Suppose that is a nice model category $\mathbf{M}$ (cofibrantly generated,...). Suppose we have a two diagrams $$F,G: \Delta^{op}\rightarrow \mathbf{M} $$ and a natural transformation $\nu: F\rightarrow G $ such that for any natural number $n$, $\nu_{n}: F(n)\rightarrow G(n) $ is a trivial cofibration. Is the induced map

$colim_{\Delta^{op}} F\rightarrow colim_{\Delta^{op}} G$ is a weak equivalence ?

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    $\begingroup$ This would hold if the colimit functor were left Quillen for the injective model structure on $M^{\Delta^{op}}$. But the colimit functor should only be expected to be left Quillen for the projective model structure (or maybe the Reedy model structure?). So presumably the answer is no, but I don't have a counterexample at hand. $\endgroup$ – Tim Campion Jun 28 '18 at 15:52
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    $\begingroup$ However, from the question title, it sounds like you're actually interested in something different -- it sounds like you'd really like to know that if $F,G$ are both projectively cofibrant and weakly equivalent, then their colimits are also weakly equivalent. This is true, by Ken Brown's lemma: a left Quillen functor preserves weak equivalences between cofibrant objects. $\endgroup$ – Tim Campion Jun 28 '18 at 16:07
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    $\begingroup$ @TimCampion, thank for your comments, in my case the diagrams are not supposed to be cofibrant (in this case it will be trivial as you noticed in your second comment) $\endgroup$ – mathphys Jun 28 '18 at 17:51

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