Definition 1.1.6.1. Let $S_{\bullet }$ be a simplicial set and let $S'_{\bullet } \subseteq S_{\bullet }$ be a simplicial subset of $S_{\bullet }$ (Remark 1.1.2.4). We will say that $S'_{\bullet }$ is a *summand* of $S_{\bullet }$ if the simplicial set $S_{\bullet }$ decomposes as a coproduct $S'_{\bullet } \coprod S''_{\bullet }$, for some other simplicial subset $S''_{\bullet } \subseteq S_{\bullet }$.

### 1.1.6 Connected Components of Simplicial Sets

In this section, we introduce the notion of a *connected* simplicial set (Definition 1.1.6.6 and show that every simplicial set $S_{\bullet }$ admits an (essentially unique) decomposition as a disjoint union of connected subsets (Proposition 1.1.6.13), indexed by a set $\pi _0( S_{\bullet } )$ which we call the *set of connected components of $S_{\bullet }$*. Moreover, we characterize the construction $S_{\bullet } \mapsto \pi _0( S_{\bullet } )$ as a left adjoint to the construction $I \mapsto \underline{I}_{\bullet }$ of Construction 1.1.4.2 (Corollary 1.1.6.21).

Remark 1.1.6.2. In the situation of Definition 1.1.6.1, if $S'_{\bullet } \subseteq S_{\bullet }$ is a summand, then the complementary summand $S''_{\bullet }$ is uniquely determined: for each $n \geq 0$, we must have $S''_{n} = S_{n} \setminus S'_{n}$. Consequently, the condition that $S'_{\bullet }$ is a summand of $S_{\bullet }$ is equivalent to the condition that the construction

is functorial: that is, that the face and degeneracy operators for the simplicial set $S_{\bullet }$ preserve the subsets $S_{n} \setminus S'_{n}$.

Remark 1.1.6.3. Let $S_{\bullet }$ be a simplicial set. Then the collection of all summands of $S_{\bullet }$ is closed under the formation of unions and intersections (this follows immediately from the criterion of Remark 1.1.6.2).

Remark 1.1.6.4 (Transitivity). Let $S_{\bullet }$ be a simplicial set. If $S'_{\bullet } \subseteq S_{\bullet }$ is a summand of $S_{\bullet }$ and $S''_{\bullet } \subseteq S'_{\bullet }$ is a summand of $S'_{\bullet }$, then $S''_{\bullet }$ is a summand of $S_{\bullet }$.

Remark 1.1.6.5. Let $f: S_{\bullet } \rightarrow T_{\bullet }$ be a map of simplicial sets and let $T'_{\bullet } \subseteq T_{\bullet }$ be a summand. Then the inverse image $f^{-1}( T'_{\bullet } ) \simeq S_{\bullet } \times _{ T_{\bullet } } T'_{\bullet }$ is a summand of $S_{\bullet }$.

Definition 1.1.6.6. Let $S_{\bullet }$ be a simplicial set. We will say that $S_{\bullet }$ is *connected* if it is nonempty and every summand $S'_{\bullet } \subseteq S_{\bullet }$ is either empty or coincides with $S_{\bullet }$.

Example 1.1.6.7. For each $n \geq 0$, the standard $n$-simplex $\Delta ^{n}$ is connected.

Definition 1.1.6.8 (Connected Components). Let $S_{\bullet }$ be a simplicial set. We will say that a simplicial subset $S'_{\bullet } \subseteq S_{\bullet }$ is a *connected component* of $S_{\bullet }$ if $S'_{\bullet }$ is a summand of $S_{\bullet }$ (Definition 1.1.6.1) and $S'_{\bullet }$ is connected (Definition 1.1.6.6). We let $\pi _0( S_{\bullet } )$ denote the set of all connected components of $S_{\bullet }$.

Warning 1.1.6.9. Let $S_{\bullet }$ be a simplicial set. As we will soon see, the set $\pi _0(S_{\bullet })$ admits many different descriptions:

We can identify $\pi _0( S_{\bullet } )$ with the set of connected components of $S_{\bullet }$ (Definition 1.1.6.8).

