# The Stacks Project

## Tag 0AMA

### 14.19. Coskeleton functors

Let $\mathcal{C}$ be a category. The coskeleton functor (if it exists) is a functor $$\text{cosk}_n : \text{Simp}_n(\mathcal{C}) \longrightarrow \text{Simp}(\mathcal{C})$$ which is right adjoint to the skeleton functor. In a formula $$\tag{14.19.0.1} \mathop{\rm Mor}\nolimits_{\text{Simp}(\mathcal{C})}(U, \text{cosk}_n V) = \mathop{\rm Mor}\nolimits_{\text{Simp}_n(\mathcal{C})}(\text{sk}_n U, V)$$ Given a $n$-truncated simplicial object $V$ we say that $\text{cosk}_nV$ exists if there exists a $\text{cosk}_nV \in \mathop{\rm Ob}\nolimits(\text{Simp}(\mathcal{C}))$ and a morphism $\text{sk}_n \text{cosk}_n V \to V$ such that the displayed formula holds, in other words if the functor $U \mapsto \mathop{\rm Mor}\nolimits_{\text{Simp}_n(\mathcal{C})}(\text{sk}_n U, V)$ is representable. If it exists it is unique up to unique isomorphism by the Yoneda lemma. See Categories, Section 4.3.

Example 14.19.1. Suppose the category $\mathcal{C}$ has finite nonempty self products. A $0$-truncated simplicial object of $\mathcal{C}$ is the same as an object $X$ of $\mathcal{C}$. In this case we claim that $\text{cosk}_0(X)$ is the simplicial object $U$ with $U_n = X^{n + 1}$ the $(n + 1)$-fold self product of $X$, and structure of simplicial object as in Example 14.3.5. Namely, a morphism $V \to U$ where $V$ is a simplicial object is given by morphisms $V_n \to X^{n + 1}$, such that all the diagrams $$\xymatrix{ V_n \ar[r] \ar[d]_{V([0] \to [n], 0 \mapsto i)} & X^{n + 1} \ar[d]^{\text{pr}_i} \\ V_0 \ar[r] & X }$$ commute. Clearly this means that the map determines and is determined by a unique morphism $V_0 \to X$. This proves that formula (14.19.0.1) holds.

Recall the category $\Delta/[n]$, see Example 14.11.4. We let $(\Delta/[n])_{\leq m}$ denote the full subcategory of $\Delta/[n]$ consisting of objects $[k] \to [n]$ of $\Delta/[n]$ with $k \leq m$. In other words we have the following commutative diagram of categories and functors $$\xymatrix{ (\Delta/[n])_{\leq m} \ar[r] \ar[d] & \Delta/[n] \ar[d] \\ \Delta_{\leq m} \ar[r] & \Delta }$$ Given a $m$-truncated simplicial object $U$ of $\mathcal{C}$ we define a functor $$U(n) : (\Delta/[n])_{\leq m}^{opp} \longrightarrow \mathcal{C}$$ by the rules \begin{eqnarray*} ([k] \to [n]) & \longmapsto & U_k \\ \psi : ([k'] \to [n]) \to ([k] \to [n]) & \longmapsto & U(\psi) : U_k \to U_{k'} \end{eqnarray*} For a given morphism $\varphi : [n] \to [n']$ of $\Delta$ we have an associated functor $$\overline{\varphi} : (\Delta/[n])_{\leq m} \longrightarrow (\Delta/[n'])_{\leq m}$$ which maps $\alpha : [k] \to [n]$ to $\varphi \circ \alpha : [k] \to [n']$. The composition $U(n') \circ \overline{\varphi}$ is equal to the functor $U(n)$.

Lemma 14.19.2. If the category $\mathcal{C}$ has finite limits, then $\text{cosk}_m$ functors exist for all $m$. Moreover, for any $m$-truncated simplicial object $U$ the simplicial object $\text{cosk}_mU$ is described by the formula $$(\text{cosk}_mU)_n = \mathop{\rm lim}\nolimits_{(\Delta/[n])_{\leq m}^{opp}} U(n)$$ and for $\varphi : [n] \to [n']$ the map $\text{cosk}_mU(\varphi)$ comes from the identification $U(n') \circ \overline{\varphi} = U(n)$ above via Categories, Lemma 4.14.8.

Proof. During the proof of this lemma we denote $\text{cosk}_mU$ the simplicial object with $(\text{cosk}_mU)_n$ equal to $\mathop{\rm lim}\nolimits_{(\Delta/[n])_{\leq m}^{opp}} U(n)$. We will conclude at the end of the proof that it does satisfy the required mapping property.

Suppose that $V$ is a simplicial object. A morphism $\gamma : V \to \text{cosk}_mU$ is given by a sequence of morphisms $\gamma_n : V_n \to (\text{cosk}_mU)_n$. By definition of a limit, this is given by a collection of morphisms $\gamma(\alpha) : V_n \to U_k$ where $\alpha$ ranges over all $\alpha : [k] \to [n]$ with $k \leq m$. These morphisms then also satisfy the rules that $$\xymatrix{ V_n \ar[r]_{\gamma(\alpha)} & U_k \\ V_{n'} \ar[r]^{\gamma(\alpha')} \ar[u]^{V(\varphi)} & U_{k'} \ar[u]_{U(\psi)} }$$ are commutative, given any $0 \leq k, k' \leq m$, $0 \leq n, n'$ and any $\psi : [k] \to [k']$, $\varphi : [n] \to [n']$, $\alpha : [k] \to [n]$ and $\alpha' : [k'] \to [n']$ in $\Delta$ such that $\varphi \circ \alpha = \alpha' \circ \psi$. Taking $n = k$, $\varphi = \alpha'$, and $\alpha = \psi = \text{id}_{[k]}$ we deduce that $\gamma(\alpha') = \gamma(\text{id}_{[k]}) \circ V(\alpha')$. In other words, the morphisms $\gamma(\text{id}_{[k]})$, $k \leq m$ determine the morphism $\gamma$. And it is easy to see that these morphisms form a morphism $\text{sk}_m V \to U$.

Conversely, given a morphism $\gamma : \text{sk}_m V \to U$, we obtain a family of morphisms $\gamma(\alpha)$ where $\alpha$ ranges over all $\alpha : [k] \to [n]$ with $k \leq m$ by setting $\gamma(\alpha) = \gamma(\text{id}_{[k]}) \circ V(\alpha)$. These morphisms satisfy all the displayed commutativity restraints pictured above, and hence give rise to a morphism $V \to \text{cosk}_m U$. $\square$

Lemma 14.19.3. Let $\mathcal{C}$ be a category. Let $U$ be an $m$-truncated simplicial object of $\mathcal{C}$. For $n \leq m$ the limit $\mathop{\rm lim}\nolimits_{(\Delta/[n])_{\leq m}^{opp}} U(n)$ exists and is canonically isomorphic to $U_n$.

