Lemma [efr-LSAI]

Let \mathcal {C} be a model category, and suppose there exists a family of functors F_i: \mathcal {C} \to \mathsf {Set}, which each preserve colimits, and so that a morphism f \in \mathcal {C} is a cofibration if and only if each F_i(f) is an injection. Then on the category of simplicial objects in \mathcal {C}, the injective and Reedy model structures coincide.

By definition, they have the same weak equivalences. Hence it suffices to prove they have the same cofibrations.

Observe that each F_i induces a functor from simplicial objects in \mathcal {C} to simplicial sets, which we also denote F_i. The latching object functors L_r commute with these F_i in that L_rF_i = F_iL_r, since the F_i preserve colimits. But recall that L_rX_\bullet for a simplicial set X is simply the subset of degenerate r-simplices.

Now suppose f: X_\bullet \to Y_\bullet is a morphism of simplicial objects in \mathcal {C}. Suppose it is an injective cofibration. This means each F_iX_n \to F_iY_n is an injection. We must show that, for each i, F_iX_n \coprod _{F_iL_nX} F_iL_nY \to F_iY_n is an injection. By assumption the map F_iX_n \to F_iY_n is injective, and the inclusion of the degenerate simplices is also an injection. So this is an injection as long as, whenever \sigma \in F_iX and \sigma ' \in F_iL_nY are sent to the same element, they are identified in the pushout. This amounts to the claim that if \sigma is sent to a degenerate simplex of F_iY, it is already degenerate in F_iX. This follows from the injectivity of the simplicial map F_iX \to F_iY assumed.

Conversely, suppose f is a Reedy cofibration. Suppose for induction that the maps F_iX_n \to F_iY_n are injective for all n<k and all i. Then the inclusion F_iX_k \coprod _{F_iL_kX} F_iL_kY is injective, since the map F_iL_kX \to F_iL_kY is the action of F_if on degenerate simplices, and hence is determined by those injective maps on lower-dimensional simplices. Hence as the composite of two injective maps, F_iX_k \to F_iY_k is also injective. (Note that L_0 = \emptyset so there is no problem starting the induction).