# Kerodon

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### 3.5.7 Truncated Kan Complexes

Let $X$ be a Kan complex. According to Proposition 3.5.2.1, $X$ is $n$-connective if and only if, for every nonnegative integer $m \leq n$, every map $\operatorname{\partial \Delta }^{m} \rightarrow X$ can be extended to an $m$-simplex of $X$. We now study a dual version of this condition.

Definition 3.5.7.1. Let $n$ be an integer. We say that a Kan complex $X$ is $n$-truncated if, for every integer $m \geq n+2$, every morphism of simplicial sets $\operatorname{\partial \Delta }^{m} \rightarrow X$ can be extended to an $m$-simplex of $X$.

Example 3.5.7.2. Let $n$ be an integer. Recall that a Kan complex $X$ is $(n+1)$-coskeletal if, for every integer $m \geq n+2$, every morphism of simplicial sets $\operatorname{\partial \Delta }^{m} \rightarrow X$ extends uniquely to an $m$-simplex of $X$ (Definition 3.5.3.1). If this condition is satisfied, then $X$ is $n$-truncated. In particular, every $n$-groupoid is $n$-truncated (Corollary 3.5.5.11). See Proposition 3.5.7.15 (or Variant 3.5.7.16) for a partial converse.

Example 3.5.7.3. For $n \leq -2$, a Kan complex $X$ is $n$-truncated if and only if it is contractible (see Theorem 3.2.4.3).

Example 3.5.7.4. A Kan complex $X$ is $(-1)$-truncated if and only if it is either empty or contractible.

Example 3.5.7.5. A Kan complex $X$ is $0$-truncated if and only if it satisfies any of the following equivalent conditions:

• Every connected component of $X$ is contractible.

• The projection map $X \twoheadrightarrow \pi _0(X)$ is a trivial Kan fibration of simplicial sets.

• The projection map $X \twoheadrightarrow \pi _0(X)$ is a homotopy equivalence.

• The Kan complex $X$ is homotopy equivalent to a discrete simplicial set.

Remark 3.5.7.6. Let $n$ be an integer. Then the collection of $n$-truncated Kan complexes is closed under products.

Proposition 3.5.7.7. Let $X$ be a Kan complex and let $n \geq 0$ be an integer. Then $X$ is $n$-truncated if and only if it satisfies the following condition for every integer $m > n$:

$(\ast _ m)$

For every vertex $x \in X$, the homotopy group $\pi _{m}(X,x)$ is trivial.

Proof. Apply Lemma 3.2.4.13. $\square$

Remark 3.5.7.8. Proposition 3.5.7.7 is also true in the case $n = -1$, provided that restate condition $(\ast _ m)$ as follows:

$(\ast '_ m)$

For every vertex $x \in X$, the set $\pi _{m}(X,x)$ consists of a single element.

Note that $(\ast '_0)$ is equivalent to the assertion that every pair of vertices of $X$ belong to the same connected component: that is, every morphism $\operatorname{\partial \Delta }^1 \rightarrow X$ can be extended to a $1$-simplex of $X$.

Remark 3.5.7.9. In the situation of Proposition 3.5.7.7, it is not necessary to verify the vanishing of the group $\pi _{m}(X,x)$ for every choice of vertex $x \in X$; it is enough to check this at one point from each connected component of $X$ (see Example 3.2.2.18). In particular, if $X$ is connected, then it is enough to check that this condition holds for any choice of vertex $x \in X$.

Example 3.5.7.10. Let $n \geq -1$ be an integer and let $A_{\ast }$ be a chain complex of abelian groups. Then the Eilenberg-MacLane space $\mathrm{K}(A_{\ast } )$ is $n$-truncated if and only if the homology groups $\mathrm{H}_{m}(A)$ vanish for $m > n$ (see Exercise 3.2.2.22).

Remark 3.5.7.11. Let $n \geq 0$ be a nonnegative integer. Then a Kan complex $X$ is $n$-truncated if and only if every connected component of $X$ is $n$-truncated.

Proof. For $n \geq 0$ this follows from the criterion of Proposition 3.5.7.7. The case $n < 0$ follows from Examples 3.5.7.4 and 3.5.7.3. $\square$

Corollary 3.5.7.13. Let $n$ be an integer and let $f: X \rightarrow Y$ be a morphism of $n$-truncated Kan complexes. Then $f$ is a homotopy equivalence if and only if it is $(n+1)$-connective.

Proof. If $n \leq -2$, then $X$ and $Y$ are contractible and there is nothing to prove. For $n \geq -1$, the desired result follows by combining Proposition 3.5.7.7 (and Remark 3.5.7.8) with Theorem 3.2.7.1. $\square$

Corollary 3.5.7.14. Let $f: X \rightarrow Y$ be a morphism of Kan complexes which is $n$-connective for some integer $n$. Then the induced map $\operatorname{cosk}_{n}(f): \operatorname{cosk}_{n}(X) \rightarrow \operatorname{cosk}_{n}(Y)$ is a homotopy equivalence.