Proof. This is true because the category $(\Delta/[n])_{\leq m}$ has an final object in this case, namely the identity map $[n] \to [n]$. $\square$

Lemma 14.19.4. Let $\mathcal{C}$ be a category with finite limits. Let $U$ be an $n$-truncated simplicial object of $\mathcal{C}$. The morphism $\text{sk}_n \text{cosk}_n U \to U$ is an isomorphism.

Proof. Combine Lemmas 14.19.2 and 14.19.3. $\square$

Let us describe a particular instance of the coskeleton functor in more detail. By abuse of notation we will denote $\text{sk}_n$ also the restriction functor $\text{Simp}_{n'}(\mathcal{C}) \to \text{Simp}_n(\mathcal{C})$ for any $n' \geq n$. We are going to describe a right adjoint of the functor $\text{sk}_n : \text{Simp}_{n + 1}(\mathcal{C}) \to \text{Simp}_n(\mathcal{C})$. For $n \geq 1$, $0 \leq i < j \leq n + 1$ define $\delta^{n + 1}_{i, j} : [n - 1] \to [n + 1]$ to be the increasing map omitting $i$ and $j$. Note that $\delta^{n + 1}_{i, j} = \delta^{n + 1}_j \circ \delta^n_i = \delta^{n + 1}_i \circ \delta^n_{j - 1}$, see Lemma 14.2.3. This motivates the following lemma.

Lemma 14.19.5. Let $n$ be an integer $\geq 1$. Let $U$ be a $n$-truncated simplicial object of $\mathcal{C}$. Consider the contravariant functor from $\mathcal{C}$ to $\textit{Sets}$ which associates to an object $T$ the set $$\{ (f_0, \ldots, f_{n + 1}) \in \mathop{\rm Mor}\nolimits_\mathcal{C}(T, U_n) \mid d^n_{j - 1} \circ f_i = d^n_i \circ f_j ~\forall~0\leq i < j\leq n + 1\}$$ If this functor is representable by some object $U_{n + 1}$ of $\mathcal{C}$, then $$U_{n + 1} = \mathop{\rm lim}\nolimits_{(\Delta/[n + 1])_{\leq n}^{opp}} U(n)$$

Proof. The limit, if it exists, represents the functor that associates to an object $T$ the set $$\{ (f_\alpha)_{\alpha : [k] \to [n + 1], k \leq n} \mid f_{\alpha \circ \psi} = U(\psi) \circ f_\alpha~\forall ~\psi : [k'] \to [k], \alpha : [k] \to [n + 1] \}.$$ In fact we will show this functor is isomorphic to the one displayed in the lemma. The map in one direction is given by the rule $$(f_\alpha)_{\alpha} \longmapsto (f_{\delta^{n + 1}_0}, \ldots, f_{\delta^{n + 1}_{n + 1}}).$$ This satisfies the conditions of the lemma because $$d^n_{j - 1} \circ f_{\delta^{n + 1}_i} = f_{\delta^{n + 1}_i \circ \delta^n_{j - 1}} = f_{\delta^{n + 1}_j \circ \delta^n_i} = d^n_i \circ f_{\delta^{n + 1}_j}$$ by the relations we recalled above the lemma. To construct a map in the other direction we have to associate to a system $(f_0, \ldots, f_{n + 1})$ as in the displayed formula of the lemma a system of maps $f_\alpha$. Let $\alpha : [k] \to [n + 1]$ be given. Since $k \leq n$ the map $\alpha$ is not surjective. Hence we can write $\alpha = \delta^{n + 1}_i \circ \psi$ for some $0 \leq i \leq n + 1$ and some $\psi : [k] \to [n]$. We have no choice but to define $$f_\alpha = U(\psi) \circ f_i.$$ Of course we have to check that this is independent of the choice of the pair $(i, \psi)$. First, observe that given $i$ there is a unique $\psi$ which works. Second, suppose that $(j, \phi)$ is another pair. Then $i \not = j$ and we may assume $i < j$. Since both $i, j$ are not in the image of $\alpha$ we may actually write $\alpha = \delta^{n + 1}_{i, j} \circ \xi$ and then we see that $\psi = \delta^n_{j - 1} \circ \xi$ and $\phi = \delta^n_i \circ \xi$. Thus \begin{eqnarray*} U(\psi) \circ f_i & = & U(\delta^n_{j - 1} \circ \xi) \circ f_i \\ & = & U(\xi) \circ d^n_{j - 1} \circ f_i \\ & = & U(\xi) \circ d^n_i \circ f_j \\ & = & U(\delta^n_i \circ \xi) \circ f_j \\ & = & U(\phi) \circ f_j \end{eqnarray*} as desired. We still have to verify that the maps $f_\alpha$ so defined satisfy the rules of a system of maps $(f_\alpha)_\alpha$. To see this suppose that $\psi : [k'] \to [k]$, $\alpha : [k] \to [n + 1]$ with $k, k' \leq n$. Set $\alpha' = \alpha \circ \psi$. Choose $i$ not in the image of $\alpha$. Then clearly $i$ is not in the image of $\alpha'$ also. Write $\alpha = \delta^{n + 1}_i \circ \phi$ (we cannot use the letter $\psi$ here because we've already used it). Then obviously $\alpha' = \delta^{n + 1}_i \circ \phi \circ \psi$. By construction above we then have $$U(\psi) \circ f_\alpha = U(\psi) \circ U(\phi) \circ f_i = U(\phi \circ \psi) \circ f_i = f_{\alpha \circ \psi} = f_{\alpha'}$$ as desired. We leave to the reader the pleasant task of verifying that our constructions are mutually inverse bijections, and are functorial in $T$. $\square$

Lemma 14.19.6. Let $n$ be an integer $\geq 1$. Let $U$ be a $n$-truncated simplicial object of $\mathcal{C}$. Consider the contravariant functor from $\mathcal{C}$ to $\textit{Sets}$ which associates to an object $T$ the set $$\{ (f_0, \ldots, f_{n + 1}) \in \mathop{\rm Mor}\nolimits_\mathcal{C}(T, U_n) \mid d^n_{j - 1} \circ f_i = d^n_i \circ f_j ~\forall~0\leq i < j\leq n + 1\}$$ If this functor is representable by some object $U_{n + 1}$ of $\mathcal{C}$, then there exists an $(n + 1)$-truncated simplicial object $\tilde U$, with $\text{sk}_n \tilde U = U$ and $\tilde U_{n + 1} = U_{n + 1}$ such that the following adjointness holds $$\mathop{\rm Mor}\nolimits_{\text{Simp}_{n + 1}(\mathcal{C})}(V, \tilde U) = \mathop{\rm Mor}\nolimits_{\text{Simp}_n(\mathcal{C})}(\text{sk}_nV, U)$$

Proof. By Lemma 14.19.3 there are identifications $$U_i = \mathop{\rm lim}\nolimits_{(\Delta/[i])_{\leq n}^{opp}} U(i)$$ for $0 \leq i \leq n$. By Lemma 14.19.5 we have $$U_{n + 1} = \mathop{\rm lim}\nolimits_{(\Delta/[n + 1])_{\leq n}^{opp}} U(n).$$ Thus we may define for any $\varphi : [i] \to [j]$ with $i, j \leq n + 1$ the corresponding map $\tilde U(\varphi) : \tilde U_j \to \tilde U_i$ exactly as in Lemma 14.19.2. This defines an $(n + 1)$-truncated simplicial object $\tilde U$ with $\text{sk}_n \tilde U = U$.

