Ken Brown's lemma

In mathematics, specifically in category theory, Ken Brown's lemma gives a sufficient condition for a functor on a category of fibrant objects to preserve weak equivalences; the sufficient condition is that acyclic fibrations go to weak equivalences. ^{[1] [2]} (There is also a co version.) The lemma or, more precisely, a result of which the lemma is a corollary, was introduced by Kenneth Brown. ^[3]

Proof

The lemma follows from the following:

Factorization lemma—Let ${\displaystyle f}$ be a morphism in a given category of fibrant objects. Then ${\displaystyle f}$ factorizes as ${\displaystyle f=g\circ j}$ where

• ${\displaystyle g}$ is a fibration,
• ${\displaystyle j}$ admits a retract that is an acyclic fibration.

To see the lemma follows from the above, let ${\displaystyle w}$ be a weak equivalence and ${\displaystyle F}$ the given functor. By the factorization lemma, we can write

${\displaystyle w=g\circ j}$

with an acyclic fibration ${\displaystyle r}$ such that ${\displaystyle r\circ j=\operatorname {id} }$. Note ${\displaystyle j}$ is a weak equivalence since ${\displaystyle r}$ is. Thus, ${\displaystyle g}$ is a weak equivalence (thus acyclic fibration) since ${\displaystyle w}$ is. So, ${\displaystyle F(g)}$ is a weak equivalence by assumption. Similarly, ${\displaystyle F(j)}$ is a weak equivalence. Hence, ${\displaystyle F(w)=F(g)\circ F(j)}$ is a weak equivalence. ${\displaystyle \square }$

Proof of factorization lemma: Let ${\displaystyle f:X\to Y}$ be the given morphism. Let

${\displaystyle e:Y^{I}\to Y\times Y}$

be the path object fibration; namely, it is obtained by factorizing the diagonal map ${\displaystyle \Delta }$ as ${\displaystyle \Delta =e\circ c}$ where ${\displaystyle c}$ is a weak equivalence.

Then let ${\displaystyle \pi :Z\to X\times Y}$ be the pull-back of ${\displaystyle e}$ along ${\displaystyle f\times \operatorname {id} }$, which is again a fibration. Then by the universal property of a pull-back, we get a map ${\displaystyle j:X\to Z}$ so that the resulting diagram with ${\displaystyle X{\overset {f}{\to }}Y{\overset {c}{\to }}Y^{I}}$ and ${\displaystyle (\operatorname {id} _{X},f)}$ commutes. Take ${\displaystyle g}$ to be ${\displaystyle Z{\overset {\pi }{\to }}X\times Y{\overset {p_{2}}{\to }}Y}$, which is a fibration since the projection ${\displaystyle p_{2}}$ is the pull-back of the fibration ${\displaystyle Y\to }$ final object.

As for ${\displaystyle j}$, let ${\displaystyle r}$ be ${\displaystyle Z{\overset {\pi }{\to }}X\times Y{\overset {p_{1}}{\to }}X}$, which is again a fibration. Note that ${\displaystyle r}$ is the pull-back of ${\displaystyle q\circ e:Y^{I}\to Y}$, ${\displaystyle q}$ a projection. Since ${\displaystyle e\circ c=\Delta }$, we have ${\displaystyle q\circ e\circ c=\operatorname {id} }$. It follows ${\displaystyle q\circ e}$ is a weak equivalence (since ${\displaystyle c}$ is) and thus ${\displaystyle r}$ is a weak equivalence. ${\displaystyle \square }$

References

1. Cisinski 2019, Proposition 7.4.13
2. Proposition 3.1. in https://ncatlab.org/nlab/show/factorization+lemma
3. Kenneth Brown, p. 421 (4 of 41) in: Abstract Homotopy Theory and Generalized Sheaf Cohomology, 1973, p. 4

• Cisinski, Denis-Charles (2019-06-30). Higher Categories and Homotopical Algebra (PDF). Cambridge University Press. ISBN   978-1108473200.

Further reading

• https://math.stackexchange.com/questions/4721727/understanding-a-proof-of-ken-browns-lemma
• https://mathoverflow.net/questions/342085/where-is-browns-lemma-from