We now articulate an important property enjoyed by simplicial sets of the form $\operatorname{Sing}_{\bullet }(X)$.
Definition 1.2.5.1. Let $S$ be a simplicial set. We will say that $S$ is a Kan complex if it satisfies the following condition:
For every pair of integers $0 \leq i \leq n$ with $n > 0$, every morphism of simplicial sets $\sigma _0: \Lambda ^{n}_{i} \rightarrow S$ can be extended to a map $\sigma : \Delta ^{n} \rightarrow S$. Here $\Lambda ^{n}_{i} \subseteq \Delta ^ n$ denotes the $i$th horn (see Construction 1.2.4.1).
Exercise 1.2.5.2. Show that for $n > 0$, the standard simplex $\Delta ^{n}$ is not a Kan complex (for a more general statement, see Proposition 1.3.5.2).
Example 1.2.5.3 (Products of Kan Complexes). Let $\{ S_{\alpha } \} _{\alpha \in A}$ be a collection of simplicial sets parametrized by a set $A$, and let $S = \prod _{\alpha \in A} S_{\alpha }$ be their product. If each $S_{\alpha }$ is a Kan complex, then $S$ is a Kan complex. The converse holds provided that each $S_{\alpha }$ is nonempty.
Example 1.2.5.4 (Coproducts of Kan Complexes). Let $\{ S_{\alpha } \} _{\alpha \in A}$ be a collection of simplicial sets parametrized by a set $A$, and let $S = \coprod _{\alpha \in A} S_{\alpha }$ be their coproduct. For every pair of integers $0 \leq i \leq n$ with $n > 0$, the restriction map can be identified with the coproduct (formed in the arrow category $\operatorname{Fun}( [1], \operatorname{Set})$) of restriction maps $\theta _{\alpha }: \operatorname{Hom}_{\operatorname{Set_{\Delta }}}( \Delta ^ n, S_{\alpha }) \rightarrow \operatorname{Hom}_{\operatorname{Set_{\Delta }}}( \Lambda ^ n_ i, S_{\alpha } )$; this follows from the connectedness of the simplicial sets $\Delta ^{n}$ and $\Lambda ^{n}_{i}$ (see Example 1.2.1.7 and Remark 1.2.4.5). It follows that $\theta $ is surjective if and only if each $\theta _{\alpha }$ is surjective. Allowing $n$ and $i$ to vary, we conclude that $S$ is a Kan complex if and only if each summand $S_{\alpha }$ is a Kan complex.
\[ \theta : \operatorname{Hom}_{\operatorname{Set_{\Delta }}}( \Delta ^ n, S ) \rightarrow \operatorname{Hom}_{\operatorname{Set_{\Delta }}}( \Lambda ^ n_ i, S ) \]
Remark 1.2.5.5. Let $S$ be a simplicial set. Combining Example 1.2.5.4 with Proposition 1.2.1.13, we deduce that $S_{\bullet }$ is a Kan complex if and only if each connected component of $S$ is a Kan complex.
Example 1.2.5.6. Let $S$ be a discrete simplicial set (Definition 1.1.5.10). Then every connected component of $S$ is isomorphic to the standard simplex $\Delta ^{0}$, which is a Kan complex. Applying Remark 1.2.5.5, we see that $S$ is a Kan complex.
Example 1.2.5.7. Let $S$ be a simplicial set of dimension exactly $1$ (that is, a simplicial set $S$ which arises from a directed graph with at least one edge). Then $S$ is not a Kan complex.
Proposition 1.2.5.8. Let $X$ be a topological space. Then the singular simplicial set $\operatorname{Sing}_{\bullet }(X)$ is a Kan complex.