To see the adjointness we argue as follows. Given any element $\gamma : \text{sk}_n V \to U$ of the right hand side of the formula consider the morphisms $f_i = \gamma_n \circ d^{n + 1}_i : V_{n + 1} \to V_n \to U_n$. These clearly satisfy the relations $d^n_{j - 1} \circ f_i = d^n_i \circ f_j$ and hence define a unique morphism $V_{n + 1} \to U_{n + 1}$ by our choice of $U_{n + 1}$. Conversely, given a morphism $\gamma' : V \to \tilde U$ of the left hand side we can simply restrict to $\Delta_{\leq n}$ to get an element of the right hand side. We leave it to the reader to show these are mutually inverse constructions. $\square$

Remark 14.19.7. Let $U$, and $U_{n + 1}$ be as in Lemma 14.19.6. On $T$-valued points we can easily describe the face and degeneracy maps of $\tilde U$. Explicitly, the maps $d^{n + 1}_i : U_{n + 1} \to U_n$ are given by $$(f_0, \ldots, f_{n + 1}) \longmapsto f_i.$$ And the maps $s^n_j : U_n \to U_{n + 1}$ are given by \begin{eqnarray*} f & \longmapsto & ( s^{n - 1}_{j - 1} \circ d^{n - 1}_0 \circ f, \\ & & s^{n - 1}_{j - 1} \circ d^{n - 1}_1 \circ f, \\ & & \ldots\\ & & s^{n - 1}_{j - 1} \circ d^{n - 1}_{j - 1} \circ f, \\ & & f, \\ & & f, \\ & & s^{n - 1}_j \circ d^{n - 1}_{j + 1} \circ f, \\ & & s^{n - 1}_j \circ d^{n - 1}_{j + 2} \circ f, \\ & & \ldots\\ & & s^{n - 1}_j \circ d^{n - 1}_n \circ f ) \end{eqnarray*} where we leave it to the reader to verify that the RHS is an element of the displayed set of Lemma 14.19.6. For $n = 0$ there is one map, namely $f \mapsto (f, f)$. For $n = 1$ there are two maps, namely $f \mapsto (f, f, s_0d_1f)$ and $f \mapsto (s_0d_0f, f, f)$. For $n = 2$ there are three maps, namely $f \mapsto (f, f, s_0d_1f, s_0d_2f)$, $f \mapsto (s_0d_0f, f, f, s_1d_2f)$, and $f \mapsto (s_1d_0f, s_1d_1f, f, f)$. And so on and so forth.

Remark 14.19.8. The construction of Lemma 14.19.6 above in the case of simplicial sets is the following. Given an $n$-truncated simplicial set $U$, we make a canonical $(n + 1)$-truncated simplicial set $\tilde U$ as follows. We add a set of $(n + 1)$-simplices $U_{n + 1}$ by the formula of the lemma. Namely, an element of $U_{n + 1}$ is a numbered collection of $(f_0, \ldots, f_{n + 1})$ of $n$-simplices, with the property that they glue as they would in a $(n + 1)$-simplex. In other words, the $i$th face of $f_j$ is the $(j-1)$st face of $f_i$ for $i < j$. Geometrically it is obvious how to define the face and degeneracy maps for $\tilde U$. If $V$ is an $(n + 1)$-truncated simplicial set, then its $(n + 1)$-simplices give rise to compatible collections of $n$-simplices $(f_0, \ldots, f_{n + 1})$ with $f_i \in V_n$. Hence there is a natural map $\mathop{\rm Mor}\nolimits(\text{sk}_nV, U) \to \mathop{\rm Mor}\nolimits(V, \tilde U)$ which is inverse to the canonical restriction mapping the other way.

Also, it is enough to do the combinatorics of the construction in the case of truncated simplicial sets. Namely, for any object $T$ of the category $\mathcal{C}$, and any $n$-truncated simplicial object $U$ of $\mathcal{C}$ we can consider the $n$-truncated simplicial set $\mathop{\rm Mor}\nolimits(T, U)$. We may apply the construction to this, and take its set of $(n + 1)$-simplices, and require this to be representable. This is a good way to think about the result of Lemma 14.19.6.

Remark 14.19.9. Inductive construction of coskeleta. Suppose that $\mathcal{C}$ is a category with finite limits. Suppose that $U$ is an $m$-truncated simplicial object in $\mathcal{C}$. Then we can inductively construct $n$-truncated objects $U^n$ as follows:

1. To start, set $U^m = U$.
2. Given $U^n$ for $n \geq m$ set $U^{n + 1} = \tilde U^n$, where $\tilde U^n$ is constructed from $U^n$ as in Lemma 14.19.6.

Since the construction of Lemma 14.19.6 has the property that it leaves the $n$-skeleton of $U^n$ unchanged, we can then define $\text{cosk}_m U$ to be the simplicial object with $(\text{cosk}_m U)_n = U^n_n = U^{n + 1}_n = \ldots$. And it follows formally from Lemma 14.19.6 that $U^n$ satisfies the formula $$\mathop{\rm Mor}\nolimits_{\text{Simp}_n(\mathcal{C})}(V, U^n) = \mathop{\rm Mor}\nolimits_{\text{Simp}_m(\mathcal{C})}(\text{sk}_mV, U)$$ for all $n \geq m$. It also then follows formally from this that $$\mathop{\rm Mor}\nolimits_{\text{Simp}(\mathcal{C})}(V, \text{cosk}_mU) = \mathop{\rm Mor}\nolimits_{\text{Simp}_m(\mathcal{C})}(\text{sk}_mV, U)$$ with $\text{cosk}_mU$ chosen as above.

Lemma 14.19.10. Let $\mathcal{C}$ be a category which has finite limits.

1. For every $n$ the functor $\text{sk}_n : \text{Simp}(\mathcal{C}) \to \text{Simp}_n(\mathcal{C})$ has a right adjoint $\text{cosk}_n$.
2. For every $n' \geq n$ the functor $\text{sk}_n : \text{Simp}_{n'}(\mathcal{C}) \to \text{Simp}_n(\mathcal{C})$ has a right adjoint, namely $\text{sk}_{n'}\text{cosk}_n$.
3. For every $m \geq n \geq 0$ and every $n$-truncated simplicial object $U$ of $\mathcal{C}$ we have $\text{cosk}_m \text{sk}_m \text{cosk}_n U = \text{cosk}_n U$.
4. If $U$ is a simplicial object of $\mathcal{C}$ such that the canonical map $U \to \text{cosk}_n \text{sk}_nU$ is an isomorphism for some $n \geq 0$, then the canonical map $U \to \text{cosk}_m \text{sk}_mU$ is an isomorphism for all $m \geq n$.

