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Dec 21, 2011 at 8:45 vote accept Adam
Dec 20, 2011 at 11:48 answer added Justin Noel timeline score: 5
Dec 20, 2011 at 8:58 comment added Justin Noel Usually the trick is to realize that the cosimplicial remplacement of a diagram satisfies $X^{n+1}\cong F^{n+1}\times M^n X$ where $F^{n+1}$ is the 'cofree' part of $X$ in degree $n+1$. This part should be explicitly computable in terms of the original diagram. Once you know this, the fibered products simplify and the matching maps should be coordinatewise fibrations, hence a fibration.
Dec 20, 2011 at 5:40 history asked Adam CC BY-SA 3.0