Lemma 115.10.1. Assumptions and notation as in Simplicial, Lemma 14.32.1. There exists a section $g : U \to V$ to the morphism $f$ and the composition $g \circ f$ is homotopy equivalent to the identity on $V$. In particular, the morphism $f$ is a homotopy equivalence.
115.10 Simplicial methods
Proof. Immediate from Simplicial, Lemmas 14.32.1 and 14.30.8. $\square$
Lemma 115.10.2. Let $\mathcal{C}$ be a category with finite coproducts and finite limits. Let $X$ be an object of $\mathcal{C}$. Let $k \geq 0$. The canonical map is an isomorphism.
\[ \mathop{\mathrm{Hom}}\nolimits (\Delta [k], X) \longrightarrow \text{cosk}_1 \text{sk}_1 \mathop{\mathrm{Hom}}\nolimits (\Delta [k], X) \]
Proof. For any simplicial object $V$ we have
\begin{eqnarray*} \mathop{\mathrm{Mor}}\nolimits (V, \text{cosk}_1 \text{sk}_1 \mathop{\mathrm{Hom}}\nolimits (\Delta [k], X)) & = & \mathop{\mathrm{Mor}}\nolimits (\text{sk}_1 V, \text{sk}_1 \mathop{\mathrm{Hom}}\nolimits (\Delta [k], X)) \\ & = & \mathop{\mathrm{Mor}}\nolimits (i_{1!} \text{sk}_1 V, \mathop{\mathrm{Hom}}\nolimits (\Delta [k], X)) \\ & = & \mathop{\mathrm{Mor}}\nolimits (i_{1!} \text{sk}_1 V \times \Delta [k], X) \end{eqnarray*}
The first equality by the adjointness of $\text{sk}$ and $\text{cosk}$, the second equality by the adjointness of $i_{1!}$ and $\text{sk}_1$, and the first equality by Simplicial, Definition 14.17.1 where the last $X$ denotes the constant simplicial object with value $X$. By Simplicial, Lemma 14.20.2 an element in this set depends only on the terms of degree $0$ and $1$ of $i_{1!} \text{sk}_1 V \times \Delta [k]$. These agree with the degree $0$ and $1$ terms of $V \times \Delta [k]$, see Simplicial, Lemma 14.21.3. Thus the set above is equal to $\mathop{\mathrm{Mor}}\nolimits (V \times \Delta [k], X) = \mathop{\mathrm{Mor}}\nolimits (V, \mathop{\mathrm{Hom}}\nolimits (\Delta [k], X))$. $\square$
Lemma 115.10.3. Let $\mathcal{C}$ be a category. Let $X$ be an object of $\mathcal{C}$ such that the self products $X \times \ldots \times X$ exist. Let $k \geq 0$ and let $C[k]$ be as in Simplicial, Example 14.5.6. With notation as in Simplicial, Lemma 14.15.2 the canonical map is identified with the map which is the projection onto the factors where $\alpha $ is a constant map.
\[ \mathop{\mathrm{Hom}}\nolimits (C[k], X)_1 \longrightarrow (\text{cosk}_0 \text{sk}_0 \mathop{\mathrm{Hom}}\nolimits (C[k], X))_1 \]
\[ \prod \nolimits _{\alpha : [k] \to [1]} X \longrightarrow X \times X \]
Proof. This is shown in the proof of Hypercoverings, Lemma 25.7.3. $\square$
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