If we are given a simplicial set $X:\Delta^{op} \to Set$, we may regard it as a \emph{simplicial} simplicial set, i.e. a bisimplicial set by composing with the "constant" inclusion $Set \to Set^{\Delta^{op}}$. We may choose to view this as a simplicial diagram of infinity groupoids. Its infinity colimit may be computed as the homotopy colimit of this diagram in the Quillen model structure on simplicial sets. A nice model for the homotopy colimit of such a diagram is its diagonal, which in this case is precisely $X$ itself. Taking $X$ to be a Kan complex implies that any infinity groupoid can be written as the colimit of a simplicial diagram of sets (in the infinity category of infinity groupoids). However, in the case of $1$-groupoids, more is true. If $\Delta\_+$ denotes the wide subcategory of $\Delta$ consisting of injective maps, if $\mathcal{G}$ is a groupoid, $\mathcal{G}$ is the weak colimit of its $\Delta\_+$-nerve regarded as a diagram in the bicategory of groupoids (in fact, we can truncate $\Delta\_+$ in this case to contain only $0$, $1$, and $2$, but that is not so important). The proof of this (which I did by pure brute force) uses the Kan condition very clearly. Hence my question is:


If $X:\Delta^{op}\to Set$ is a Kan complex, and $$X^{+}:\left(\Delta\_{+}\right)^{op} \to Set^{\Delta^{op}}$$ is the associated diagram of simplicial sets, is $X$ still the homotopy colimit of $X^{+}$? Is this written anywhere?



Writing an infinity groupoid as a colimit of sets
For the fat realization and the realization of simplicial spaces (where space here means simplicial set ;) to be weakly equivalent you need the simplicial diagramm to be Reedy cofibrant. This is e.g. the case if all degeneracies are cofibrations which is true here (if I understand your construction correct). 


