## 14.21 Left adjoints to the skeleton functors

In this section we construct a left adjoint $i_{m!}$ of the skeleton functor $\text{sk}_ m$ in certain cases. The adjointness formula is

\[ \mathop{Mor}\nolimits _{\text{Simp}_ m(\mathcal{C})}(U, \text{sk}_ mV) = \mathop{Mor}\nolimits _{\text{Simp}(\mathcal{C})}(i_{m!}U, V). \]

It turns out that this left adjoint exists when the category $\mathcal{C}$ has finite colimits.

We use a similar construction as in Section 14.12. Recall the category $[n]/\Delta $ of objects under $[n]$, see Categories, Example 4.2.14. Its objects are morphisms $\alpha : [n] \to [k]$ and its morphisms are commutative triangles. We let $([n]/\Delta )_{\leq m}$ denote the full subcategory of $[n]/\Delta $ consisting of objects $[n] \to [k]$ with $k \leq m$. Given a $m$-truncated simplicial object $U$ of $\mathcal{C}$ we define a functor

\[ U(n) : ([n]/\Delta )_{\leq m}^{opp} \longrightarrow \mathcal{C} \]

by the rules

\begin{eqnarray*} ([n] \to [k]) & \longmapsto & U_ k \\ \psi : ([n] \to [k']) \to ([n] \to [k]) & \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

\[ \underline{\varphi } : ([n']/\Delta )_{\leq m} \longrightarrow ([n]/\Delta )_{\leq m} \]

which maps $\alpha : [n'] \to [k]$ to $\varphi \circ \alpha : [n] \to [k]$. The composition $U(n) \circ \underline{\varphi }$ is equal to the functor $U(n')$.

Lemma 14.21.1. Let $\mathcal{C}$ be a category which has finite colimits. The functors $i_{m!}$ exist for all $m$. Let $U$ be an $m$-truncated simplicial object of $\mathcal{C}$. The simplicial object $i_{m!}U$ is described by the formula

\[ (i_{m!}U)_ n = \mathop{\mathrm{colim}}\nolimits _{([n]/\Delta )_{\leq m}^{opp}} U(n) \]

and for $\varphi : [n] \to [n']$ the map $i_{m!}U(\varphi )$ comes from the identification $U(n) \circ \underline{\varphi } = U(n')$ above via Categories, Lemma 4.14.8.

**Proof.**
In this proof we denote $i_{m!}U$ the simplicial object whose $n$th term is given by the displayed formula of the lemma. We will show it satisfies the adjointness property.

Let $V$ be a simplicial object of $\mathcal{C}$. Let $\gamma : U \to \text{sk}_ mV$ be given. A morphism

\[ \mathop{\mathrm{colim}}\nolimits _{([n]/\Delta )_{\leq m}^{opp}} U(n) \to T \]

is given by a compatible system of morphisms $f_\alpha : U_ k \to T$ where $\alpha : [n] \to [k]$ with $k \leq m$. Certainly, we have such a system of morphisms by taking the compositions

\[ U_ k \xrightarrow {\gamma _ k} V_ k \xrightarrow {V(\alpha )} V_ n. \]

Hence we get an induced morphism $(i_{m!}U)_ n \to V_ n$. We leave it to the reader to see that these form a morphism of simplicial objects $\gamma ' : i_{m!}U \to V$.

Conversely, given a morphism $\gamma ' : i_{m!}U \to V$ we obtain a morphism $\gamma : U \to \text{sk}_ m V$ by setting $\gamma _ i : U_ i \to V_ i$ equal to the composition

\[ U_ i \xrightarrow {\text{id}_{[i]}} \mathop{\mathrm{colim}}\nolimits _{([i]/\Delta )_{\leq m}^{opp}} U(i) \xrightarrow {\gamma '_ i} V_ i \]

for $0 \leq i \leq n$. We leave it to the reader to see that this is the inverse of the construction above.
$\square$

Lemma 14.21.2. Let $\mathcal{C}$ be a category. Let $U$ be an $m$-truncated simplicial object of $\mathcal{C}$. For any $n \leq m$ the colimit

\[ \mathop{\mathrm{colim}}\nolimits _{([n]/\Delta )_{\leq m}^{opp}} U(n) \]

exists and is equal to $U_ n$.