Proof. Let $\sigma _0: \Lambda ^{n}_{i} \rightarrow \operatorname{Sing}_{\bullet }(X)$ be a map of simplicial sets for $n > 0$; we wish to show that $\sigma _0$ can be extended to an $n$-simplex of $X$. Using the geometric realization functor, we can identify $\sigma _0$ with a continuous map of topological spaces $f_0: | \Lambda ^{n}_{i} | \rightarrow X$; we wish to show that $f_0$ factors as a composition
Using Remark 1.2.4.6, we can identify $| \Lambda ^{n}_{i} |$ with the subset
In this case, we can take $f$ to be the composition $f_0 \circ r$, where $r$ is any continuous retraction of $| \Delta ^{n} |$ onto the subset $| \Lambda ^{n}_{i} |$. For example, we can take $r$ to be the map given by the formula
Algebra furnishes another rich supply of examples of Kan complexes:
Proposition 1.2.5.9. Let $G_{\bullet }$ be a simplicial group (that is, a simplicial object of the category of groups). Then (the underlying simplicial set of) $G_{\bullet }$ is a Kan complex.
Proof. Let $n$ be a positive integer and $\vec{\sigma }: \Lambda ^{n}_{i} \rightarrow G_{\bullet }$ be a map of simplicial sets for some $0 \leq i \leq n$, which we will identify with a tuple $( \sigma _0, \sigma _1, \ldots , \sigma _{i-1}, \bullet , \sigma _{i+1}, \ldots , \sigma _ n)$ of elements of the group $G_{n-1}$ (Proposition 1.2.4.7). We wish to prove that there exists an element $\tau \in G_{n}$ satisfying $d^{n}_{j} \tau = \sigma _ j$ for $j \neq i$. Let $e$ denote the identity element of $G_{n-1}$. We first treat the special case where $\sigma _{i+1} = \cdots = \sigma _{n} = e$. If, in addition, we have $\sigma _{0} = \sigma _1 = \cdots = \sigma _{i-1} = e$, then we can take $\tau $ to be the identity element of $G_{n}$. Otherwise, there exists some smallest integer $j < i$ such that $\sigma _{j} \neq e$. We proceed by descending induction on $j$. Set $\tau '' = s^{n-1}_ j \sigma _ j \in G_{n}$, and consider the map $\vec{\sigma }': \Lambda ^{n}_{i} \rightarrow G_{\bullet }$ given by the tuple $( \sigma '_0, \sigma '_1, \ldots , \sigma '_{i-1}, \bullet , \sigma '_{i+1}, \ldots , \sigma '_ n)$ with $\sigma '_ k = \sigma _ k (d^{n}_ k \tau '')^{-1}$. We then have $\sigma '_0 = \sigma '_1 = \cdots = \sigma '_ j = e$ and $\sigma '_{i+1} = \cdots = \sigma '_{n} = e$. Invoking our inductive hypothesis we conclude that there exists an element $\tau ' \in G_{n}$ satisfying $d^{n}_ k \tau ' = \sigma '_{k}$ for $k \neq i$. We can then complete the proof by taking $\tau $ to be the product $\tau ' \tau ''$.