Proof. The existence in (1) follows from Lemma 14.19.2 above. Parts (2) and (3) follow from the discussion in Remark 14.19.9. After this (4) is obvious. $\square$

Remark 14.19.11. We do not need all finite limits in order to be able to define the coskeleton functors. Here are some remarks

1. We have seen in Examples 14.19.1 that if $\mathcal{C}$ has products of pairs of objects then $\text{cosk}_0$ exists.
2. For $k > 0$ the functor $\text{cosk}_k$ exists if $\mathcal{C}$ has finite connected limits.

This is clear from the inductive procedure of constructing coskeleta (Remarks 14.19.8 and 14.19.9) but it also follows from the fact that the categories $(\Delta/[n])_{\leq k}$ for $k \geq 1$ and $n \geq k + 1$ used in Lemma 14.19.2 are connected. Observe that we do not need the categories for $n \leq k$ by Lemma 14.19.3 or Lemma 14.19.4. (As $k$ gets higher the categories $(\Delta/[n])_{\leq k}$ for $k \geq 1$ and $n \geq k + 1$ are more and more connected in a topological sense.)

Lemma 14.19.12. Let $U$, $V$ be $n$-truncated simplicial objects of a category $\mathcal{C}$. Then $$\text{cosk}_n (U \times V) = \text{cosk}_nU \times \text{cosk}_nV$$ whenever the left and right hand sides exist.

Proof. Let $W$ be a simplicial object. We have \begin{eqnarray*} \mathop{\rm Mor}\nolimits(W, \text{cosk}_n (U \times V)) & = & \mathop{\rm Mor}\nolimits(\text{sk}_n W, U \times V) \\ & = & \mathop{\rm Mor}\nolimits(\text{sk}_n W, U) \times \mathop{\rm Mor}\nolimits(\text{sk}_nW, V) \\ & = & \mathop{\rm Mor}\nolimits(W, \text{cosk}_n U) \times \mathop{\rm Mor}\nolimits(W, \text{cosk}_n V) \\ & = & \mathop{\rm Mor}\nolimits(W, \text{cosk}_n U \times \text{cosk}_n V) \end{eqnarray*} The lemma follows. $\square$

Lemma 14.19.13. Assume $\mathcal{C}$ has fibre products. Let $U, V, W$ be $n$-truncated simplicial objects of the category $\mathcal{C}$. Then $$\text{cosk}_n (V \times_U W) = \text{cosk}_nU \times_{\text{cosk}_n U} \text{cosk}_nV$$ whenever the left and right hand side exist.

Proof. Omitted, but very similar to the proof of Lemma 14.19.12 above. $\square$

Lemma 14.19.14. Let $\mathcal{C}$ be a category with finite limits. Let $X \in \mathop{\rm Ob}\nolimits(\mathcal{C})$. The functor $\mathcal{C}/X \to \mathcal{C}$ commutes with the coskeleton functors $\text{cosk}_k$ for $k \geq 1$.

Proof. The statement means that if $U$ is a simplicial object of $\mathcal{C}/X$ which we can think of as a simplicial object of $\mathcal{C}$ with a morphism towards the constant simplicial object $X$, then $\text{cosk}_k U$ computed in $\mathcal{C}/X$ is the same as computed in $\mathcal{C}$. This follows for example from Categories, Lemma 4.16.2 because the categories $(\Delta/[n])_{\leq k}$ for $k \geq 1$ and $n \geq k + 1$ used in Lemma 14.19.2 are connected. Observe that we do not need the categories for $n \leq k$ by Lemma 14.19.3 or Lemma 14.19.4. $\square$

Lemma 14.19.15. The canonical map $\Delta[n] \to \text{cosk}_1 \text{sk}_1 \Delta[n]$ is an isomorphism.

Proof. Consider a simplicial set $U$ and a morphism $f : U \to \Delta[n]$. This is a rule that associates to each $u \in U_i$ a map $f_u : [i] \to [n]$ in $\Delta$. Furthermore, these maps should have the property that $f_u \circ \varphi = f_{U(\varphi)(u)}$ for any $\varphi : [j] \to [i]$. Denote $\epsilon^i_j : [0] \to [i]$ the map which maps $0$ to $j$. Denote $F : U_0 \to [n]$ the map $u \mapsto f_u(0)$. Then we see that $$f_u(j) = F(\epsilon^i_j(u))$$ for all $0 \leq j \leq i$ and $u \in U_i$. In particular, if we know the function $F$ then we know the maps $f_u$ for all $u\in U_i$ all $i$. Conversely, given a map $F : U_0 \to [n]$, we can set for any $i$, and any $u \in U_i$ and any $0 \leq j \leq i$ $$f_u(j) = F(\epsilon^i_j(u))$$ This does not in general define a morphism $f$ of simplicial sets as above. Namely, the condition is that all the maps $f_u$ are nondecreasing. This clearly is equivalent to the condition that $F(\epsilon^i_j(u)) \leq F(\epsilon^i_{j'}(u))$ whenever $0 \leq j \leq j' \leq i$ and $u \in U_i$. But in this case the morphisms $$\epsilon^i_j, \epsilon^i_{j'} : [0] \to [i]$$ both factor through the map $\epsilon^i_{j, j'} : [1] \to [i]$ defined by the rules $0 \mapsto j$, $1 \mapsto j'$. In other words, it is enough to check the inequalities for $i = 1$ and $u \in X_1$. In other words, we have $$\mathop{\rm Mor}\nolimits(U, \Delta[n]) = \mathop{\rm Mor}\nolimits(\text{sk}_1 U, \text{sk}_1 \Delta[n])$$ as desired. $\square$

The code snippet corresponding to this tag is a part of the file simplicial.tex and is located in lines 2130–2761 (see updates for more information).

