# simplicial objects in a model category

Suppose that we have a (combinatorial if necessary) model category $$M$$, and let $$F:\Delta^{op}\rightarrow M$$ a simplicial object in $$M$$, such that for any natural number $$n$$, $$F([n])$$ is a fibrant object in $$M$$. We define a new object $$X= colim_{n} F([n])$$. Is it true that $$X$$ is a fibrant object ?

If what you mean by $$\rm colim_n$$ is the actual colimit of $$F$$ as a diagram of shape $$\Delta$$, then this colimit is isomorphic to the coequalizer of the two maps $$F([1]) \rightrightarrows F([0])$$. Coequalizers rarely preserve fibrancy, and with a little thought we can think of a counterexample that extends to a simplicial object. Let $$M$$ be a model category in which coproducts of fibrant objects are fibrant (e.g. simplicial sets), and let $$G:D\to M$$ be any diagram of fibrant objects whose colimit is not fibrant (such as a cospan $$X \leftarrow \Delta[0] \to Y$$ in simplicial sets for almost any Kan complexes $$X$$ and $$Y$$). Let $$F$$ be the simplicial bar construction of $$G$$, with $$F([n]) = \coprod_{d_0 \to \cdots \to d_n} G(d_0)$$. Then each $$F([n])$$ is fibrant by our assumptions on $$M$$ and $$G$$, but the colimit of $$F$$ is the coequalizer of $$\coprod_{d_0\to d_1} G(d_0) \rightrightarrows \coprod_d G(d)$$, which is just the colimit of $$G$$ (it is the usual computation of colimits in terms of coproducts and coequalizers).
If instead what you mean by $$\rm colim_n$$ is actually the geometric realization, then there is an even easier counterexample. Let $$M$$ be simplicial sets, let $$X$$ be a simplicial set that is not a Kan complex, and let $$F([n]) = X_n$$ regarded as a discrete simplicial set. Discrete simplicial sets are Kan complexes, so each $$F([n])$$ is fibrant, but the geometric realization of $$F$$ is just $$X$$ itself, which is not fibrant.
• I see, do you think it is still false if we impose that the two maps $F([1]) \rightrightarrows F([0])$ are fibrations ? – Paris Feb 21 at 20:36