If not all of the equalities $\sigma _{i+1} = \cdots = \sigma _{n} = e$ hold, then there exists some largest integer $j > i$ such that $\sigma _ j \neq e$. We now proceed by ascending induction on $j$. Set $\tau '' = s^{n-1}_{j-1} \sigma _ j$ and let $\vec{\sigma }': \Lambda ^{n}_{i} \rightarrow G_{\bullet }$ be the map given by the tuple $( \sigma '_0, \sigma '_1, \ldots , \sigma '_{i-1}, \bullet , \sigma '_{i+1}, \ldots , \sigma '_ n)$ with $\sigma '_ k = \sigma _ k (d^{n}_ k \tau '')^{-1}$, as above. We then have $\sigma '_{j} = \sigma '_{j+1} = \cdots = \sigma '_{n} = e$, so the inductive hypothesis guarantees the existence of an element $\tau ' \in G_{n}$ satisfying $d^{n}_ k \tau ' = \sigma '_{k}$ for $k \neq i$. As before, we complete the proof by setting $\tau = \tau ' \tau ''$. $\square$
Let $S = S_{\bullet }$ be a simplicial set. According to Remark 1.2.1.23, we can identify the set of connected components $\pi _0( S )$ with the quotient $S_0 / \sim $, where $\sim $ is the equivalence relation generated by the image of the map $(d^{1}_0, d^{1}_1): S_1 \rightarrow S_0 \times S_0$. In the special case where $S= \operatorname{Sing}_{\bullet }(X)$ is the singular simplicial set of a topological space $X$, this description simplifies: the image of the map $(d^{1}_0, d^{1}_1): \operatorname{Sing}_{1}(X) \rightarrow \operatorname{Sing}_0(X) \times \operatorname{Sing}_0(X) = X \times X$ is already an equivalence relation, and $\pi _0( S_{\bullet } )$ can be identified with the set of path components $\pi _0(X)$ (Remark 1.2.2.5). A similar phenomenon occurs for any Kan complex:
Proposition 1.2.5.10. Let $S$ be a Kan complex and let $x$ and $y$ be vertices of $S$. Then $x$ and $y$ belong to the same connected component of $S$ if and only if there exists an edge $e$ of $S$ having source $x$ and target $y$.
Proof. Let $S_0$ denote the set of vertices of $S$. Let $R$ be the collection of pairs $(x,y) \in S_0$ for which there exists an edge $e$ of $S$ having source $x$ and target $y$. Using Remark 1.2.1.23, we can identify $\pi _0( S )$ with the quotient of $S_{0}$ by the equivalence relation generated by $R$. It will therefore suffice to show that $R$ is already an equivalence relation on $S_0$. To prove this, we must verify three things:
The relation $R$ is reflexive. This follows from the observation that for every vertex $x \in S_0$, the degenerate edge $\operatorname{id}_{x}$ has source $x$ and target $x$.
The relation $R$ is symmetric. Suppose that $(x,y) \in R$: that is, there exists an edge $e$ of $S$ having source $x$ and target $y$. Then the tuple $(\bullet , \operatorname{id}_ x, e)$ determines a map of simplicial sets $\sigma _0: \Lambda ^{2}_{0} \rightarrow S$ (see Proposition 1.2.4.7), which we depict as a diagram
Since $S$ is a Kan complex, we can complete this diagram to a $2$-simplex $\sigma : \Delta ^2 \rightarrow S$. Then $e' = d^{2}_0(\sigma )$ is an edge of $S$ having source $y$ and target $x$, so the pair $(y,x)$ also belongs to $R$.
The relation $R$ is transitive. Suppose that we are given vertices $x,y,z \in S_0$ with $(x,y) \in R$ and $(y,z) \in R$; we wish to show that $(x,z) \in R$. Let $e$ be an edge of $S$ having source $x$ and target $y$, and let $e'$ be an edge of $S$ having source $y$ and target $z$. Then the tuple $(e', \bullet , e)$ determines a map of simplicial sets $\tau _0: \Lambda ^2_1 \rightarrow S$ (see Proposition 1.2.4.7), which we depict as a diagram
Our assumption that $S$ is a Kan complex guarantees that we can extend $\tau _0$ to a $2$-simplex $\tau : \Delta ^2 \rightarrow S$. Then $e'' = d^{2}_1(\tau )$ is an edge of $S$ having source $x$ and target $z$, so that $(x,z)$ belongs to $R$.
Corollary 1.2.5.11. Let $\{ S_{\alpha } \} _{\alpha \in A}$ be a collection of Kan complexes parametrized by a set $A$, and let $S = \prod _{\alpha \in A} S_{\alpha \bullet }$ denote their product. Then the canonical map is bijective. In particular, $S$ is connected if and only if each factor $S_{\alpha }$ is connected.
\[ \pi _0( S ) \rightarrow \prod _{\alpha \in A} \pi _0 ( S_{\alpha } ) \]