\section{Coskeleton functors}
\label{section-coskeleton}

\noindent
Let $\mathcal{C}$ be a category.
The {\it coskeleton functor} (if it exists) is a functor
$$\text{cosk}_n : \text{Simp}_n(\mathcal{C}) \longrightarrow \text{Simp}(\mathcal{C})$$
which is right adjoint to the skeleton functor. In a formula

\label{equation-cosk}
\Mor_{\text{Simp}(\mathcal{C})}(U, \text{cosk}_n V)
=
\Mor_{\text{Simp}_n(\mathcal{C})}(\text{sk}_n U, V)

Given a $n$-truncated simplicial object $V$ we
say that {\it $\text{cosk}_nV$ exists} if there
exists a $\text{cosk}_nV \in \Ob(\text{Simp}(\mathcal{C}))$
and a morphism $\text{sk}_n \text{cosk}_n V \to V$
such that the displayed formula holds, in other words
if the functor
$U \mapsto \Mor_{\text{Simp}_n(\mathcal{C})}(\text{sk}_n U, V)$
is representable. If it exists it
is unique up to unique isomorphism by the Yoneda lemma.
See Categories, Section \ref{categories-section-opposite}.

\begin{example}
\label{example-cosk0}
Suppose the category $\mathcal{C}$ has finite nonempty self products.
A $0$-truncated simplicial object of $\mathcal{C}$ is the same
as an object $X$ of $\mathcal{C}$. In this case
we claim that $\text{cosk}_0(X)$ is the simplicial
object $U$ with $U_n = X^{n + 1}$ the $(n + 1)$-fold self
product of $X$, and structure of simplicial object
as in Example \ref{example-fibre-products-simplicial-object}.
Namely, a morphism $V \to U$ where $V$ is a simplicial
object is given by morphisms $V_n \to X^{n + 1}$, such
that all the diagrams
$$\xymatrix{ V_n \ar[r] \ar[d]_{V([0] \to [n], 0 \mapsto i)} & X^{n + 1} \ar[d]^{\text{pr}_i} \\ V_0 \ar[r] & X }$$
commute. Clearly this means that the map determines and is determined
by a unique morphism $V_0 \to X$. This proves that formula
(\ref{equation-cosk}) holds.
\end{example}

\noindent
Recall the category $\Delta/[n]$, see Example \ref{example-simplex-category}.
We let $(\Delta/[n])_{\leq m}$ denote the full subcategory
of $\Delta/[n]$ consisting of objects $[k] \to [n]$
of $\Delta/[n]$ with $k \leq m$. In other words we have
the following commutative diagram of categories and functors
$$\xymatrix{ (\Delta/[n])_{\leq m} \ar[r] \ar[d] & \Delta/[n] \ar[d] \\ \Delta_{\leq m} \ar[r] & \Delta }$$
Given a $m$-truncated
simplicial object $U$ of $\mathcal{C}$
we define a functor
$$U(n) : (\Delta/[n])_{\leq m}^{opp} \longrightarrow \mathcal{C}$$
by the rules
\begin{eqnarray*}
([k] \to [n]) & \longmapsto & U_k \\
\psi : ([k'] \to [n]) \to ([k] \to [n]) &
\longmapsto &
U(\psi) : U_k \to U_{k'}
\end{eqnarray*}
For a given morphism $\varphi : [n] \to [n']$ of $\Delta$
we have an associated functor
$$\overline{\varphi} : (\Delta/[n])_{\leq m} \longrightarrow (\Delta/[n'])_{\leq m}$$
which maps $\alpha : [k] \to [n]$ to
$\varphi \circ \alpha : [k] \to [n']$.
The composition $U(n') \circ \overline{\varphi}$ is
equal to the functor $U(n)$.

\begin{lemma}
\label{lemma-existence-cosk}
If the category $\mathcal{C}$ has finite limits, then
$\text{cosk}_m$ functors exist for all $m$. Moreover,
for any $m$-truncated simplicial object $U$ the
simplicial object $\text{cosk}_mU$ is described
by the formula
$$(\text{cosk}_mU)_n = \lim_{(\Delta/[n])_{\leq m}^{opp}} U(n)$$
and for $\varphi : [n] \to [n']$ the map
$\text{cosk}_mU(\varphi)$ comes from the
identification $U(n') \circ \overline{\varphi} = U(n)$ above
via Categories, Lemma \ref{categories-lemma-functorial-limit}.
\end{lemma}

\begin{proof}
During the proof of this lemma we denote $\text{cosk}_mU$ the
simplicial object with $(\text{cosk}_mU)_n$ equal to
$\lim_{(\Delta/[n])_{\leq m}^{opp}} U(n)$.
We will conclude at the end of the proof that it does
satisfy the required mapping property.

\medskip\noindent
Suppose that $V$ is a simplicial object.
A morphism $\gamma : V \to \text{cosk}_mU$ is given by a sequence
of morphisms $\gamma_n : V_n \to (\text{cosk}_mU)_n$.
By definition of a limit, this is given by a
collection of morphisms $\gamma(\alpha) : V_n \to U_k$
where $\alpha$ ranges over all $\alpha : [k] \to [n]$
with $k \leq m$. These morphisms then also satisfy
the rules that
$$\xymatrix{ V_n \ar[r]_{\gamma(\alpha)} & U_k \\ V_{n'} \ar[r]^{\gamma(\alpha')} \ar[u]^{V(\varphi)} & U_{k'} \ar[u]_{U(\psi)} }$$
are commutative, given any $0 \leq k, k' \leq m$, $0 \leq n, n'$
and any $\psi : [k] \to [k']$, $\varphi : [n] \to [n']$,
$\alpha : [k] \to [n]$ and $\alpha' : [k'] \to [n']$ in $\Delta$
such that $\varphi \circ \alpha = \alpha' \circ \psi$.
Taking $n = k$, $\varphi = \alpha'$, and $\alpha = \psi = \text{id}_{[k]}$
we deduce that $\gamma(\alpha') = \gamma(\text{id}_{[k]}) \circ V(\alpha')$.
In other words, the morphisms $\gamma(\text{id}_{[k]})$, $k \leq m$
determine the morphism $\gamma$. And it is easy to see that these
morphisms form a morphism $\text{sk}_m V \to U$.

\medskip\noindent
Conversely, given a morphism $\gamma : \text{sk}_m V \to U$,
we obtain a family of morphisms $\gamma(\alpha)$
where $\alpha$ ranges over all $\alpha : [k] \to [n]$
with $k \leq m$ by setting $\gamma(\alpha) = \gamma(\text{id}_{[k]}) \circ V(\alpha)$. These morphisms
satisfy all the displayed commutativity restraints pictured
above, and hence give rise to a morphism $V \to \text{cosk}_m U$.
\end{proof}

\begin{lemma}
\label{lemma-trivial-cosk}
Let $\mathcal{C}$ be a category.
Let $U$ be an $m$-truncated simplicial object of $\mathcal{C}$.
For $n \leq m$ the limit $\lim_{(\Delta/[n])_{\leq m}^{opp}} U(n)$
exists and is canonically isomorphic to $U_n$.
\end{lemma}

\begin{proof}
This is true because the category $(\Delta/[n])_{\leq m}$
has an final object in this case, namely the identity
map $[n] \to [n]$.
\end{proof}

\begin{lemma}
\label{lemma-recover-cosk}
Let $\mathcal{C}$ be a category with finite limits.
Let $U$ be an $n$-truncated simplicial object of $\mathcal{C}$.
The morphism $\text{sk}_n \text{cosk}_n U \to U$
is an isomorphism.
\end{lemma}

\begin{proof}
Combine Lemmas \ref{lemma-existence-cosk} and \ref{lemma-trivial-cosk}.
\end{proof}