We can identify $\pi _0( S_{\bullet } )$ with a colimit of the diagram $\operatorname{{\bf \Delta }}^{\operatorname{op}} \rightarrow \operatorname{Set_{\Delta }}$ given by the simplicial set $S_{\bullet }$ (Remark 1.1.6.20).

We can identify $\pi _0( S_{\bullet } )$ with the quotient of the set of vertices $S_{0}$ by an equivalence relation $\sim $ generated by the set of edges $S_{1}$ (Remark 1.1.6.23).

We can identify $\pi _0( S_{\bullet } )$ with the set of connected components of the directed graph $\mathrm{Gr}( S_{\bullet } )$ (Variant 1.1.6.24).

When $S_{\bullet }$ is a Kan complex, we can identify $\pi _0( S_{\bullet } )$ as the set of isomorphism classes of objects in the fundamental groupoid $\pi _{\leq 1}(S_{\bullet })$ (Remark 1.3.6.15).

Because of this abundance of perspectives, it often will be convenient to view $I = \pi _0( S_{\bullet } )$ as an abstract index set which is equipped with a bijection

rather than as the set of connected components itself.

Example 1.1.6.10. Let $I$ be a set and let $\underline{I}_{\bullet }$ be the constant simplicial set associated to $I$ (Construction 1.1.4.2). Then the connected components of $\underline{I}_{\bullet }$ are exactly the simplicial subsets of the form $\{ i \} = \underline{ \{ i \} }_{\bullet }$ for $i \in I$. In particular, we have a canonical bijection $I \simeq \pi _0( \underline{I}_{\bullet } )$.

Proposition 1.1.6.11. Let $f: S_{\bullet } \rightarrow T_{\bullet }$ be a map of simplicial sets, and suppose that $S_{\bullet }$ is connected. Then there is a unique connected component $T'_{\bullet } \subseteq T_{\bullet }$ such that $f( S_{\bullet } ) \subseteq T'_{\bullet }$.

**Proof.**
Let $T'_{\bullet }$ be the smallest summand of $T_{\bullet }$ which contains the image of $f$ (the existence of $T'_{\bullet }$ follows from Remark 1.1.6.3: we can take $T'_{\bullet }$ to be the intersection of all those summands of $T_{\bullet }$ which contain the image of $f$). We will complete the proof by showing that $T'_{\bullet }$ is connected. Since $S_{\bullet }$ is nonempty, $T'_{\bullet }$ must be nonempty. Let $T''_{\bullet } \subseteq T'_{\bullet }$ be a summand; we wish to show that $T''_{\bullet } = T'_{\bullet }$ or $T''_{\bullet } = \emptyset $. Note that $f^{-1}(T''_{\bullet } )$ is a summand of $S_{\bullet }$ (Remark 1.1.6.5). Since $S_{\bullet }$ is connected, we must have $f^{-1}( T''_{\bullet } ) = S_{\bullet }$ or $f^{-1}( T''_{\bullet } ) = \emptyset $. Replacing $T''_{\bullet }$ by its complement if necessary, we may assume that $f^{-1}( T''_{\bullet } ) = S_{\bullet }$, so that $f$ factors through $T''_{\bullet }$. Since $T''_{\bullet }$ is a summand of $T_{\bullet }$ (Remark 1.1.6.4), the minimality of $T'_{\bullet }$ guarantees that $T''_{\bullet } = T'_{\bullet }$, as desired.
$\square$

Corollary 1.1.6.12. Let $S_{\bullet }$ be a simplicial set. The following conditions are equivalent:

- $(a)$
The simplicial set $S_{\bullet }$ is connected.

- $(b)$
For every set $I$, the canonical map

\[ I \simeq \operatorname{Hom}_{\operatorname{Set_{\Delta }}}( \Delta ^0, \underline{I}_{\bullet } ) \rightarrow \operatorname{Hom}_{\operatorname{Set_{\Delta }}}( S_{\bullet }, \underline{I}_{\bullet } ) \]is bijective.

