Lemma 14.21.10. Let $\mathcal{A}$ be an abelian category. Let $U$ be a simplicial object of $\mathcal{A}$. Let $n \geq 0$ be an integer. The morphism $i_{n!} \text{sk}_ n U \to U$ identifies $i_{n!} \text{sk}_ n U$ with the simplicial subobject $U' \subset U$ defined in Lemma 14.18.9.

**Proof.**
By Lemma 14.21.9 we have $N(i_{n!} \text{sk}_ n U)_ i = 0$ for $i > n$. The map $i_{n!} \text{sk}_ n U \to U$ is an isomorphism in degrees $\leq n$, see Lemma 14.21.3. Combined we conclude that the map $i_{n!} \text{sk}_ n U \to U$ induces injective maps $N(i_{n!} \text{sk}_ n U)_ i \to N(U)_ i$ for all $i$. Hence Lemma 14.18.7 applies. Thus $i_{n!} \text{sk}_ n U \to U$ is injective. The result follows easily from this.
$\square$

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