The Stacks project

Proof. Pick $m$, $i \leq n$ and some $\varphi : [m] \to [i]$. The image under $U(\psi )$ of $\mathop{\mathrm{Im}}(U(\varphi ))$ for any $\psi : [m'] \to [m]$ is equal to the image of $U(\varphi \circ \psi )$ and $\varphi \circ \psi : [m'] \to [i]$. Hence $U'$ is a simplicial object. Pick $m > n$. We have to show $N(U'_ m) = 0$. By definition of $N(U_ m)$ and $N(U'_ m)$ we have $N(U'_ m) = U'_ m \cap N(U_ m)$ (intersection of subobjects). Since $U$ is split by Lemma 14.18.6, it suffices to show that $U'_ m$ is contained in the sum

By the splitting each $U_{m'}$ is the sum of images of $N(U_{m''})$ via $U(\psi )$ for surjective maps $\psi : [m'] \to [m'']$. Hence the displayed sum above is the same as

Clearly $U'_ m$ is contained in this by the simple fact that any $\varphi : [m] \to [i]$, $i \leq n$ occurring in the definition of $U'_ m$ may be factored as $[m] \to [m'] \to [i]$ with $[m] \to [m']$ surjective and $m' < m$ as in the last displayed sum above. $\square$