## 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

14.19.0.1
\begin{equation} \label{simplicial-equation-cosk} \mathop{Mor}\nolimits _{\text{Simp}(\mathcal{C})}(U, \text{cosk}_ n V) = \mathop{Mor}\nolimits _{\text{Simp}_ n(\mathcal{C})}(\text{sk}_ n U, V) \end{equation}

Given a $n$-truncated simplicial object $V$ we say that *$\text{cosk}_ nV$ exists* if there exists a $\text{cosk}_ nV \in \mathop{\mathrm{Ob}}\nolimits (\text{Simp}(\mathcal{C}))$ and a morphism $\text{sk}_ n \text{cosk}_ n V \to V$ such that the displayed formula holds, in other words if the functor $U \mapsto \mathop{Mor}\nolimits _{\text{Simp}_ n(\mathcal{C})}(\text{sk}_ n U, V)$ 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

\[ (\text{cosk}_ mU)_ n = \mathop{\mathrm{lim}}\nolimits _{(\Delta /[n])_{\leq m}^{opp}} U(n) \]

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 4.14.9.

**Proof.**
During the proof of this lemma we denote $\text{cosk}_ mU$ the simplicial object with $(\text{cosk}_ mU)_ n$ equal to $\mathop{\mathrm{lim}}\nolimits _{(\Delta /[n])_{\leq m}^{opp}} U(n)$. 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 $\mathop{\mathrm{lim}}\nolimits _{(\Delta /[n])_{\leq m}^{opp}} U(n)$ 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

\[ \{ (f_0, \ldots , f_{n + 1}) \in \mathop{Mor}\nolimits _\mathcal {C}(T, U_ n) \mid d^ n_{j - 1} \circ f_ i = d^ n_ i \circ f_ j \ \forall \ 0\leq i < j\leq n + 1\} \]

If this functor is representable by some object $U_{n + 1}$ of $\mathcal{C}$, then

\[ U_{n + 1} = \mathop{\mathrm{lim}}\nolimits _{(\Delta /[n + 1])_{\leq n}^{opp}} U(n) \]

**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

\[ \{ (f_0, \ldots , f_{n + 1}) \in \mathop{Mor}\nolimits _\mathcal {C}(T, U_ n) \mid d^ n_{j - 1} \circ f_ i = d^ n_ i \circ f_ j \ \forall \ 0\leq i < j\leq n + 1\} \]

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

\[ \mathop{Mor}\nolimits _{\text{Simp}_{n + 1}(\mathcal{C})}(V, \tilde U) = \mathop{Mor}\nolimits _{\text{Simp}_ n(\mathcal{C})}(\text{sk}_ nV, U) \]

**Proof.**
By Lemma 14.19.3 there are identifications

\[ U_ i = \mathop{\mathrm{lim}}\nolimits _{(\Delta /[i])_{\leq n}^{opp}} U(i) \]

for $0 \leq i \leq n$. By Lemma 14.19.5 we have

\[ U_{n + 1} = \mathop{\mathrm{lim}}\nolimits _{(\Delta /[n + 1])_{\leq n}^{opp}} U(n). \]

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.
$\square$

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

For every $n$ the functor $\text{sk}_ n : \text{Simp}(\mathcal{C}) \to \text{Simp}_ n(\mathcal{C})$ has a right adjoint $\text{cosk}_ n$.

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 $\text{sk}_{n'}\text{cosk}_ n$.

For every $m \geq n \geq 0$ and every $n$-truncated simplicial object $U$ of $\mathcal{C}$ we have $\text{cosk}_ m \text{sk}_ m \text{cosk}_ n U = \text{cosk}_ n U$.

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.
$\square$

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{Mor}\nolimits (W, \text{cosk}_ n (U \times V)) & = & \mathop{Mor}\nolimits (\text{sk}_ n W, U \times V) \\ & = & \mathop{Mor}\nolimits (\text{sk}_ n W, U) \times \mathop{Mor}\nolimits (\text{sk}_ nW, V) \\ & = & \mathop{Mor}\nolimits (W, \text{cosk}_ n U) \times \mathop{Mor}\nolimits (W, \text{cosk}_ n V) \\ & = & \mathop{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 $U \to V$ and $W \to V$ be morphisms of $n$-truncated simplicial objects of the category $\mathcal{C}$. Then

\[ \text{cosk}_ n (U \times _ V W) = \text{cosk}_ nU \times _{\text{cosk}_ n V} \text{cosk}_ nW \]

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 $X \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$. 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

\[ \mathop{Mor}\nolimits (U, \Delta [n]) = \mathop{Mor}\nolimits (\text{sk}_1 U, \text{sk}_1 \Delta [n]) \]

as desired.
$\square$

## Comments (2)

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