**Proof.**
The implication $(a) \Rightarrow (b)$ follows from Propoposition 1.1.6.11 and Example 1.1.6.10. Conversely, suppose that $(b)$ is satisfied. Applying $(b)$ in the case $I = \emptyset $, we conclude that there are no maps from $S_{\bullet }$ to the empty simplicial set, so that $S_{\bullet }$ is nonempty. If $S_{\bullet }$ is a disjoint union of simplicial subsets $S'_{\bullet }, S''_{\bullet } \subseteq S_{\bullet }$, then we obtain a map of simplicial sets

and assumption $(b)$ guarantees that this map factors through one of the summands on the right hand side; it follows that either $S'_{\bullet }$ or $S''_{\bullet }$ is empty. $\square$

Proposition 1.1.6.13. Let $S_{\bullet }$ be a simplicial set. Then $S_{\bullet }$ is the disjoint union of its connected components.

**Proof.**
Let $\sigma $ be an $n$-simplex of $S_{\bullet }$; we wish to show that there is a unique connected component of $S_{\bullet }$ which contains $\sigma $. This follows from Proposition 1.1.6.11, applied to the map $\Delta ^ n \rightarrow S_{\bullet }$ classified by $\sigma $ (since the standard $n$-simplex $\Delta ^ n$ is connected; see Example 1.1.6.7).
$\square$

Corollary 1.1.6.14. Let $S_{\bullet }$ be a simplicial set. Then $S_{\bullet }$ is empty if and only if $\pi _0( S_{\bullet } )$ is empty.

Corollary 1.1.6.15. Let $S_{\bullet }$ be a simplicial set. Then $S_{\bullet }$ is connected if and only if $\pi _0( S_{\bullet })$ has exactly one element.

Exercise 1.1.6.16 (Classification of Summands). Let $S_{\bullet }$ be a simplicial set. Show that a simplicial subset $S'_{\bullet } \subseteq S_{\bullet }$ is a summand if and only if it can be written as a union of connected components of $S_{\bullet }$. Consequently, we have a canonical bijection

Remark 1.1.6.17 (Functoriality of $\pi _0$). Let $f: S_{\bullet } \rightarrow T_{\bullet }$ be a map of simplicial sets. It follows from Proposition 1.1.6.11 that for each connected component $S'_{\bullet } \subseteq S_{\bullet }$, there is a unique connected component $T'_{\bullet } \subseteq T_{\bullet }$ such that $f(S'_{\bullet } ) \subseteq T'_{\bullet }$. The construction $S'_{\bullet } \mapsto T'_{\bullet }$ then determines a map of sets $\pi _0(f): \pi _0(S_{\bullet }) \rightarrow \pi _{0}( T_{\bullet } )$. This construction is compatible with composition, and therefore allows us to view the construction $S_{\bullet } \mapsto \pi _0( S_{\bullet } )$ as a functor $\pi _0: \operatorname{Set_{\Delta }}\rightarrow \operatorname{Set}$ from the category of simplicial sets to the category of sets.

We now show that the connected component functor $\pi _0: \operatorname{Set_{\Delta }}\rightarrow \operatorname{Set}$ can be characterized by a universal property.

Construction 1.1.6.18 (The Component Map). Let $S_{\bullet }$ be a simplicial set. For every $n$-simplex $\sigma $ of $S_{\bullet }$, Proposition 1.1.6.13 implies that there is a unique connected component $S'_{\bullet } \subseteq S_{\bullet }$ which contains $\sigma $. The construction $\sigma \mapsto S'_{\bullet }$ then determines a map of simplicial sets

where $\underline{ \pi _0( S_{\bullet })}_{\bullet }$ denotes the constant simplicial set associated to $\pi _0(S_{\bullet })$ (Construction 1.1.4.2). We will refer to $u$ as the *component map*.

