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1.1.3 The Skeletal Filtration

Roughly speaking, one can think of the simplicial sets $\Delta ^ n$ of Construction 1.1.2.1 as elementary building blocks out of which more complicated simplicial sets can be constructed. In this section, we make this idea more precise by introducing the skeletal filtration of a simplicial set. This filtration allows us to write every simplicial set $S_{\bullet }$ as the union of an increasing sequence of simplicial subsets

$\operatorname{sk}_0( S_{\bullet } ) \subseteq \operatorname{sk}_1( S_{\bullet } ) \subseteq \operatorname{sk}_2( S_{\bullet } ) \subseteq \operatorname{sk}_3( S_{\bullet } ) \subseteq \cdots ,$

where each $\operatorname{sk}_{n}( S_{\bullet } )$ is obtained from $\operatorname{sk}_{n-1}( S_{\bullet } )$ by attaching copies of $\Delta ^ n$ (see Proposition 1.1.3.11 below for a precise statement). We will need some preliminaries.

Proposition 1.1.3.1. Let $S_{\bullet }$ be a simplicial set and let $\sigma \in S_{n}$ be an $n$-simplex of $S_{\bullet }$ for some $n > 0$, which we will identify with a map of simplicial sets $\sigma : \Delta ^{n} \rightarrow S_{\bullet }$. The following conditions are equivalent:

$(1)$

The simplex $\sigma$ belongs to the image of the degeneracy map $s_{i}: S_{n-1} \rightarrow S_{n}$ for some $0 \leq i \leq n-1$ (see Notation 1.1.1.9).

$(2)$

The map $\sigma$ factors as a composition $\Delta ^{n} \xrightarrow {f} \Delta ^{n-1} \rightarrow S_{\bullet }$, where $f$ corresponds to a surjective map of linearly ordered sets $[n] \rightarrow [n-1]$.

$(3)$

The map $\sigma$ factors as a composition $\Delta ^{n} \xrightarrow {f} \Delta ^{m} \rightarrow S_{\bullet }$, where $m < n$ and $f$ corresponds to a surjective map of linearly ordered sets $[n] \rightarrow [m]$.

$(4)$

The map $\sigma$ factors as a composition $\Delta ^{n} \rightarrow \Delta ^{m} \rightarrow S_{\bullet }$, where $m < n$.

Proof. The implications $(1) \Leftrightarrow (2) \Rightarrow (3) \Rightarrow (4)$ are immediate. We will complete the proof by showing that $(4)$ implies $(1)$. Assume that $\sigma$ factors as a composition $\Delta ^{n} \xrightarrow {f} \Delta ^{m} \xrightarrow {\sigma '} S_{\bullet }$, where $m < n$. Let us abuse notation by identifying $f$ with a map of linearly ordered sets $[n] \rightarrow [m]$. Since $m < n$, this map cannot be injective. It follows that we can find some $i < n$ such that $f(i) = f(i+1)$. It follows that $f$ factors through the map $\sigma ^{i}: [n] \rightarrow [n-1]$ of Notation 1.1.1.9, so that $\sigma$ belongs to the image of the degeneracy map $s_{i}$. $\square$

Definition 1.1.3.2. Let $S_{\bullet }$ be a simplicial set and let $\sigma : \Delta ^{n} \rightarrow S_{\bullet }$ be an $n$-simplex of $S_{\bullet }$. We will say that $\sigma$ is degenerate if $n > 0$ and $\sigma$ satisfies the equivalent conditions of Proposition 1.1.3.1. We say that $\sigma$ is nondegenerate if it is not degenerate (in particular, every $0$-simplex of $S_{\bullet }$ is nondegenerate).

Remark 1.1.3.3. Let $f: S_{\bullet } \rightarrow T_{\bullet }$ be a map of simplicial sets. If $\sigma$ is a degenerate $n$-simplex of $S_{\bullet }$, then $f(\sigma )$ is a degenerate $n$-simplex of $T_{\bullet }$. The converse holds if $f$ is a monomorphism of simplicial sets (for example, if $S_{\bullet }$ is a simplicial subset of $T_{\bullet }$).

Proposition 1.1.3.4. Let $\sigma : \Delta ^{n} \rightarrow S_{\bullet }$ be a map of simplicial sets. Then $\sigma$ can be factored as a composition

$\Delta ^{n} \xrightarrow {\alpha } \Delta ^{m} \xrightarrow { \tau } S_{\bullet },$

where $\alpha$ corresponds to a surjective map of linearly ordered sets $[n] \rightarrow [m]$ and $\tau$ is a nondegenerate $m$-simplex of $S_{\bullet }$. Moreover, this factorization is unique.

