Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I just came across a strange property of morphisms that are preserved under pullbacks, and it made me wonder. Consider a model category $\mathcal{M}$.

Because the fibrations are exactly the maps that have the right lifting property (with respect to the acyclic cofibrations), it is easy to see that they are closed by pullbacks. Moreover, monomorphisms are also closed by pullbacks. Similarly, both cofibrations and epimorphisms are closed by pushouts.

However, in my mind, I always thought that cofibrations and monomorphisms would be closed under the same type of diagrams, and dually that fibrations and epimorphisms would be closed under the same type of diagrams. I realize that its silly and it may be dangerous to associate cofibrations to monomorphisms, but this association is more of a "mental trick to remember things" than anything.

I don't know if this is a reasonnable question but here I go :

Is there a more rational explanation about why cofibrations and epi are "together", and similarly for fibrations and mono ?

For example on Set, I think there is a model structure in which cofibrations are epi and fibrations are mono. The factorization of $X \stackrel{f}{\to} Y$ would be through the image $X \stackrel{f}{\twoheadrightarrow} \text{im}(f) \hookrightarrow Y$, and so, this forces mono to be closed under pullbacks and epi under pushouts, since fibrations and cofibrations are.

Is there such a model on any complete and cocomplete category for example ?

Thank you. Bogdan

share|cite|improve this question
is this the bogdan I know? – Sean Tilson Jun 11 '12 at 21:14
Yes that's me ! Finishing the writing of the preliminary part of my project, introducing model categories (pre-tty bo-ring) – Bogdan Jun 12 '12 at 8:27
up vote 2 down vote accepted

There is a common explanation. As you already mentioned, in order to show that pullbacks preserve fibrations one does not use all the axioms of the model category structure but only the lifting properties. Given two maps $f\colon A\to B$ and $g\colon X\to Y$, write $f \mathrel{\Box} g$ if every square $$ \matrix{ A & \longrightarrow & X \cr {\scriptstyle f} \big\downarrow {\ } & & {\ } \big\downarrow {\scriptstyle g} \cr B & \longrightarrow & Y } $$ has a (not necessarily unique) diagonal $d\colon B\to X$ making the two resulting triangles commutative.

Now let ${\cal H}$ be a class of maps in a category ${\cal K}$ and define $$ {\cal H}^\Box = \{ g \in {\cal K} \mid \forall h\in {\cal H}: h \mathrel{\Box} g \} $$ and $$ {}^\Box{\cal H} = \{ f \in {\cal K} \mid \forall h\in {\cal H}: f \mathrel{\Box} h \} $$

Then we have that ${\cal H}^\Box$ is always stable under pullbacks and dually ${}^\Box{\cal H}$ is always stable under pushouts.

In a model category with cofibrations ${\cal C}$, fibrations ${\cal F}$ and weak equivalences ${\cal W}$ one has ${\cal F} = ({\cal C}\cap {\cal W})^\Box$.

For the case of monomorphisms, suppose that the underlying category ${\cal K}$ has binary copowers and consider the class ${\cal H} = \{ (id_K|id_K)\colon K+K \to K \mid K \in {\cal K} \}$ of all codiagonal maps. Then ${\cal H}^\Box$ is exactly the class of all monomorphisms of ${\cal K}$

Of course, since the usual proof that pullbacks preserve monomorphisms does not need any assumption about the underlying category, this is a bit overkill.

share|cite|improve this answer
I was reading a bit about factorization systems and stuff, but didn't think of it. Thank you so much. Your answer is exactly the sort of answer I was looking for ! Dually, in a category with binary powers, the epimorphisms are the morphisms with the left lifting property with respect to all diagonal maps ? I should definitely try to check it myself... – Bogdan Jun 13 '12 at 21:01

I think there is a model structure in which cofibrations are epi and fibrations are mono.

This is false because in a model cat $(\mathbf{C},\cal,W,F,C)$, you can prove that $\mathcal F = (\mathcal C\cap \mathcal W)^\square$ and $\mathcal C = {}^\square(\mathcal F\cap\mathcal W)$, against the characterization Mark offered you.

share|cite|improve this answer
Actually, there are model structures where the cofibrations are the epimorphisms and the fibrations are the monomorphisms, e.g. on $\mathbf{Set}$. Of course in this example every map is a weak equivalence, but never mind. – Marc Olschok Feb 24 '13 at 22:22

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.