\noindent
Let us describe a particular instance of the coskeleton functor in more detail.
By abuse of notation we will denote $\text{sk}_n$
also the restriction functor
$\text{Simp}_{n'}(\mathcal{C}) \to \text{Simp}_n(\mathcal{C})$
for any $n' \geq n$. We are going to describe a right adjoint
of the functor
$\text{sk}_n : \text{Simp}_{n + 1}(\mathcal{C}) \to \text{Simp}_n(\mathcal{C})$.
For $n \geq 1$, $0 \leq i < j \leq n + 1$
define $\delta^{n + 1}_{i, j} : [n - 1] \to [n + 1]$
to be the increasing map omitting $i$ and $j$.
Note that
$\delta^{n + 1}_{i, j} = \delta^{n + 1}_j \circ \delta^n_i = \delta^{n + 1}_i \circ \delta^n_{j - 1}$, see
Lemma \ref{lemma-relations-face-degeneracy}. This motivates
the following lemma.

\begin{lemma}
\label{lemma-formula-limit}
Let $n$ be an integer $\geq 1$.
Let $U$ be a $n$-truncated simplicial object of $\mathcal{C}$.
Consider the contravariant functor from $\mathcal{C}$ to
$\textit{Sets}$ which associates to an object $T$ the set
$$\{ (f_0, \ldots, f_{n + 1}) \in \Mor_\mathcal{C}(T, U_n) \mid d^n_{j - 1} \circ f_i = d^n_i \circ f_j \ \forall\ 0\leq i < j\leq n + 1\}$$
If this functor is representable by some object $U_{n + 1}$
of $\mathcal{C}$, then
$$U_{n + 1} = \lim_{(\Delta/[n + 1])_{\leq n}^{opp}} U(n)$$
\end{lemma}

\begin{proof}
The limit, if it exists, represents the functor
that associates to an object $T$ the set
$$\{ (f_\alpha)_{\alpha : [k] \to [n + 1], k \leq n} \mid f_{\alpha \circ \psi} = U(\psi) \circ f_\alpha\ \forall \ \psi : [k'] \to [k], \alpha : [k] \to [n + 1] \}.$$
In fact we will show this functor is isomorphic to the
one displayed in the lemma. The map in one direction
is given by the rule
$$(f_\alpha)_{\alpha} \longmapsto (f_{\delta^{n + 1}_0}, \ldots, f_{\delta^{n + 1}_{n + 1}}).$$
This satisfies the conditions of the lemma because
$$d^n_{j - 1} \circ f_{\delta^{n + 1}_i} = f_{\delta^{n + 1}_i \circ \delta^n_{j - 1}} = f_{\delta^{n + 1}_j \circ \delta^n_i} = d^n_i \circ f_{\delta^{n + 1}_j}$$
by the relations we recalled above the lemma. To construct a map
in the other direction we have to associate to a system
$(f_0, \ldots, f_{n + 1})$ as in the displayed formula
of the lemma a system of maps $f_\alpha$. Let $\alpha : [k] \to [n + 1]$
be given. Since $k \leq n$ the map $\alpha$ is not surjective.
Hence we can write $\alpha = \delta^{n + 1}_i \circ \psi$
for some $0 \leq i \leq n + 1$ and some
$\psi : [k] \to [n]$. We have no choice but to define
$$f_\alpha = U(\psi) \circ f_i.$$
Of course we have to check that this is independent of the
choice of the pair $(i, \psi)$. First, observe that given $i$
there is a unique $\psi$ which works. Second, suppose that $(j, \phi)$ is
another pair. Then $i \not = j$ and we may assume $i < j$. Since
both $i, j$ are not in the image of $\alpha$ we may actually
write $\alpha = \delta^{n + 1}_{i, j} \circ \xi$ and then
we see that $\psi = \delta^n_{j - 1} \circ \xi$ and
$\phi = \delta^n_i \circ \xi$. Thus
\begin{eqnarray*}
U(\psi) \circ f_i & = & U(\delta^n_{j - 1} \circ \xi) \circ f_i \\
& = & U(\xi) \circ d^n_{j - 1} \circ f_i \\
& = & U(\xi) \circ d^n_i \circ f_j \\
& = & U(\delta^n_i \circ \xi) \circ f_j \\
& = & U(\phi) \circ f_j
\end{eqnarray*}
as desired. We still have to verify that the maps
$f_\alpha$ so defined satisfy the rules of a system
of maps $(f_\alpha)_\alpha$. To see this suppose that
$\psi : [k'] \to [k]$, $\alpha : [k] \to [n + 1]$ with
$k, k' \leq n$. Set $\alpha' = \alpha \circ \psi$.
Choose $i$ not in the image of $\alpha$. Then clearly
$i$ is not in the image of $\alpha'$ also. Write
$\alpha = \delta^{n + 1}_i \circ \phi$ (we cannot use the letter $\psi$ here
because we've already used it). Then obviously
$\alpha' = \delta^{n + 1}_i \circ \phi \circ \psi$. By construction above
we then have
$$U(\psi) \circ f_\alpha = U(\psi) \circ U(\phi) \circ f_i = U(\phi \circ \psi) \circ f_i = f_{\alpha \circ \psi} = f_{\alpha'}$$
as desired. We leave to the reader the pleasant task of verifying
that our constructions are mutually inverse bijections, and are
functorial in $T$.
\end{proof}

\begin{lemma}
\label{lemma-work-out}
Let $n$ be an integer $\geq 1$. Let $U$ be a $n$-truncated
simplicial object of $\mathcal{C}$. Consider the
contravariant functor from $\mathcal{C}$ to $\textit{Sets}$
which associates to an object $T$ the set
$$\{ (f_0, \ldots, f_{n + 1}) \in \Mor_\mathcal{C}(T, U_n) \mid d^n_{j - 1} \circ f_i = d^n_i \circ f_j \ \forall\ 0\leq i < j\leq n + 1\}$$
If this functor is representable by some object $U_{n + 1}$
of $\mathcal{C}$, then there exists an $(n + 1)$-truncated
simplicial object $\tilde U$, with $\text{sk}_n \tilde U = U$
and $\tilde U_{n + 1} = U_{n + 1}$ such that the following
$$\Mor_{\text{Simp}_{n + 1}(\mathcal{C})}(V, \tilde U) = \Mor_{\text{Simp}_n(\mathcal{C})}(\text{sk}_nV, U)$$
\end{lemma}

\begin{proof}
By Lemma \ref{lemma-trivial-cosk} there are identifications
$$U_i = \lim_{(\Delta/[i])_{\leq n}^{opp}} U(i)$$
for $0 \leq i \leq n$. By Lemma \ref{lemma-formula-limit}
we have
$$U_{n + 1} = \lim_{(\Delta/[n + 1])_{\leq n}^{opp}} U(n).$$
Thus we may define for any $\varphi : [i] \to [j]$
with $i, j \leq n + 1$ the corresponding map
$\tilde U(\varphi) : \tilde U_j \to \tilde U_i$ exactly as
in Lemma \ref{lemma-existence-cosk}. This defines
an $(n + 1)$-truncated simplicial object $\tilde U$
with $\text{sk}_n \tilde U = U$.

