We now formulate a relative version of Definition 3.5.7.1.
Definition 3.5.9.1. Let $f: X \rightarrow Y$ be a morphism of Kan complexes and let $n \geq -1$ be an integer. We say that $f$ is $n$-truncated if, for every vertex $x \in X$ having image $y = f(x)$, the induced map is injective for $m = n+1$ and bijective for $m > n+1$. If $n \leq -2$, we say that $f$ is $n$-truncated if it is $(-1)$-truncated and the map $\pi _0(f): \pi _0(X) \rightarrow \pi _0(Y)$ is surjective.
\[ \pi _{m}(f): \pi _{m}(X,x) \rightarrow \pi _{m}(Y,y) \]
Example 3.5.9.2. For $n \leq -2$, a morphism of Kan complexes $f: X \rightarrow Y$ is $n$-truncated if and only if it is a homotopy equivalence. This is a reformulation of Theorem 3.2.7.1.
Example 3.5.9.3. A morphism of Kan complexes $f: X \rightarrow Y$ is $(-1)$-truncated if and only if it induces a homotopy equivalence from $X$ to a summand of $Y$.
Example 3.5.9.4. Let $X$ be a Kan complex and let $n$ be an integer. Then $X$ is $n$-truncated (in the sense of Definition 3.5.7.1) if and only if the projection map $X \rightarrow \Delta ^0$ is $n$-truncated (in the sense of Definition 3.5.9.1). For $n \geq 0$, this is a restatement of Proposition 3.5.7.7.
Remark 3.5.9.5 (Homotopy Invariance). Let $f,f': X \rightarrow Y$ be morphisms of Kan complexes which are homotopic. Then $f$ is $n$-truncated if and only if $f'$ is $n$-truncated.
Remark 3.5.9.6 (Monotonicity). Let $f: X \rightarrow Y$ be a morphism of Kan complexes which is $m$-truncated for some integer $m$. Then $f$ is also $n$-truncated for every integer $n \geq m$.
Remark 3.5.9.7 (Symmetry). Let $f: X \rightarrow Y$ be a morphism of Kan complexes. Then $f$ is $n$-truncated if and only if the opposite morphism $f^{\operatorname{op}}: X^{\operatorname{op}} \rightarrow Y^{\operatorname{op}}$ is $n$-truncated. See Remark 3.2.2.20.
Proposition 3.5.9.8. Let $f: X \rightarrow Y$ be a Kan fibration between Kan complexes and let $n$ be an integer. Then $f$ is $n$-truncated (in the sense of Definition 3.5.9.1) if and only if, for each vertex $y \in Y$, the Kan complex $X_{y} = \{ y\} \times _{Y} X$ is $n$-truncated (in the sense of Definition 3.5.7.1).
Proof. For $n \geq 0$, this follows from Corollary 3.2.6.8. This extends to the case $n = -1$ by virtue of Variant 3.2.6.9, and to the case $n \leq -2$ by virtue of Corollary 3.2.6.3. $\square$
Remark 3.5.9.9. In the situation of Proposition 3.5.9.8, it is not necessary to verify that the fiber $X_{y}$ is $n$-truncated for every vertex $y \in Y$; it is enough to check this condition at one vertex from each connected component of $Y$ (see Remark 3.3.7.3). In particular, if $Y$ is connected, then it is enough to check that the fiber $X_{y}$ is $n$-truncated for any choice of vertex $y \in Y$.
Variant 3.5.9.10. Suppose we are given a commutative diagram of Kan complexes where the vertical maps are Kan fibrations. Then $f$ is $n$-truncated if and only if, for every vertex $z \in Z$, the induced map $f_{z}: X_{z} \rightarrow Y_{z}$ is $n$-truncated. To prove this, we can use Proposition 3.1.7.1 to reduce to the case where $f$ is a Kan fibration. In this case, the desired result follows from the criterion of Proposition 3.5.9.8 (since a Kan complex can be realized as a fiber of $f$ if and only if it can be realized as a fiber of $f_{z}$ for some vertex $z \in Z$).
\[ \xymatrix@R =50pt@C=50pt{ X \ar [rr]^{f} \ar [dr] & & Y \ar [dl] \\ & Z, & } \]
Corollary 3.5.9.11. Let $n$ be an integer and suppose we are given a homotopy pullback diagram of Kan complexes If $f'$ is $n$-truncated, then $f$ is also $n$-truncated. The converse holds if $\pi _0(g)$ is surjective.