Proof. We have a commutative diagram of Kan complexes

$\xymatrix@R =50pt@C=50pt{ X \ar [r] \ar [d]^{f} & \operatorname{cosk}_{n}(X) \ar [d]^{\operatorname{cosk}_{n}(f) } \\ Y \ar [r] & \operatorname{cosk}_{n}(Y), }$

where the horizontal maps are $n$-connective (Remark 3.5.3.22). Applying Corollary 3.5.1.28, we deduce that $\operatorname{cosk}_{n}(f)$ is $n$-connective. Since $\operatorname{cosk}_{n}(X)$ and $\operatorname{cosk}_{n}(Y)$ are $(n-1)$-truncated (Example 3.5.7.2), Corollary 3.5.7.13 guarantees that $\operatorname{cosk}_{n}(f)$ is a homotopy equivalence. $\square$

Proposition 3.5.7.15. Let $X$ be a Kan complex and let $n$ be an integer. The following conditions are equivalent:

$(1)$

The Kan complex $X$ is $n$-truncated.

$(2)$

There exists an $n$-truncated Kan complex $Y$ which is homotopy equivalent to $X$.

$(3)$

There exists an $(n+1)$-coskeletal Kan complex $Y$ which is homotopy equivalent to $X$.

$(4)$

The tautological map $X \rightarrow \operatorname{cosk}_{n+1}(X)$ is a homotopy equivalence.

Proof. The implication $(2) \Rightarrow (1)$ follows from Corollary 3.5.7.12, the implication $(3) \Rightarrow (2)$ from Example 3.5.7.2, and the implication $(4) \Rightarrow (3)$ from the observation that $\operatorname{cosk}_{n+1}(X)$ is a Kan complex (Proposition 3.5.3.23). We will complete the proof by showing that $(1)$ implies $(4)$. Assume that $X$ is $n$-truncated; we wish to show that the tautological map $u: X \rightarrow \operatorname{cosk}_{n+1}(X)$ is a homotopy equivalence. Since $\operatorname{cosk}_{n+1}(X)$ is also $n$-truncated (Example 3.5.7.2), it will suffice to show that $u$ is $(n+1)$-connective (Corollary 3.5.7.13). This is a special case of Remark 3.5.3.22. $\square$

Variant 3.5.7.16. Let $X$ be a Kan complex and let $n \geq 0$. The following conditions are equivalent:

$(1)$

The Kan complex $X$ is $n$-truncated.

$(2)$

There exists a homotopy equivalence $X \rightarrow Y$, where $Y$ is an $n$-groupoid.

$(3)$

The tautological map $X \rightarrow \pi _{\leq n}(X)$ is a homotopy equivalence.

Proof. The implication $(3) \Rightarrow (2)$ follows from Corollary 3.5.6.16 and the implication $(2) \Rightarrow (1)$ follows from Example 3.5.7.2. The implication $(3) \Rightarrow (1)$ follows from Proposition 3.5.7.15, since the tautological map $X \rightarrow \pi _{\leq n}(X)$ factors as a composition

$X \rightarrow \operatorname{cosk}_{n+1}(X) \xrightarrow {q} \pi _{\leq n}(X),$

where $q$ is a trivial Kan fibration (Corollary 3.5.6.15). $\square$

Example 3.5.7.17. Let $X$ be a Kan complex. The following conditions are equivalent:

• The Kan complex $X$ is $1$-truncated.

• For every vertex $x \in X$, the homotopy groups $\pi _{n}(X,x)$ are trivial for $n \geq 2$.

• There exists a groupoid $\mathcal{G}$ and a homotopy equivalence $X \xrightarrow {\sim } \operatorname{N}_{\bullet }( \mathcal{G} )$.

• The tautological map $X \rightarrow \pi _{\leq 1}(X)$ is a homotopy equivalence.

We now give the proof of Proposition 3.5.0.1:

Corollary 3.5.7.18. Let $X$ be a Kan complex and let $n \geq 0$ be an integer. The following conditions are equivalent:

$(1)$

There exists a homotopy equivalence of Kan complexes $X \rightarrow \mathrm{K}(G,n)$. Here $G$ is a set if $n=0$, a group if $n =1$, and an abelian group if $n \geq 2$ (see Construction 2.5.6.9).

$(2)$

The Kan complex $X$ is $n$-truncated and $n$-connective.