\medskip\noindent
To see the adjointness we argue as follows. Given any element
$\gamma : \text{sk}_n V \to U$ of the right hand side of the formula
consider the morphisms
$f_i = \gamma_n \circ d^{n + 1}_i : V_{n + 1} \to V_n \to U_n$.
These clearly satisfy the relations $d^n_{j - 1} \circ f_i = d^n_i \circ f_j$
and hence define a unique morphism $V_{n + 1} \to U_{n + 1}$
by our choice of $U_{n + 1}$.
Conversely, given a morphism $\gamma' : V \to \tilde U$
of the left hand side we can simply restrict to
$\Delta_{\leq n}$ to get an element of the right hand side.
We leave it to the reader to show these are mutually inverse
constructions.
\end{proof}

\begin{remark}
\label{remark-explicit-face-degeneracy}
Let $U$, and $U_{n + 1}$ be as in Lemma \ref{lemma-work-out}.
On $T$-valued points we can easily describe the face
and degeneracy maps of $\tilde U$.
Explicitly, the maps $d^{n + 1}_i : U_{n + 1} \to U_n$
are given by
$$(f_0, \ldots, f_{n + 1}) \longmapsto f_i.$$
And the maps $s^n_j : U_n \to U_{n + 1}$ are given by
\begin{eqnarray*}
f & \longmapsto & (
s^{n - 1}_{j - 1} \circ d^{n - 1}_0 \circ f, \\
& &
s^{n - 1}_{j - 1} \circ d^{n - 1}_1 \circ f, \\
& &
\ldots\\
& &
s^{n - 1}_{j - 1} \circ d^{n - 1}_{j - 1} \circ f, \\
& &
f, \\
& &
f, \\
& &
s^{n - 1}_j \circ d^{n - 1}_{j + 1} \circ f, \\
& &
s^{n - 1}_j \circ d^{n - 1}_{j + 2} \circ f, \\
& &
\ldots\\
& &
s^{n - 1}_j \circ d^{n - 1}_n \circ f
)
\end{eqnarray*}
where we leave it to the reader to verify that the RHS
is an element of the displayed set of Lemma \ref{lemma-work-out}.
For $n = 0$ there is one map, namely $f \mapsto (f, f)$.
For $n = 1$ there are two maps, namely
$f \mapsto (f, f, s_0d_1f)$ and
$f \mapsto (s_0d_0f, f, f)$.
For $n = 2$ there are three maps, namely
$f \mapsto (f, f, s_0d_1f, s_0d_2f)$,
$f \mapsto (s_0d_0f, f, f, s_1d_2f)$, and
$f \mapsto (s_1d_0f, s_1d_1f, f, f)$.
And so on and so forth.
\end{remark}

\begin{remark}
\label{remark-cosk-simplicial-sets}
The construction of Lemma \ref{lemma-work-out}
above in the case of simplicial
sets is the following. Given an $n$-truncated simplicial
set $U$, we make a canonical $(n + 1)$-truncated simplicial
set $\tilde U$ as follows. We add a set of $(n + 1)$-simplices
$U_{n + 1}$ by the formula of the lemma. Namely,
an element of $U_{n + 1}$ is a numbered collection of
$(f_0, \ldots, f_{n + 1})$ of $n$-simplices,
with the property that they glue
as they would in a $(n + 1)$-simplex. In other words,
the $i$th face of $f_j$ is the $(j-1)$st face of $f_i$
for $i < j$. Geometrically it is obvious how to define the
face and degeneracy maps for $\tilde U$.
If $V$ is an $(n + 1)$-truncated simplicial set,
then its $(n + 1)$-simplices give rise to compatible collections
of $n$-simplices $(f_0, \ldots, f_{n + 1})$ with $f_i \in V_n$.
Hence there is a natural map
$\Mor(\text{sk}_nV, U) \to \Mor(V, \tilde U)$
which is inverse to the canonical restriction mapping
the other way.

\medskip\noindent
Also, it is enough to do the combinatorics of the
construction in the case of truncated simplicial sets.
Namely, for any object $T$ of the category $\mathcal{C}$,
and any $n$-truncated simplicial object $U$ of $\mathcal{C}$
we can consider the $n$-truncated simplicial set
$\Mor(T, U)$. We may apply the construction to this,
and take its set of $(n + 1)$-simplices, and require this to be
representable. This is a good way to think about
the result of Lemma \ref{lemma-work-out}.
\end{remark}

\begin{remark}
\label{remark-inductive-coskeleton}
{\it Inductive construction of coskeleta.}
Suppose that $\mathcal{C}$ is a category with
finite limits. Suppose that $U$ is an $m$-truncated
simplicial object in $\mathcal{C}$. Then we can
inductively construct $n$-truncated objects $U^n$ as
follows:
\begin{enumerate}
\item To start, set $U^m = U$.
\item Given $U^n$ for $n \geq m$ set $U^{n + 1} = \tilde U^n$,
where $\tilde U^n$ is constructed from $U^n$ as in Lemma
\ref{lemma-work-out}.
\end{enumerate}
Since the construction of Lemma \ref{lemma-work-out} has
the property that it leaves the $n$-skeleton of $U^n$
unchanged, we can then define $\text{cosk}_m U$ to be
the simplicial object with
$(\text{cosk}_m U)_n = U^n_n = U^{n + 1}_n = \ldots$.
And it follows formally from Lemma \ref{lemma-work-out}
that $U^n$ satisfies the formula
$$\Mor_{\text{Simp}_n(\mathcal{C})}(V, U^n) = \Mor_{\text{Simp}_m(\mathcal{C})}(\text{sk}_mV, U)$$
for all $n \geq m$. It also then follows formally
from this that
$$\Mor_{\text{Simp}(\mathcal{C})}(V, \text{cosk}_mU) = \Mor_{\text{Simp}_m(\mathcal{C})}(\text{sk}_mV, U)$$
with $\text{cosk}_mU$ chosen as above.
\end{remark}