3.78
\begin{equation} \begin{gathered}\label{equation:homotopy-pullback-square-truncatedness} \xymatrix@R =50pt@C=50pt{ X \ar [r] \ar [d]^{f} & X' \ar [d]^{f'} \\ Y \ar [r]^-{g} & Y'. } \end{gathered} \end{equation}
Proof. Using Proposition 3.1.7.1, we can reduce to the case where $f$ and $f'$ are Kan fibrations. In this case, our assumption that (3.78) is a homotopy pullback square guarantees that for each vertex $y \in Y$, the induced map of fibers $X_{y} \rightarrow X'_{g(y)}$ is a homotopy equivalence (Example 3.4.1.4) In particular, $X_{y}$ is $n$-truncated if and only if $X'_{g(y)}$ is $n$-truncated (Corollary 3.5.7.12). The desired result now follows from the criterion of Proposition 3.5.9.8 (together with Remark 3.5.9.9). $\square$
Corollary 3.5.9.12. Let $f: X \rightarrow Y$ be a morphism of Kan complexes and let $n$ be an integer. The following conditions are equivalent:
The morphism $f$ is $n$-truncated.
For every morphism of Kan complexes $Y' \rightarrow Y$, the projection map $Y' \times _{Y}^{\mathrm{h}} X \rightarrow Y'$ is $n$-truncated.
For every vertex $y \in Y$, the homotopy fiber $\{ y\} \times ^{\mathrm{h}}_{Y} X$ is $n$-truncated.
Proof. Using Proposition 3.4.0.9, we can reduce to the case where $f$ is a Kan fibration. In this case, we can use Proposition 3.4.0.7 to reformulate conditions $(2)$ and $(3)$ as follows:
For every morphism of Kan complexes $Y' \rightarrow Y$, the projection map $Y' \times _{Y} X \rightarrow Y'$ is $n$-truncated.
For every vertex $y \in Y$, the fiber $\{ y\} \times _{Y} X$ is $n$-truncated.
The equivalence $(1) \Leftrightarrow (3')$ now follows from Proposition 3.5.9.8, and the equivalence $(1) \Leftrightarrow (2')$ from Corollary 3.5.9.11. $\square$
Proposition 3.5.9.13. Let $f: X \rightarrow Y$ be a morphism of Kan complexes and let $n$ be an integer. Then:
If $Y$ is $n$-truncated and $f$ is $n$-truncated, then $X$ is $n$-truncated.
If $X$ is $n$-truncated and $Y$ is $(n+1)$-truncated, then $f$ is $n$-truncated.
If $X$ is $n$-truncated, $f$ is $(n-1)$-truncated, and $\pi _0(f)$ is surjective, then $Y$ is $n$-truncated.
Proof. For every integer $m \geq 0$ and every vertex $x \in X$ having image $y = f(x)$, we make the following observations:
If the morphism $\pi _{m}(f): \pi _{m}(X,x) \rightarrow \pi _{m}(Y,y)$ is injective and $\pi _{m}(Y,y)$ is a singleton, then $\pi _{m}(X,x)$ is also a singleton.
If the sets $\pi _{m}(X,x)$ and $\pi _{m+1}(Y,y)$ are singletons, then $\pi _{m}(f)$ is injective and $\pi _{m+1}(f)$ is surjective.
If $\pi _{m}(f)$ is surjective and $\pi _{m}(X,x)$ is a singleton, then $\pi _{m}(Y,y)$ is a singleton.
