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It is claimed that in the model category of simplicial sets (with usual model structure), a trivial fibration $X \to Y$ has a section, which is a trivial cofibration.

Now, I see that there is a section, by using lifting property with $\emptyset \to Y$ on the left (all objects are cofibrant!). I see from 2-out-of-3 that this section is an equivalence. How can I see that it is a cofibration?

Thank you, Sasha

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The usual way to show something is a cofibration is to show it has the left left property with respect to acyclic fibrations – Juan S Apr 1 '12 at 12:29
    
Yes, but I can't figure out what to do in this specific case. – Sasha Apr 1 '12 at 12:44
    
If $f:A \to B$ is an acyclic fibration and $g:Y \to X$ the section, then $g$ will be a cofibration if there is a map $X \to A$ making the usual square commute. But we have the map $X \to Y$ and then composing this with the map $Y \to A$ should give the lifting. Does that work? Seems too easy (it's very late where I am, so I'm probably not thinking straight!) – Juan S Apr 1 '12 at 13:19
    
Seems not to work formally, since your $Y \to A$ does not make the lower triangle commutative. – Sasha Apr 1 '12 at 14:17
5  
OK, I now see how to show it. It works only in our specific model category of simplicial sets, though. Indeed, I build a section as above; and it is a weak equivalence. That it is a cofibration just follows from it being an inverse (from the correct direction), since it implies that it is a monomorphism, hence a cofibration (since in our model structure cofibrations are exactly monomorphisms). Why I have not succedded: Because I wanted to much generality. – Sasha Apr 1 '12 at 14:35

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