Proof. Let $m$ be the smallest nonnegative integer for which $\sigma$ can be factored as a composition $\Delta ^{n} \xrightarrow {\alpha } \Delta ^{m} \xrightarrow {\tau } S_{\bullet }$. It follows from the minimality of $m$ that $\alpha$ must induce a surjection of linearly ordered sets $[n] \rightarrow [m]$ (otherwise, we could replace $[m]$ by the image of $\alpha$) and that the $m$-simplex $\tau$ is nondegenerate. This proves the existence of the desired factorization.

To establish uniqueness, let us suppose we are given another factorization of $\sigma$ as a composition $\Delta ^{n} \xrightarrow {\alpha '} \Delta ^{m'} \xrightarrow {\tau '} S_{\bullet }$. By assumption, $\alpha$ and $\alpha '$ determine surjections of linearly ordered sets $[n] \rightarrow [m]$ and $[n] \rightarrow [m']$, and therefore admit sections which we will denote by $\beta$ and $\beta '$, respectively. The equality $\sigma = \tau \circ \alpha$ then gives

$\tau = \sigma \circ \beta = \tau ' \circ \alpha ' \circ \beta .$

Our assumption that $\tau$ is nondegenerate then guarantees that the map $\alpha ' \circ \beta : [m] \rightarrow [m']$ is injective, so that $m \leq m'$. The same argument shows that $m' \leq m$, so we must have $m = m'$. Since the only nondecreasing injection from $[m]$ to itself is the identity map, we conclude that $\alpha ' \circ \beta = \operatorname{id}_{[m]}$. The desired uniqueness now follows from the calculations

$\tau = \tau ' \circ \alpha ' \circ \beta = \tau ' \quad \quad \alpha = \alpha ' \circ \beta \circ \alpha = \alpha '.$
$\square$

Construction 1.1.3.5. Let $S_{\bullet }$ be a simplicial set, let $k \geq -1$ be an integer, and let $\sigma : \Delta ^{n} \rightarrow S_{\bullet }$ be an $n$-simplex of $S_{\bullet }$. The proof of Proposition 1.1.3.4 shows that the following conditions are equivalent:

$(a)$

Let $\Delta ^{n} \xrightarrow {\alpha } \Delta ^{m} \xrightarrow {\tau } S_{\bullet }$ be the factorization of Proposition 1.1.3.4 (so that $\alpha$ induces a surjection $[n] \rightarrow [m]$, the map $\tau$ is nondegenerate, and $\sigma = \tau \circ \alpha$). Then $m \leq k$.

$(b)$

There exists a factorization $\Delta ^{n} \rightarrow \Delta ^{m'} \rightarrow S_{\bullet }$ of $\sigma$ for which $m' \leq k$.

For each $n \geq 0$, we let $\operatorname{sk}_{k}( S_{n} )$ denote the subset of $S_{n}$ consisting of those $n$-simplices which satisfy conditions $(a)$ and $(b)$. From characterization $(b)$, we see that the collection of subsets $\{ \operatorname{sk}_ k(S_ n) \subseteq S_ n \} _{n \geq 0}$ is stable under the face and degeneracy operators of $S_{\bullet }$, and therefore determine a simplicial subset of $S_{\bullet }$ (Remark 1.1.2.4). We will denote this simplicial subset by $\operatorname{sk}_{k}( S_{\bullet } )$ and refer to it as the $k$-skeleton of $S_{\bullet }$.

Remark 1.1.3.6. Let $S_{\bullet }$ be a simplicial set and let $k \geq -1$. If $n \leq k$, then $\operatorname{sk}_{k}( S_{\bullet } )$ contains every $n$-simplex of $S_{\bullet }$. In particular, the union $\bigcup _{k \geq -1} \operatorname{sk}_{k}( S_{\bullet } )$ is equal to $S_{\bullet }$.

Remark 1.1.3.7. Let $S_{\bullet }$ be a simplicial set and let $\sigma$ be a nondegenerate $n$-simplex of $S_{\bullet }$. Then $\sigma$ is contained in $\operatorname{sk}_{k}( S_{\bullet } )$ if and only if $n \leq k$.

Example 1.1.3.8. For any simplicial set $S_{\bullet }$, the $(-1)$-skeleton $\operatorname{sk}_{-1}( S_{\bullet } )$ is empty.

We now show that the $k$-skeleton of a simplicial set $S_{\bullet }$ can be characterized by a universal property.

Definition 1.1.3.9. Let $S_{\bullet }$ be a simplicial set and let $k \geq -1$ be an integer. We will say that $S_{\bullet }$ has dimension $\leq k$ if, for $n > k$, every $n$-simplex of $S_{\bullet }$ is degenerate. If $k \geq 0$, we say that $S_{\bullet }$ has dimension $k$ if it has dimension $\leq k$ but does not have dimension $\leq k-1$.