If $n \geq -1$, then Proposition 3.5.9.13 follows by combining these observations with Proposition 3.5.7.7 (together with Remarks 3.5.7.8 and 3.5.7.9), by allowing $m$ to range over integers $> n$ and allowing the vertex $x$ to vary. The case $n \leq -2$ then follows from the following additional observations:
If the morphism $\pi _0(f)$ is surjective and $\pi _0(Y)$ is nonempty, then $\pi _0(X)$ is also nonempty.
If $\pi _0(X)$ is nonempty and $\pi _0(Y)$ has at most one element, then $\pi _0(f)$ is surjective.
If $\pi _0(X)$ is nonempty, then $\pi _0(Y)$ is also nonempty.
Corollary 3.5.9.14 (Transitivity). Let $f: X \rightarrow Y$ and $g: Y \rightarrow Z$ be morphisms of Kan complexes and let $n$ be an integer. Then:
If the morphisms $f$ and $g$ are $n$-truncated, then the composition $(g \circ f): X \rightarrow Z$ is $n$-truncated.
If $(g \circ f)$ is $n$-truncated and $g$ is $(n+1)$-truncated, then $f$ is $n$-truncated.
If $(g \circ f)$ is $n$-truncated, $f$ is $(n-1)$-truncated, and $\pi _0(f)$ is surjective, then $g$ is $n$-truncated.
Proof. Using Proposition 3.1.7.1, we can reduce to the case where $Z$ is a Kan complex and the morphisms $f$ and $g$ are Kan fibrations. Using the criterion of Proposition 3.5.9.8, we can further reduce to the case $Z = \Delta ^0$. In this case, Corollary 3.5.9.14 is a restatement of Proposition 3.5.9.13. $\square$
Proposition 3.5.9.15. Let $X$ be a Kan complex, let $n$ be an integer, and let $k$ be a nonnegative integer. If $X$ is $n$-truncated, then the diagonal map $\delta : X \rightarrow \operatorname{Fun}( \operatorname{\partial \Delta }^{k}, X )$ is $(n-k)$-truncated. The converse holds if $k \leq n+2$.
Proof. We proceed by induction on $k$. If $k = 0$, the result is a reformulation of Example 3.5.9.4. Let us therefore assume that $k > 0$. Note that $\delta $ factors as a composition $X \hookrightarrow \operatorname{Fun}( \Delta ^ k, X ) \xrightarrow { R_ k } \operatorname{Fun}( \operatorname{\partial \Delta }^{k}, X)$, where the first map is a homotopy equivalence (see Example 3.2.4.2) and $R_{k}$ is a Kan fibration (Corollary 3.1.3.3). Consequently, $\delta $ is $(n-k)$-truncated if and only if the morphism $R_{k}$ is $(n-k)$-truncated. To carry out the inductive step, it will suffice to prove the following:
If $R_{k-1}: \operatorname{Fun}( \Delta ^{k-1}, X) \rightarrow \operatorname{Fun}( \operatorname{\partial \Delta }^{k-1}, X)$ is $m$-truncated, then $R_{k}: \operatorname{Fun}( \Delta ^{k}, X) \rightarrow \operatorname{Fun}( \operatorname{\partial \Delta }^{k}, X)$ is $(m-1)$-truncated. The converse holds for $m \geq -1$.
Assume first that $R_{k-1}$ is $m$-truncated. Note that we have a pullback diagram of restriction maps
Applying the criterion of Proposition 3.5.9.8, we conclude that $T$ is also $m$-truncated. Note that the composition $(T \circ R_{k} ): \operatorname{Fun}( \Delta ^{k}, X) \rightarrow \operatorname{Fun}( \Lambda ^{k}_{k}, X)$ is given by precomposition with the horn inclusion $\Lambda ^{k}_{k} \hookrightarrow \Delta ^{k}$, and is therefore a trivial Kan fibration (Corollary 3.1.3.6). In particular, $T \circ R_{k}$ is $(m-1)$-truncated, so Corollary 3.5.9.14 guarantees that $R_{k}$ is $(m-1)$-truncated by virtue of Corollary 3.5.9.14.
