1.1. TH E DEFINITION OF A MODEL CATEGORY 5
C into the pointed category C*. If C is already pointed, these functors define an
equivalence of categories between C and C*.
1.1.8. Suppose 6 is a model category. Define a map f in C*
to be a cofibration (fibration, weak equivalence) if and only if U f is a cofibration
(fibration, weak equivalence) in C. Then C* is a model category.
It is clear that weak equivalences in C* satisfy the two out of three
property, and that cofibrations, fibrations, and weak equivalences are closed under
retracts. Suppose i is a cofibration in C* and p is a trivial fibration. Then Ui
has the left lifting property with respect to Up\ it follows that i has the left lifting
property with respect to p, since any lift must automatically preserve the basepoint.
Similarly, trivial cofibrations have the left lifting property with respect to fibrations.
If / = /3(f) o a(f) is a functorial factorization in C, then it is also a functorial
factorization in C*; we give the codomain of a(f) the basepoint inherited from a,
and then (3(f) is forced to preserve the basepoint. Thus the factorization axiom
also holds and so C* is a model category. •
Note that we could replace the terminal object * by any object A of C, to obtain
the model category of objects under A. In fact, we could also consider the category
of objects over A, whose objects consist of pairs (X, / ) , where / : X —- A is a map
in C. A similar proof as in Proposition 1.1.8 shows that this also forms a model
category. Finally, we could iterate these constructions to form the model category
of objects under A and over B. We leave the exact statements and proofs to the
Note that by applying the functors (3 and a to the map from the initial ob-
ject to X, we get a functor X i— • QX such that QX is cofibrant, and a natural
transformation QX -^ X that is a trivial fibration. We refer to Q as the cofibrant
replacement functor of C. Similarly, there is a fibrant replacement functor RX
together with a natural trivial cofibration X — RX.
The following lemma is often useful when dealing with model categories.
1.1.9 (The Retract Argument). Suppose we have a factorization f =
pi in a category G, and suppose that f has the left lifting property with respect to p.
Then f is a retract of i. Dually, if f has the right lifting property with respect to i,
then f is a retract of p.
First suppose / has the left lifting property with respect to p. Write
/ : A — C and i: A~^B. Then we have a lift r: C — » B in the diagram below.
A —*—* B
c == c
Then the diagram below
A = A = A
The proof when / has the right lifting property with
displays / as a retract of i.
respect to i is similar.