Proposition 1.1.6.19. Let $S_{\bullet }$ be a simplicial set and let $u: S_{\bullet } \rightarrow \underline{ \pi _0( S_{\bullet })}_{\bullet }$ be the component map of Construction 1.1.6.18. For every set $J$, composition with $u$ induces a bijection

**Proof.**
Decomposing $S_{\bullet }$ as the union of its connected components, we can reduce to the case where $S_{\bullet }$ is connected, in which case the desired result is a reformulation of Corollary 1.1.6.12.
$\square$

Remark 1.1.6.20 ($\pi _0$ as a Colimit). Let $S_{\bullet }$ be a simplicial set. It follows from Proposition 1.1.6.19 that the component map $u: S_{\bullet } \rightarrow \underline{ \pi _0( S_{\bullet })}_{\bullet }$ exhibits $\pi _0( S_{\bullet } )$ as the colimit of the diagram $\operatorname{{\bf \Delta }}^{\operatorname{op}} \rightarrow \operatorname{Set}$ determined by $S_{\bullet }$.

Corollary 1.1.6.21. The connected component functor

of Remark 1.1.6.17 is left adjoint to the constant simplicial set functor

of Construction 1.1.4.2. More precisely, the construction $S_{\bullet } \mapsto (u: S_{\bullet } \rightarrow \underline{ \pi _0( S_{\bullet })}_{\bullet })$ is the unit of an adjunction.

We now make Remark 1.1.6.20 more concrete.

Proposition 1.1.6.22. Let $S_{\bullet }$ be a simplicial set, and let $u_0: S_0 \rightarrow \pi _0( S_{\bullet } )$ be the map of sets given by the component map of Construction 1.1.6.18. Then $u_0$ exhibits $\pi _0(S_{\bullet } )$ as the coequalizer of the face maps $d_0, d_1: S_{1} \rightrightarrows S_0$.

Remark 1.1.6.23. Let $S_{\bullet }$ be a simplicial set. Proposition 1.1.6.22 supplies a coequalizer diagram of sets

In other words, it allows us to identify $\pi _0( S_{\bullet } )$ with the quotient of $S_0 / \sim $, where $\sim $ is the equivalence relation generated by the set of edges of $S_{\bullet }$ (that is, the smallest equivalence relation with the property that $d_0(e) \sim d_1(e)$, for every edge $e \in S_1$). In particular, the set $\pi _0( S_{\bullet } )$ depends only on the $1$-skeleton of $S_{\bullet }$.

Variant 1.1.6.24. Let $S_{\bullet }$ be a simplicial set. Then the set of connected components $\pi _0( S_{\bullet } )$ can also be described as the coequalizer of the pair of maps $d_0, d_1: S_{1}^{ \mathrm{nd} } \rightrightarrows S_0$, where $S_{1}^{\mathrm{nd} } \subseteq S_{1}$ denotes the set of nondegenerate edges of $S_{\bullet }$ (since every degenerate edge $e \in S_{1}$ automatically satisfies $d_0(e) = d_1(e)$). We therefore have a coequalizer diagram of sets

where $G = \mathrm{Gr}(S_{\bullet } )$ is the directed graph of Example 1.1.5.4. In other words, we can identify $\pi _0( S_{\bullet } )$ with the set of connected components of $G$, in the usual graph-theoretic sense.

**Proof of Proposition 1.1.6.22.**
Let $I$ be a set and let $f: S_0 \rightarrow I$ be a function satisfying $f \circ d_0 = f \circ d_1$ (as functions from $S_1$ to $I$). We wish to show that $f$ factors uniquely as a composition

By virtue of Proposition 1.1.6.19, this is equivalent to the assertion that there is a unique map of simplicial sets $F: S_{\bullet } \rightarrow \underline{I}_{\bullet }$ which coincides with $f$ on simplices of degree zero. Let $\sigma $ be an $n$-simplex of $S_{\bullet }$, which we identify with a map of simplicial sets $\sigma : \Delta ^ n \rightarrow S_{\bullet }$. For $0 \leq i \leq n$, we regard $\sigma (i)$ as a vertex of $S_{\bullet }$. Note that if $0 \leq i \leq j \leq n$, then we have $f( \sigma (i) ) = f ( \sigma (j) )$: to prove this, we can assume without loss of generality that $i=0$ and $j = n = 1$, in which case it follows from our hypothesis that $f \circ d_0 = f \circ d_1$. It follows that there is a unique element $F(\sigma ) \in I$ such that $F(\sigma ) = f( \sigma (i) )$ for each $0 \leq i \leq n$. The construction $\sigma \mapsto F(\sigma )$ defines a map of simplicial sets $F: S_{\bullet } \rightarrow \underline{I}_{\bullet }$ with the desired properties. $\square$

Proposition 1.1.6.25. The collection of connected simplicial sets is closed under finite products.