Proof. We will show that $(2) \Rightarrow (1)$ (the reverse implication is clear). Assume that $X$ is $n$-truncated and $n$-connective. By virtue of Proposition 3.5.2.9 (and Remark 3.5.2.10), we can assume that $X$ has a single $m$-simplex for each $m < n$. Our assumption that $X$ is $n$-truncated guarantees that the tautological map $X \rightarrow \pi _{\leq n}(X)$ is a homotopy equivalence (Variant 3.5.7.16). It will therefore suffice to show that $\pi _{\leq n}(X)$ is an Eilenberg-MacLane space $\mathrm{K}(G,n)$, which follows from the criterion of Proposition 3.5.5.16. $\square$

Definition 3.5.7.19. Let $f: X \rightarrow Y$ be a morphism of Kan complexes and let $n$ be an integer. We say that $f$ exhibits $Y$ as an $n$-truncation of $Y$ if $Y$ is $n$-truncated and $f$ is $(n+1)$-connective. We say that $Y$ is an $n$-truncation of $X$ if there exists a morphism $f: X \rightarrow Y$ which exhibits $Y$ as an $n$-truncation of $X$.

Remark 3.5.7.20. Let $f: X \rightarrow Y$ be a morphism of Kan complexes and let $n \geq 0$. Then $f$ exhibits $Y$ as an $n$-truncation of $X$ if and only if the following conditions are satisfied:

• The morphism $f$ induces a bijection from $\pi _0(X)$ to $\pi _0(Y)$.

• For every vertex $x \in X$ having image $y = f(x)$, the map of homotopy groups $\pi _{m}(X,x) \rightarrow \pi _{m}(Y,y)$ is a bijection for $0 < m \leq n$.

• For each vertex $y \in Y$ and every integer $m > n$, the homotopy group $\pi _{m}(Y,y)$ vanishes.

See Proposition 3.5.7.7.

Remark 3.5.7.21. Let $f: X \rightarrow Y$ be a morphism of Kan complexes. The condition that $f$ exhibits $Y$ as an $n$-truncation of $X$ depends only on the homotopy class $[f]$ (see Remark 3.5.1.20).

Remark 3.5.7.22. Let $f: X \rightarrow Y$ and be a morphism of Kan complexes, and let $g: Y \rightarrow Z$ be a homotopy equivalence of Kan complexes. Then $f$ exhibits $Y$ as an $n$-truncation of $X$ if and only if $g \circ f$ exhibits $Z$ as an $n$-truncation of $X$. See Corollaries 3.5.1.28 and 3.5.7.12.

Example 3.5.7.23. Let $X$ be a Kan complex and let $n$ be an integer. The coskeleton $\operatorname{cosk}_{n+1}(X)$ is a Kan complex (Proposition 3.5.3.23) which is $(n+1)$-coskeletal, and therefore $n$-truncated (Example 3.5.7.2). Remark 3.5.3.22 guarantees that the tautological map $f: X \rightarrow \operatorname{cosk}_{n+1}(X)$ is $(n+1)$-connective. It follows that $f$ exhibits $\operatorname{cosk}_{n+1}(X)$ as an $n$-truncation of $X$.

Example 3.5.7.24. Let $X$ be a Kan complex and let $n$ be an integer. Then the tautological map $X \rightarrow \operatorname{cosk}^{\circ }_{n}(X)$ exhibits the weak coskeleton $\operatorname{cosk}^{\circ }_{n}(X)$ as an $n$-truncation of $X$. This follows from Example 3.5.7.23 and Remark 3.5.7.22, since the quotient map $\operatorname{cosk}_{n+1}(X) \twoheadrightarrow \operatorname{cosk}_{n}^{\circ }(X)$ is a trivial Kan fibration (Proposition 3.5.4.22). Alternatively, it can be deduced directly from Remark 3.5.4.16.

Example 3.5.7.25. Let $X$ be a Kan complex and let $n$ be a nonnegative integer. Then the tautological map $f: X \rightarrow \pi _{\leq n}(X)$ exhibits the fundamental $n$-groupoid $\pi _{\leq n}(X)$ as an $n$-truncation of $X$. This follows from Example 3.5.7.23 and Remark 3.5.7.22, since the quotient map $\operatorname{cosk}_{n+1}(X) \twoheadrightarrow \pi _{\leq n}(X)$ is a trivial Kan fibration (Corollary 3.5.6.15).

Example 3.5.7.26. Let $f: X \rightarrow Y$ be a morphism of Kan complexes. Then $f$ exhibits $Y$ as a $(-1)$-truncation of $X$ if and only if one of the following two conditions is satisfied:

• Both $X$ and $Y$ are empty.

• The Kan complex $X$ is nonempty and $Y$ is contractible.

See Example 3.5.7.4.