**Proof.**
This is so because the category $([n]/\Delta )_{\leq m}$ has an initial object, namely $\text{id} : [n] \to [n]$.
$\square$

Lemma 14.21.3. Let $\mathcal{C}$ be a category which has finite colimits. Let $U$ be an $m$-truncated simplicial object of $\mathcal{C}$. The map $U \to \text{sk}_ m i_{m!}U$ is an isomorphism.

**Proof.**
Combine Lemmas 14.21.1 and 14.21.2.
$\square$

Lemma 14.21.4. If $U$ is an $m$-truncated simplicial set and $n > m$ then all $n$-simplices of $i_{m!}U$ are degenerate.

**Proof.**
This can be seen from the construction of $i_{m!}U$ in Lemma 14.21.1, but we can also argue directly as follows. Write $V = i_{m!}U$. Let $V' \subset V$ be the simplicial subset with $V'_ i = V_ i$ for $i \leq m$ and all $i$ simplices degenerate for $i > m$, see Lemma 14.18.4. By the adjunction formula, since $\text{sk}_ m V' = U$, there is an inverse to the injection $V' \to V$. Hence $V' = V$.
$\square$

Lemma 14.21.5. Let $U$ be a simplicial set. 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 set $U' \subset U$ defined in Lemma 14.18.4.

**Proof.**
By Lemma 14.21.4 the only nondegenerate simplices of $i_{n!} \text{sk}_ n U$ are in degrees $\leq n$. The map $i_{n!} \text{sk}_ n U \to U$ is an isomorphism in degrees $\leq n$. Combined we conclude that the map $i_{n!} \text{sk}_ n U \to U$ maps nondegenerate simplices to nondegenerate simplices and no two nondegenerate simplices have the same image. Hence Lemma 14.18.3 applies. Thus $i_{n!} \text{sk}_ n U \to U$ is injective. The result follows easily from this.
$\square$

Lemma 14.21.7. Let $U \subset V$ be simplicial sets. Suppose $n \geq 0$ and $x \in V_ n$, $x \not\in U_ n$ are such that

$V_ i = U_ i$ for $i < n$,

$V_ n = U_ n \cup \{ x\} $,

any $z \in V_ j$, $z \not\in U_ j$ for $j > n$ is degenerate.

Let $\Delta [n] \to V$ be the unique morphism mapping the nondegenerate $n$-simplex of $\Delta [n]$ to $x$. In this case the diagram

\[ \xymatrix{ \Delta [n] \ar[r] & V \\ i_{(n - 1)!} \text{sk}_{n - 1} \Delta [n] \ar[r] \ar[u] & U \ar[u] } \]

is a pushout diagram.

**Proof.**
Let us denote $\partial \Delta [n] = i_{(n - 1)!} \text{sk}_{n - 1} \Delta [n]$ for convenience. There is a natural map $U \amalg _{\partial \Delta [n]} \Delta [n] \to V$. We have to show that it is bijective in degree $j$ for all $j$. This is clear for $j \leq n$. Let $j > n$. The third condition means that any $z \in V_ j$, $z \not\in U_ j$ is a degenerate simplex, say $z = s^{j - 1}_ i(z')$. Of course $z' \not\in U_{j - 1}$. By induction it follows that $z'$ is a degeneracy of $x$. Thus we conclude that all $j$-simplices of $V$ are either in $U$ or degeneracies of $x$. This implies that the map $U \amalg _{\partial \Delta [n]} \Delta [n] \to V$ is surjective. Note that a nondegenerate simplex of $U \amalg _{\partial \Delta [n]} \Delta [n]$ is either the image of a nondegenerate simplex of $U$, or the image of the (unique) nondegenerate $n$-simplex of $\Delta [n]$. Since clearly $x$ is nondegenerate we deduce that $U \amalg _{\partial \Delta [n]} \Delta [n] \to V$ maps nondegenerate simplices to nondegenerate simplices and is injective on nondegenerate simplices. Hence it is injective, by Lemma 14.18.3.
$\square$

Lemma 14.21.8. Let $U \subset V$ be simplicial sets, with $U_ n, V_ n$ finite nonempty for all $n$. Assume that $U$ and $V$ have finitely many nondegenerate simplices. Then there exists a sequence of sub simplicial sets

\[ U = W^0 \subset W^1 \subset W^2 \subset \ldots W^ r = V \]

such that Lemma 14.21.7 applies to each of the inclusions $W^ i \subset W^{i + 1}$.