**Proof.**
Since the final object $\Delta ^{0} \in \operatorname{Set_{\Delta }}$ is connected (Example 1.1.6.7), it will suffice to show that the collection of connected simplicial sets is closed under pairwise products. Let $S_{\bullet }$ and $T_{\bullet }$ be connected simplicial sets; we wish to show that $S_{\bullet } \times T_{\bullet }$ is connected. Equivalently, we wish to show that $\pi _0( S_{\bullet } \times T_{\bullet } )$ consists of a single element (Corollary 1.1.6.15). By virtue of Proposition 1.1.6.22, the component map supplies a surjection

It will therefore suffice to show that for every pair of vertices $(s,t), (s', t') \in S_0 \times T_0$ belong to the same connected component of $S_{\bullet } \times T_{\bullet }$. Let $K_{\bullet } \subseteq S_{\bullet } \times T_{\bullet }$ be the connected component which contains the vertex $(s', t )$. Since $S_{\bullet }$ is connected, the map

factors through a unique connected component of $S_{\bullet } \times T_{\bullet }$, which must be equal to $K_{\bullet }$. It follows that $K_{\bullet }$ contains the vertex $(s,t)$. A similar argument (with the roles of $S_{\bullet }$ and $T_{\bullet }$ reversed) shows that $K_{\bullet }$ contains $(s',t')$. $\square$

Corollary 1.1.6.26. The functor $\pi _0: \operatorname{Set_{\Delta }}\rightarrow \operatorname{Set}$ preserves finite products.

**Proof.**
Since $\pi _0( \Delta ^0)$ is a singleton (Example 1.1.6.7), it will suffice to show that for every pair of simplicial sets $S_{\bullet }$ and $T_{\bullet }$, the canonical map

is bijective. Writing $S_{\bullet }$ and $T_{\bullet }$ as a disjoint union of connected components (Proposition 1.1.6.13), we can reduce to the case where $S_{\bullet }$ and $T_{\bullet }$ are connected, in which case the desired result follows from Proposition 1.1.6.25. $\square$

Warning 1.1.6.27. The collection of connected simplicial sets is not closed under infinite products (so the functor $\pi _0: \operatorname{Set_{\Delta }}\rightarrow \operatorname{Set}$ does not commute with infinite products). For example, let $G$ be the directed graph with vertex set $\operatorname{Vert}(G) = \operatorname{\mathbf{Z}}_{\geq 0} = \operatorname{Edge}(G)$, with source and target maps

More informally, $G$ is the graph depicted in the diagram

The associated $1$-dimensional simplicial set $G_{\bullet }$ is connected. However, the infinite product $S_{\bullet } = \prod _{n \in \operatorname{\mathbf{Z}}_{\geq 0}} G_{\bullet }$ is not connected. By definition, the vertices of $S_{\bullet }$ can be identified with functions $f: \operatorname{\mathbf{Z}}_{\geq 0} \rightarrow \operatorname{\mathbf{Z}}_{\geq 0}$. It is not difficult to see that two such functions $f,g: \operatorname{\mathbf{Z}}_{\geq 0} \rightarrow \operatorname{\mathbf{Z}}_{\geq 0}$ belong to the same connected component of $S_{\bullet }$ if and only if the function $n \mapsto | f(n) - g(n) |$ is bounded. In particular, the identity function $n \mapsto n$ and the zero function $n \mapsto 0$ do not belong to the same connected component of $S_{\bullet }$.