# Kerodon

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### 1.1.9 Kan Complexes

We now articulate an important property enjoyed by simplicial sets of the form $\operatorname{Sing}_{\bullet }(X)$.

Definition 1.1.9.1. Let $S_{\bullet }$ be a simplicial set. We will say that $S_{\bullet }$ is a Kan complex if it satisfies the following condition:

$(\ast )$

For $n > 0$ and $0 \leq i \leq n$, any map of simplicial sets $\sigma _0: \Lambda ^{n}_{i} \rightarrow S_{\bullet }$ can be extended to a map $\sigma : \Delta ^{n} \rightarrow S_{\bullet }$. Here $\Lambda ^{n}_{i} \subseteq \Delta ^ n$ denotes the $i$th horn (see Construction 1.1.2.9).

Exercise 1.1.9.2. Show that for $n > 0$, the standard simplex $\Delta ^{n}$ is not a Kan complex (for a more general statement, see Proposition 1.2.4.2).

Example 1.1.9.3. Let $S_{\bullet }$ be a simplicial set of dimension exactly $1$ (that is, a simplicial set $S_{\bullet }$ which arises from a directed graph with at least one edge). Then $S_{\bullet }$ is not Kan complex.

Example 1.1.9.4 (Products of Kan Complexes). Let $\{ S_{\alpha \bullet } \} _{\alpha \in A}$ be a collection of simplicial sets parametrized by a set $A$, and let $S_{\bullet } = \prod _{\alpha \in A} S_{\alpha \bullet }$ be their product. If each $S_{\alpha \bullet }$ is a Kan complex, then $S_{\bullet }$ is a Kan complex. The converse holds provided that each $S_{\alpha \bullet }$ is nonempty.

Example 1.1.9.5 (Coproducts of Kan Complexes). Let $\{ S_{\alpha \bullet } \} _{\alpha \in A}$ be a collection of simplicial sets parametrized by a set $A$, and let $S_{\bullet } = \coprod _{\alpha \in A} S_{\alpha \bullet }$ be their coproduct. For each $0 \leq i \leq n$, the restriction map

$\theta : \operatorname{Hom}_{\operatorname{Set_{\Delta }}}( \Delta ^ n, S_{\bullet } ) \rightarrow \operatorname{Hom}_{\operatorname{Set_{\Delta }}}( \Lambda ^ n_ i, S_{\bullet } )$

can be identified with the coproduct (formed in the arrow category $\operatorname{Fun}( [1], \operatorname{Set})$) of restriction maps $\theta _{\alpha }: \operatorname{Hom}_{\operatorname{Set_{\Delta }}}( \Delta ^ n, S_{\alpha \bullet }) \rightarrow \operatorname{Hom}_{\operatorname{Set_{\Delta }}}( \Lambda ^ n_ i, S_{\alpha \bullet } )$ (this follows from the observation that the simplicial sets $\Delta ^ n$ and $\Lambda ^{n}_{i}$ are connected). It follows that $\theta$ is surjective if and only if each $\theta _{\alpha }$ is surjective. Allowing $n$ and $i$ to vary, we conclude that $S_{\bullet }$ is a Kan complex if and only if each summand $S_{\alpha \bullet }$ is a Kan complex.

Remark 1.1.9.6. Let $S_{\bullet }$ be a simplicial set. Combining Example 1.1.9.5 with Proposition 1.1.6.13, we deduce that $S_{\bullet }$ is a Kan complex if and only if each connected component of $S_{\bullet }$ is a Kan complex.

Example 1.1.9.7. Let $S$ be a set and let $\underline{S}_{\bullet }$ denote the associated constant simplicial set (Construction 1.1.4.2). Then $\underline{S}_{\bullet }$ is a Kan complex (this follows from Remark 1.1.9.6, since each connected component of $\underline{S}_{\bullet }$ is isomorphic to $\Delta ^0$; see Example 1.1.6.10).

Proposition 1.1.9.8. 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

$| \Lambda ^{n}_{i} | \rightarrow | \Delta ^{n} | \xrightarrow {f} X.$

Using Example 1.1.8.13, we can identify $| \Lambda ^{n}_{i} |$ with the subset

$\{ (t_0, \ldots , t_ n) \in | \Delta ^{n} |: t_ j = 0 \text{ for some j \neq i} \} \subseteq | \Delta ^{n} |.$

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

$r( t_0, \ldots , t_ n ) = (t_0 - c, \ldots , t_{i-1} - c, t_{i} + nc, t_{i+1} - c, \ldots , t_ n - c)$

$c = \min \{ t_0, \ldots , t_{i-1}, t_{i+1}, \ldots , t_ n \} .$
$\square$

Algebra furnishes another rich supply of examples:

Proposition 1.1.9.9. 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 $G_{n-1}$ (Exercise 1.1.2.14). We wish to prove that there exists an element $\tau \in G_{n}$ satisfying $d_{j} \tau = \sigma _ j$ 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 $\tau '' = s_ j \sigma _ j \in G_{n}$, 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 $\sigma '_ k = \sigma _ k (d_ k \tau '')^{-1}$. 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 $d_ k \tau ' = \sigma '_{k}$ 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 $\tau '' = s_{j-1} \sigma _ j$ 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 $\sigma '_ k = \sigma _ k (d_ k \tau '')^{-1}$, 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 $d_ k \tau ' = \sigma '_{k}$ for $k \neq i$. As before, we complete the proof by setting $\tau = \tau ' \tau ''$. $\square$

Let $S_{\bullet }$ be a simplicial set. According to Remark 1.1.6.23, we can identify the set of connected components $\pi _0( S_{\bullet } )$ with the quotient $S_0 / \sim$, where $\sim$ is the equivalence relation generated by the image of the map $(d_0, d_1): S_1 \rightarrow S_0 \times S_0$. In the special case where $S_{\bullet } = \operatorname{Sing}_{\bullet }(X)$ is the singular simplicial set of a topological space $X$, this description simplifies: the image of the map $(d_0, d_1): \operatorname{Sing}_{1}(X) \rightarrow \operatorname{Sing}_0(X) \times \operatorname{Sing}_0(X) = X \times X$ is already an equivalence relation, and $\pi _0( S_{\bullet } )$ can be identified with the set of path components $\pi _0(X)$ (Remark 1.1.7.4). A similar phenomenon occurs for any Kan complex:

Proposition 1.1.9.10. Let $S_{\bullet }$ be a Kan complex containing a pair of vertices $x,y \in S_0$. Then $x$ and $y$ belong to the same connected component of $S_{\bullet }$ if and only if there exists an edge $e \in S_1$ satisfying $d_0(e) = x$ and $d_1(e) = y$.

Proof. Let $R$ denote the image of the map $(d_0, d_1): S_1 \rightarrow S_0 \times S_0$. According to Remark 1.1.6.23, we can identify $\pi _0( S_{\bullet } )$ 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 map $(d_0, d_1)$ carries the degenerate edge $s_0(x)$ to the pair $(x,x) \in S_0 \times S_0$.

• The relation $R$ is symmetric. Suppose that $(x,y) \in R$: that is, there exists an edge $e \in S_1$ satisfying $d_0(e) = x$ and $d_1(e) = y$. Then the tuple $(e, s_0(x), \bullet )$ determines a map of simplicial sets $\sigma _0: \Lambda ^{2}_{2} \rightarrow S_{\bullet }$ (see Exercise 1.1.2.14), which we depict as a diagram

$\xymatrix { & y \ar [dr]^{e} & \\ x \ar@ {-->}[ur] \ar [rr]^{s_0(x)} & & x. }$

Since $S_{\bullet }$ is a Kan complex, we can complete this diagram to a $2$-simplex $\sigma : \Delta ^2 \rightarrow S_{\bullet }$. Then $e' = d_2(\sigma )$ is an edge of $S_{\bullet }$ satisfying $d_0(e') = y$ and $d_1(e') = x$, which proves that the pair $(y,x)$ 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 $(x,z) \in R$. Choose edges $e, e' \in S_1$ satisfying $d_0(e) = x$, $d_1(e) = y = d_0(e')$, and $d_1(e') = z$. Then the tuple $(e', \bullet , e)$ determines a map of simplicial sets $\tau _0: \Lambda ^2_1 \rightarrow S_{\bullet }$ (see Exercise 1.1.2.14), which we depict as a diagram

$\xymatrix { & y \ar [dr]^{e} & \\ z \ar [ur]^{e'} \ar@ {-->}[rr] & & x. }$

Our assumption that $S_{\bullet }$ is a Kan complex guarantees that we can extend $\tau _0$ to a $2$-simplex $\tau : \Delta ^2 \rightarrow S_{\bullet }$. Then $e'' = d_1(\tau )$ is an edge of $S_{\bullet }$ satisfying $d_0( e'' ) = x$ and $d_1(e'') = z$, which proves that $(x,z) \in R$.

$\square$

Corollary 1.1.9.11. Let $\{ S_{\alpha \bullet } \} _{\alpha \in A}$ be a collection of Kan complexes parametrized by a set $A$, and let $S_{\bullet } = \prod _{\alpha \in A} S_{\alpha \bullet }$ denote their product. Then the canonical map

$\pi _0( S_{\bullet } ) \rightarrow \prod _{\alpha \in A} \pi _0 ( S_{\alpha \bullet } )$

is bijective. In particular, $S_{\bullet }$ is connected if and only if each factor $S_{\alpha \bullet }$ is connected.