Denote $Y = K_k$. Let $C[k]$ be the cosimplicial set defined in Simplicial, Example \ref{simplicial-example-simplex-cosimplicial-set}. We will use the description of $\Hom(C[k], Y)$ and $\Hom(C[k], Z)$ given in Simplicial, Lemma \ref{simplicial-lemma-morphism-into-product}. There is a canonical morphism $K \to \Hom(C[k], Y)$ corresponding to $\text{id} : K_k = Y \to Y$. Consider the morphism $\Hom(C[k], Z) \to \Hom(C[k], Y)$ which on degree $n$ terms is the morphism $$ \prod\nolimits_{\alpha : [k] \to [n]} Z \longrightarrow \prod\nolimits_{\alpha : [k] \to [n]} Y $$ using the given morphism $Z \to Y$ on each factor. Set $$ L = K \times_{\Hom(C[k], Y)} \Hom(C[k], Z). $$ The morphism $L_k \to K_k$ sits in to a commutative diagram $$ \xymatrix{ L_k \ar[r] \ar[d] & \prod_{\alpha : [k] \to [k]} Z \ar[r]^-{\text{pr}_{\text{id}_{[k]}}} \ar[d] & Z \ar[d] \\ K_k \ar[r] & \prod_{\alpha : [k] \to [k]} Y \ar[r]^-{\text{pr}_{\text{id}_{[k]}}} & Y } $$ Since the composition of the two bottom arrows is the identity we conclude that we have the desired factorization. \medskip\noindent We still have to show that $L$ is a hypercovering of $X$. To see this we will use Lemma \ref{lemma-funny-gamma}. Condition (1) is satisfied by assumption. For (2), the morphism $$ \Hom(C[k], Z)_0 \to \Hom(C[k], Y)_0 $$ is a covering because it is isomorphic to $Z \to Y$ as there is only one morphism $[k] \to [0]$. \medskip\noindent Let us consider condition (3) for $n = 0$. Then, since $(\text{cosk}_0 T)_1 = T \times T$ (Simplicial, Example \ref{simplicial-example-cosk0}) and since $\Hom(C[k], Z)_1 = \prod_{\alpha : [k] \to [1]} Z$ we obtain the diagram $$ \xymatrix{ \prod\nolimits_{\alpha : [k] \to [1]} Z \ar[r] \ar[d] & Z \times Z \ar[d] \\ \prod\nolimits_{\alpha : [k] \to [1]} Y \ar[r] & Y \times Y } $$ with horizontal arrows corresponding to the projection onto the factors corresponding to the two nonsurjective $\alpha$. Thus the arrow $\gamma$ is the morphism $$ \prod\nolimits_{\alpha : [k] \to [1]} Z \longrightarrow \prod\nolimits_{\alpha : [k] \to [1]\text{ not onto}} Z \times \prod\nolimits_{\alpha : [k] \to [1]\text{ onto}} Y $$ which is a product of coverings and hence a covering by Lemma \ref{lemma-covering-permanence}. \medskip\noindent Let us consider condition (3) for $n > 0$. We claim there is an injective map $\tau : S' \to S$ of finite sets, such that for any object $T$ of $\text{SR}(\mathcal{C}, X)$ the morphism \begin{equation} \label{equation-map} \Hom(C[k], T)_{n + 1} \to (\text{cosk}_n \text{sk}_n \Hom(C[k], T))_{n + 1} \end{equation} is isomorphic to the projection $\prod_{s \in S} T \to \prod_{s' \in S'} T$ functorially in $T$. If this is true, then we see, arguing as in the previous paragraph, that the arrow $\gamma$ is the morphism $$ \prod\nolimits_{s \in S} Z \longrightarrow \prod\nolimits_{s \in S'} Z \times \prod\nolimits_{s \not\in \tau(S')} Y $$ which is a product of coverings and hence a covering by Lemma \ref{lemma-covering-permanence}. By construction, we have $\Hom(C[k], T)_{n + 1} = \prod_{\alpha : [k] \to [n + 1]} T$ (see Simplicial, Lemma \ref{simplicial-lemma-morphism-into-product}). Correspondingly we take $S = \text{Map}([k], [n + 1])$. On the other hand, Simplicial, Lemma \ref{simplicial-lemma-formula-limit}, provides a description of points of $(\text{cosk}_n \text{sk}_n \Hom(C[k], T))_{n + 1}$ as sequences $(f_0, \ldots, f_{n + 1})$ of points of $\Hom(C[k], T)_n$ satisfying $d^n_{j - 1} f_i = d^n_i f_j$ for $0 \leq i < j \leq n + 1$. We can write $f_i = (f_{i, \alpha})$ with $f_{i, \alpha}$ a point of $T$ and $\alpha \in \text{Map}([k], [n])$. The conditions translate into $$ f_{i, \delta^n_{j - 1} \circ \beta} = f_{j, \delta_i^n \circ \beta} $$ for any $0 \leq i < j \leq n + 1$ and $\beta : [k] \to [n - 1]$. Thus we see that $$ S' = \{0, \ldots, n + 1\} \times \text{Map}([k], [n]) / \sim $$ where the equivalence relation is generated by the equivalences $$ (i, \delta^n_{j - 1} \circ \beta) \sim (j, \delta_i^n \circ \beta) $$ for $0 \leq i < j \leq n + 1$ and $\beta : [k] \to [n - 1]$. A computation (omitted) shows that the morphism (\ref{equation-map}) corresponds to the map $S' \to S$ which sends $(i, \alpha)$ to $\delta^{n + 1}_i \circ \alpha \in S$. (It may be a comfort to the reader to see that this map is well defined by part (1) of Simplicial, Lemma \ref{simplicial-lemma-relations-face-degeneracy}.) To finish the proof it suffices to show that if $\alpha, \alpha' : [k] \to [n]$ and $0 \leq i < j \leq n + 1$ are such that $$ \delta^{n + 1}_i \circ \alpha = \delta^{n + 1}_j \circ \alpha' $$ then we have $\alpha = \delta^n_{j - 1} \circ \beta$ and $\alpha' = \delta_i^n \circ \beta$ for some $\beta : [k] \to [n - 1]$. This is easy to see and omitted.