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*.

PROPOSITION

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.

PROOF.

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

reader.

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.

LEMMA

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.

PROOF.

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

'{

i

i'

C

— •

B C

r p

The proof when / has the right lifting property with

•

displays / as a retract of i.

respect to i is similar.