**Proof.**
Let $n$ be the smallest integer such that $V$ has a nondegenerate simplex that does not belong to $U$. Let $x \in V_ n$, $x\not\in U_ n$ be such a nondegenerate simplex. Let $W \subset V$ be the set of elements which are either in $U$, or are a (repeated) degeneracy of $x$ (in other words, are of the form $V(\varphi )(x)$ with $\varphi : [m] \to [n]$ surjective). It is easy to see that $W$ is a simplicial set. The inclusion $U \subset W$ satisfies the conditions of Lemma 14.21.7. Moreover the number of nondegenerate simplices of $V$ which are not contained in $W$ is exactly one less than the number of nondegenerate simplices of $V$ which are not contained in $U$. Hence we win by induction on this number.
$\square$

Lemma 14.21.9. Let $\mathcal{A}$ be an abelian category Let $U$ be an $m$-truncated simplicial object of $\mathcal{A}$. For $n > m$ we have $N(i_{m!}U)_ n = 0$.

**Proof.**
Write $V = i_{m!}U$. Let $V' \subset V$ be the simplicial subobject of $V$ with $V'_ i = V_ i$ for $i \leq m$ and $N(V'_ i) = 0$ for $i > m$, see Lemma 14.18.9. By the adjunction formula, since $\text{sk}_ m V' = U$, there is an inverse to the injection $V' \to V$. Hence $V' = V$.
$\square$

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.
$\square$

Here is another way to think about the coskeleton functor using the material above.

Lemma 14.21.11. Let $\mathcal{C}$ be a category with finite coproducts and finite limits. Let $V$ be a simplicial object of $\mathcal{C}$. In this case

\[ (\text{cosk}_ n \text{sk}_ n V)_{n + 1} = \mathop{\mathrm{Hom}}\nolimits (i_{n !}\text{sk}_ n \Delta [n + 1], V)_0. \]

**Proof.**
By Lemma 14.13.4 the object on the left represents the functor which assigns to $X$ the first set of the following equalities

\begin{eqnarray*} \mathop{Mor}\nolimits (X \times \Delta [n + 1], \text{cosk}_ n \text{sk}_ n V) & = & \mathop{Mor}\nolimits (X \times \text{sk}_ n \Delta [n + 1], \text{sk}_ n V) \\ & = & \mathop{Mor}\nolimits (X \times i_{n !} \text{sk}_ n \Delta [n + 1], V). \end{eqnarray*}

The object on the right in the formula of the lemma is represented by the functor which assigns to $X$ the last set in the sequence of equalities. This proves the result.

In the sequence of equalities we have used that $\text{sk}_ n (X \times \Delta [n + 1]) = X \times \text{sk}_ n \Delta [n + 1]$ and that $i_{n!}(X \times \text{sk}_ n \Delta [n + 1]) = X \times i_{n !} \text{sk}_ n \Delta [n + 1]$. The first equality is obvious. For any (possibly truncated) simplicial object $W$ of $\mathcal{C}$ and any object $X$ of $\mathcal{C}$ denote temporarily $\mathop{Mor}\nolimits _\mathcal {C}(X, W)$ the (possibly truncated) simplicial set $[n] \mapsto \mathop{Mor}\nolimits _\mathcal {C}(X, W_ n)$. From the definitions it follows that $\mathop{Mor}\nolimits (U \times X, W) = \mathop{Mor}\nolimits (U, \mathop{Mor}\nolimits _\mathcal {C}(X, W))$ for any (possibly truncated) simplicial set $U$. Hence

\begin{eqnarray*} \mathop{Mor}\nolimits (X \times i_{n !} \text{sk}_ n \Delta [n + 1], W) & = & \mathop{Mor}\nolimits (i_{n !} \text{sk}_ n \Delta [n + 1], \mathop{Mor}\nolimits _\mathcal {C}(X, W)) \\ & = & \mathop{Mor}\nolimits (\text{sk}_ n \Delta [n + 1], \text{sk}_ n\mathop{Mor}\nolimits _\mathcal {C}(X, W)) \\ & = & \mathop{Mor}\nolimits (X \times \text{sk}_ n \Delta [n + 1], \text{sk}_ nW) \\ & = & \mathop{Mor}\nolimits (i_{n!}(X \times \text{sk}_ n \Delta [n + 1]), W). \end{eqnarray*}

This proves the second equality used, and ends the proof of the lemma.
$\square$

## Comments (0)