisms with free action on B − im(A).
Example 7.10. C is the category of simplicial sets with G-action, W is the subcategory of
weak equivalences, COF is the subcategory of monomorphisms, FIB = (COF ∩W ) ⊥ .
8. July 25 (Schlank)
If A, B ⊆ C are subcategories, write C = B ◦ A if every morphism of C can be factored
as a morphism in A followed by a morphism in B (not necessarily uniquely, or functorially).
Lemma 8.1.
(a) COF = ⊥ (FIB ∩W )
(b) FIB = (COF ∩W ) ⊥
(c) COF ∩W = ⊥ FIB
(d) FIB ∩W = COF ⊥ .
Proof. We prove the first statement (the others are similar). Suppose that f : A → B is in
⊥ (FIB ∩W ). It can be factored as A ↩→ C ∼ ↠ B. Make a commutative square:
A C
l
B B,
∼
Then
A
A
A
B l C ∼ B
Then f is in R(A ↩→ C), and A ↩→ C is in COF, so f is in COF.
f
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f
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