\begin{lemma}
\label{lemma-cosk-up}
Let $\mathcal{C}$ be a category which has finite limits.
\begin{enumerate}
\item For every $n$ the functor $\text{sk}_n : \text{Simp}(\mathcal{C}) \to \text{Simp}_n(\mathcal{C})$ has a right adjoint $\text{cosk}_n$.
\item For every $n' \geq n$ the functor
$\text{sk}_n : \text{Simp}_{n'}(\mathcal{C}) \to \text{Simp}_n(\mathcal{C})$
has a right adjoint, namely $\text{sk}_{n'}\text{cosk}_n$.
\item For every $m \geq n \geq 0$ and every $n$-truncated simplicial
object $U$ of $\mathcal{C}$ we have
$\text{cosk}_m \text{sk}_m \text{cosk}_n U = \text{cosk}_n U$.
\item If $U$ is a simplicial object of $\mathcal{C}$ such that
the canonical map
$U \to \text{cosk}_n \text{sk}_nU$
is an isomorphism for some $n \geq 0$, then the canonical map
$U \to \text{cosk}_m \text{sk}_mU$
is an isomorphism for all $m \geq n$.
\end{enumerate}
\end{lemma}

\begin{proof}
The existence in (1) follows from Lemma \ref{lemma-existence-cosk} above.
Parts (2) and (3) follow from the discussion
in Remark \ref{remark-inductive-coskeleton}. After this (4) is obvious.
\end{proof}

\begin{remark}
\label{remark-existence-cosk}
We do not need all finite limits in order to be able to define
the coskeleton functors. Here are some remarks
\begin{enumerate}
\item We have seen in Examples \ref{example-cosk0} that if $\mathcal{C}$
has products of pairs of objects then $\text{cosk}_0$ exists.
\item For $k > 0$ the functor $\text{cosk}_k$ exists if
$\mathcal{C}$ has finite connected limits.
\end{enumerate}
This is clear from the inductive procedure of constructing coskeleta
(Remarks \ref{remark-cosk-simplicial-sets} and
\ref{remark-inductive-coskeleton}) but it also follows from the fact that
the categories $(\Delta/[n])_{\leq k}$ for $k \geq 1$ and
$n \geq k + 1$ used in Lemma \ref{lemma-existence-cosk}
are connected. Observe that we do not need the categories
for $n \leq k$ by Lemma \ref{lemma-trivial-cosk} or
Lemma \ref{lemma-recover-cosk}. (As $k$ gets higher the categories
$(\Delta/[n])_{\leq k}$ for $k \geq 1$ and $n \geq k + 1$ are more
and more connected in a topological sense.)
\end{remark}

\begin{lemma}
\label{lemma-cosk-product}
Let $U$, $V$ be $n$-truncated simplicial objects of a
category $\mathcal{C}$. Then
$$\text{cosk}_n (U \times V) = \text{cosk}_nU \times \text{cosk}_nV$$
whenever the left and right hand sides exist.
\end{lemma}

\begin{proof}
Let $W$ be a simplicial object. We have
\begin{eqnarray*}
\Mor(W, \text{cosk}_n (U \times V))
& = &
\Mor(\text{sk}_n W, U \times V) \\
& = &
\Mor(\text{sk}_n W, U)
\times
\Mor(\text{sk}_nW, V) \\
& = &
\Mor(W, \text{cosk}_n U)
\times
\Mor(W, \text{cosk}_n V) \\
& = &
\Mor(W, \text{cosk}_n U \times \text{cosk}_n V)
\end{eqnarray*}
The lemma follows.
\end{proof}

\begin{lemma}
\label{lemma-cosk-fibre-product}
Assume $\mathcal{C}$ has fibre products.
Let $U, V, W$ be $n$-truncated simplicial objects of the
category $\mathcal{C}$. Then
$$\text{cosk}_n (V \times_U W) = \text{cosk}_nU \times_{\text{cosk}_n U} \text{cosk}_nV$$
whenever the left and right hand side exist.
\end{lemma}

\begin{proof}
Omitted, but very similar to the proof of
Lemma \ref{lemma-cosk-product} above.
\end{proof}

\begin{lemma}
\label{lemma-cosk-above-object}
Let $\mathcal{C}$ be a category with finite limits.
Let $X \in \Ob(\mathcal{C})$.
The functor $\mathcal{C}/X \to \mathcal{C}$ commutes with
the coskeleton functors $\text{cosk}_k$ for $k \geq 1$.
\end{lemma}

\begin{proof}
The statement means that if $U$ is a simplicial object of $\mathcal{C}/X$
which we can think of as a simplicial object of $\mathcal{C}$ with a morphism
towards the constant simplicial object $X$, then $\text{cosk}_k U$
computed in $\mathcal{C}/X$ is the same as computed in $\mathcal{C}$.
This follows for example from
Categories, Lemma \ref{categories-lemma-connected-limit-over-X}
because the categories $(\Delta/[n])_{\leq k}$ for $k \geq 1$ and
$n \geq k + 1$ used in Lemma \ref{lemma-existence-cosk}
are connected. Observe that we do not need the categories
for $n \leq k$ by Lemma \ref{lemma-trivial-cosk} or
Lemma \ref{lemma-recover-cosk}.
\end{proof}

\begin{lemma}
\label{lemma-simplex-cosk}
The canonical map
$\Delta[n] \to \text{cosk}_1 \text{sk}_1 \Delta[n]$
is an isomorphism.
\end{lemma}

\begin{proof}
Consider a simplicial set $U$ and a morphism
$f : U \to \Delta[n]$. This is a rule that associates
to each $u \in U_i$ a map $f_u : [i] \to [n]$ in $\Delta$.
Furthermore, these maps should have the property that
$f_u \circ \varphi = f_{U(\varphi)(u)}$ for any
$\varphi : [j] \to [i]$. Denote $\epsilon^i_j : [0] \to [i]$
the map which maps $0$ to $j$. Denote $F : U_0 \to [n]$
the map $u \mapsto f_u(0)$. Then we see that
$$f_u(j) = F(\epsilon^i_j(u))$$
for all $0 \leq j \leq i$ and $u \in U_i$.
In particular, if we know the function $F$
then we know the maps $f_u$ for all $u\in U_i$ all $i$.
Conversely, given a map $F : U_0 \to [n]$,
we can set for any $i$, and any $u \in U_i$
and any $0 \leq j \leq i$
$$f_u(j) = F(\epsilon^i_j(u))$$
This does not in general define a morphism $f$ of simplicial sets
as above. Namely, the condition is that all the maps $f_u$ are
nondecreasing. This clearly is equivalent to the condition
that $F(\epsilon^i_j(u)) \leq F(\epsilon^i_{j'}(u))$
whenever $0 \leq j \leq j' \leq i$ and $u \in U_i$. But in this
case the morphisms
$$\epsilon^i_j, \epsilon^i_{j'} : [0] \to [i]$$
both factor through the map
$\epsilon^i_{j, j'} : [1] \to [i]$ defined by the rules
$0 \mapsto j$, $1 \mapsto j'$.
In other words, it is enough to check the inequalities for
$i = 1$ and $u \in X_1$. In other words, we have
$$\Mor(U, \Delta[n]) = \Mor(\text{sk}_1 U, \text{sk}_1 \Delta[n])$$
as desired.
\end{proof}

There are no comments yet for this tag.

## Add a comment on tag 0AMA

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the lower-right corner).