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Define a model structure on $\bf Cat$ by the following rules:

  1. A weak equivalence is an equivalence of categories;
  2. A cofibration is a functor which is injective on objects;
  3. A fibration is a functor $F\colon \bf C\to D$ such that for all $C\in \bf C$, for all isomorphism $f\colon F(C)\cong D$, there exists an object $C'\in \bf C$ and an isomorphism $f'\colon C\cong C'$ such that $F(f')=f$, $F(C')=D$.

I'm stuck in proving that these condition really define a model structure on $\bf Cat$, and in particular I'm not able to show that in a diagram $$ \begin{array}{ccc} \bf C &\xrightarrow{U}& \bf K \\ F\downarrow&&\downarrow G \\ \bf D &\xrightarrow[V]{}& \bf L \end{array} $$

  1. (LLP) if $G$ is an acyclic fibration and $F$ a cofibration, then there exists a filling arrow $W\colon \bf D\to K$ making the diagram commute.

  2. (RLP) if $G$ is a fibration and $F$ an acyclic cofibration, then there exists a filling arrow $W\colon \bf D\to K$ making the diagram commute.

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No, I don't agree. Why should an acyclic cofibration be strictly surjective on objects? – Zhen Lin Oct 20 '12 at 15:56
Yes, you're right. What about the proof I'm looking for? – Fosco Loregian Oct 20 '12 at 16:56
Just as a comment: every functor from a category to the one object-one morphism category is a fibration but it doesn't reflect iso, generally. – Giorgio Mossa Oct 20 '12 at 17:59
@Zhen Lin: Am I wrong again or acyclic fibrations are in fact strictly surjective on objects? This should conclude that LLP holds... – Fosco Loregian Oct 21 '12 at 11:38
Acyclic fibrations are indeed strictly surjective on objects. – Zhen Lin Oct 21 '12 at 12:02

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