Proposition 1.1.3.10. Let $S_{\bullet }$ be a simplicial set and let $k \geq -1$ be an integer. Then:

$(a)$

The simplicial set $\operatorname{sk}_{k}( S_{\bullet } )$ has dimension $\leq k$.

$(b)$

For every simplicial set $T_{\bullet }$ of dimension $\leq k$, composition with the inclusion map $\operatorname{sk}_{k}( S_{\bullet } ) \hookrightarrow S_{\bullet }$ induces a bijection

$\operatorname{Hom}_{\operatorname{Set_{\Delta }}}( T_{\bullet }, \operatorname{sk}_{k}( S_{\bullet } ) ) \rightarrow \operatorname{Hom}_{ \operatorname{Set_{\Delta }}}( T_{\bullet }, S_{\bullet } ).$

In other words, the image of any map $T_{\bullet } \rightarrow S_{\bullet }$ is contained in $\operatorname{sk}_{k}( S_{\bullet } )$.

Proof. Assertion $(a)$ follows from Remark 1.1.3.7. To prove $(b)$, suppose that $f: T_{\bullet } \rightarrow S_{\bullet }$ is a map of simplicial sets, where $T_{\bullet }$ has dimension $\leq k$. We wish to show that $f$ carries every $n$-simplex $\sigma$ of $T_{\bullet }$ to an $n$-simplex of $\operatorname{sk}_{k}( S_{\bullet } )$. Using Proposition 1.1.3.4, we can reduce to the case where $\sigma$ is a nondegenerate $n$-simplex of $T_{\bullet }$. In this case, our assumption that $T_{\bullet }$ has dimension $\leq k$ guarantees that $n \leq k$, so that $f( \sigma )$ belongs to $\operatorname{sk}_{k}( S_{\bullet } )$ by virtue of Remark 1.1.3.6. $\square$

Let $S_{\bullet }$ be a simplicial set. For each $k \geq 0$, we let $S_{k}^{\mathrm{nd}}$ denote the collection of all nondegenerate $k$-simplices of $S_{\bullet }$. Every element $\sigma \in S_{k}^{\mathrm{nd}}$ determines a map of simplicial sets $\Delta ^{k} \rightarrow \operatorname{sk}_{k}( S_{\bullet } )$. Since the boundary $\operatorname{\partial \Delta }^ k \subseteq \Delta ^{k}$ has dimension $\leq k-1$, this map carries $\operatorname{\partial \Delta }^{k}$ into the $(k-1)$-skeleton $\operatorname{sk}_{k-1}( S_{\bullet } )$.

Proposition 1.1.3.11. Let $S_{\bullet }$ be a simplicial set and let $k \geq 0$. Then the construction outlined above determines a pushout square

$\xymatrix { \underset { \sigma \in S_{k}^{\mathrm{nd}} }{\coprod } \operatorname{\partial \Delta }^{k} \ar [r] \ar [d] & \underset { \sigma \in S_{k}^{\mathrm{nd}} }{\coprod } \Delta ^{k} \ar [d] \\ \operatorname{sk}_{k-1}( S_{\bullet } ) \ar [r] & \operatorname{sk}_{k}( S_{\bullet } ) }$

in the category $\operatorname{Set_{\Delta }}$ of simplicial sets.

Proof. Unwinding the definitions, we must prove the following:

$(\ast )$

Let $\tau$ be an $n$-simplex of $\operatorname{sk}_{k}( S_{\bullet } )$ which is not contained in $\operatorname{sk}_{k-1}(S_{\bullet } )$. Then $\tau$ factors uniquely as a composition

$\Delta ^{n} \xrightarrow {\alpha } \Delta ^{k} \xrightarrow {\sigma } S_{\bullet },$

where $\sigma$ is a nondegenerate simplex of $S_{\bullet }$ and $\alpha$ does not factor through the boundary $\operatorname{\partial \Delta }^{k}$ (in other words, $\alpha$ induces a surjection of linearly ordered sets $[n] \rightarrow [k]$).

Proposition 1.1.3.4 implies that any $n$-simplex of $S_{\bullet }$ admits a unique factorization $\Delta ^{n} \xrightarrow {\alpha } \Delta ^{m} \xrightarrow {\sigma } S_{\bullet }$, where $\alpha$ is surjective and $\sigma$ is nondegenerate. Our assumption that $\tau$ belongs to the $\operatorname{sk}_{k}(S_{\bullet })$ guarantees that $m \leq k$, and our assumption that $\tau$ does not belong to $\operatorname{sk}_{k-1}( S_{\bullet } )$ guarantees that $m \geq k$. $\square$