Example 3.5.7.27. Let $f: X \rightarrow Y$ be a morphism of Kan complexes. For $n \leq -2$, $f$ exhibits $Y$ as an $n$-truncation of $X$ if and only if $Y$ is contractible. See Example 3.5.7.3.

Example 3.5.7.28. Let $X$ be a Kan complex and let $n$ be an integer. Then the projection map $X \rightarrow \Delta ^0$ exhibits $\Delta ^0$ as an $n$-truncation of $X$ if and only if $X$ is $(n+1)$-connective.

Let $X$ be a Kan complex and let $n$ be an integer. It follows from Example 3.5.7.23 that there exists a morphism of Kan complexes $f: X \rightarrow Y$ which exhibits $Y$ as an $n$-truncation of $X$. We now show that this property characterizes $Y$ up to homotopy equivalence. This is a consequence of the following universal mapping property:

Proposition 3.5.7.29. Let $n$ be an integer and let $f: X \rightarrow Y$ be a morphism of Kan complexes, where $Y$ is $n$-truncated. The following conditions are equivalent:

$(1)$

The morphism $f$ exhibits $Y$ as an $n$-truncation of $X$: that is, $f$ is $(n+1)$-connective.

$(2)$

For every $n$-truncated Kan complex $Z$, composition with $f$ induces a homotopy equivalence of Kan complexes $\operatorname{Fun}(Y, Z ) \rightarrow \operatorname{Fun}(X,Z )$.

$(3)$

For every $n$-truncated Kan complex $Z$, composition with the homotopy class $[f]$ induces a bijection $\pi _0( \operatorname{Fun}(Y,Z) ) \rightarrow \pi _0( \operatorname{Fun}(X, Z ) )$.

Proof. We first show that $(1)$ implies $(2)$. Let $Z$ be an $n$-truncated simplicial set; we wish to show that composition with $f$ induces a homotopy equivalence $\theta : \operatorname{Fun}(Y, Z ) \rightarrow \operatorname{Fun}(X, Z)$. By virtue of Proposition 3.5.7.15, we may assume without loss of generality that $Z$ is $(n+1)$-coskeletal. In this case, we can use Proposition 3.5.3.17 to identify $\theta$ with the map

$\operatorname{Fun}( \operatorname{cosk}_{n+1}(Y), Z ) \rightarrow \operatorname{Fun}( \operatorname{cosk}_{n+1}(X), Z )$

given by precomposition with $\operatorname{cosk}_{n+1}(f)$. If $f$ is $(n+1)$-connective, then Corollary 3.5.7.14 guarantees that $\operatorname{cosk}_{n+1}(f)$ is a homotopy equivalence, so that $\theta$ is also a homotopy equivalence.

The implication $(2) \Rightarrow (3)$ follows from Remark 3.1.6.5. We will complete the proof by showing that $(3)$ implies $(1)$. Let $u: X \rightarrow \operatorname{cosk}_{n+1}(X)$ be the tautological map. Then $u$ exhibits $\operatorname{cosk}_{n+1}(X)$ as an $n$-truncation of $X$ (Example 3.5.7.23). In particular, $\operatorname{cosk}_{n+1}(X)$ is an $n$-truncated Kan complex, so condition $(3)$ guarantees that there exists a morphism $g: Y \rightarrow \operatorname{cosk}_{n+1}(X)$ such that $g \circ f$ is homotopic to $u$. For every $n$-truncated Kan complex $Z$, we have a commutative diagram

$\xymatrix@R =50pt@C=50pt{ \operatorname{Hom}_{\mathrm{h} \mathit{\operatorname{Kan}}}( \operatorname{cosk}_{n+1}(X), Z) ) \ar [rr]^{ \circ [g] } \ar [dr]_{ \circ [u] } & & \operatorname{Hom}_{\mathrm{h} \mathit{\operatorname{Kan}}}( Y, Z ) \ar [dl]^{ \circ [f] } \\ & \operatorname{Hom}_{\mathrm{h} \mathit{\operatorname{Kan}}}( X, Z), & }$

where the vertical maps are bijective. It follows that $g$ is a homotopy equivalence, so that $f$ exhibits $Y$ as an $n$-truncation of $X$ by virtue of Remarks 3.5.7.22 and 3.5.7.21. $\square$

Corollary 3.5.7.30. Let $n$ be an integer and let $\mathrm{h} \mathit{\operatorname{Kan}}^{\leq n}$ denote the full subcategory of the homotopy category $\mathrm{h} \mathit{\operatorname{Kan}}$ spanned by the $n$-truncated Kan complexes. Then the inclusion map $\mathrm{h} \mathit{\operatorname{Kan}}^{\leq n} \hookrightarrow \mathrm{h} \mathit{\operatorname{Kan}}$ admits a left adjoint, given by the construction $X \mapsto \operatorname{cosk}_{n+1}(X)$.