We now prove the converse. Assume that $R_{k}$ is $(m-1)$-truncated and that $m \geq -1$; we wish to show that $R_{k-1}$ is $m$-truncated. Let $\operatorname{Fun}'( \operatorname{\partial \Delta }^{k}, X )$ denote the summand of $\operatorname{Fun}( \operatorname{\partial \Delta }^{k}, X)$ whose vertices are nullhomotopic maps $\operatorname{\partial \Delta }^{k} \rightarrow X$, and let $T': \operatorname{Fun}'( \operatorname{\partial \Delta }^{k}, X) \rightarrow \operatorname{Fun}( \Lambda ^{k}_{k}, X)$ be the restriction map. As above, the composition $T' \circ R_{k}$ is a trivial Kan fibration, and therefore $m$-truncated Applying Corollary 3.5.9.14, we conclude that $T'$ is $m$-truncated.
Fix a morphism $\sigma _0: \operatorname{\partial \Delta }^{k-1} \rightarrow X$, and set $Y = \{ \sigma _0 \} \times _{ \operatorname{Fun}( \operatorname{\partial \Delta }^{k-1}, X) } \operatorname{Fun}( \Delta ^{k-1}, X)$; by virtue of Proposition 3.5.9.8, it will suffice to show that $Y$ is $m$-truncated. We first consider the case $m \geq 0$. By virtue of Remark 3.5.7.11, it will suffice to show that every connected component $Z \subseteq Y$ is $m$-truncated. Fix a vertex of $Z$, corresponding to a map $\sigma : \Delta ^{k=1} \rightarrow X$ extending $\sigma _0$. Choose an extension of $\sigma $ to a $k$-simplex $\tau : \Delta ^{k} \rightarrow X$ (for example, we can take $\tau $ to be the degenerate $k$-simplex $s^{k-1}_{k-1}(\sigma )$), and set $\tau _0 = \tau |_{ \Lambda ^{k}_{k} }$. Since (3.79) is a pullback square, it induces an isomorphism from $Y$ to the fiber $T^{-1} \{ \tau _0 \} = \{ \tau _0 \} \times _{ \operatorname{Fun}( \Lambda ^{k}_{k}, X) } \operatorname{Fun}( \operatorname{\partial \Delta }^{k}, X)$. By construction, this isomorphism identifies $Z$ to a connected component of the fiber $T'^{-1} \{ \tau _0 \} = \times _{ \operatorname{Fun}( \Lambda ^{k}_{k}, X) } \operatorname{Fun}( \operatorname{\partial \Delta }^{k}, X)$. Our assumption that $T'$ is $m$-truncated guarantees that this fiber $T'^{-1} \{ \tau _0 \} $ is $m$-truncated (Proposition 3.5.9.8), so that $Z$ is also $m$-truncated (Remark 3.5.7.11).
We now treat the case $m = -1$: in this case, we wish to show that $Y$ is either empty or contractible. Let us assume that $Y$ is nonempty: that is, $\sigma _0$ can be extended to a $(k-1)$-simplex $\sigma : \Delta ^{k-1} \rightarrow X$. Define $\tau $ and $\tau _0$ as above, so that we can identify $Y$ with the fiber $T^{-1} \{ \tau _0 \} $. We will complete the proof by showing that $T$ is a trivial Kan fibration. Since $T$ is a Kan fibration, it will suffice to show that it is a homotopy equivalence (Proposition 3.2.7.2). Since $T \circ R_{k}$ is a homotopy equivalence, we are reduced to showing that $R_{k}$ is a homotopy equivalence. This is a reformulation of our hypothesis that $R_{k}$ is $(m-1)$-truncated (see Example 3.5.9.2). $\square$
Corollary 3.5.9.16. Let $f: X \rightarrow Y$ be a Kan fibration between Kan complexes, let $n$ be an integer, and let $k$ be a nonnegative integer. If $f$ is $n$-truncated, then the relative diagonal map is $(n-k)$-truncated. The converse holds if $k \geq n+2$.
\[ \delta : X \rightarrow Y \times _{ \operatorname{Fun}( \operatorname{\partial \Delta }^{k}, Y) } \operatorname{Fun}( \operatorname{\partial \Delta }^{k}, X) \]
Proof. We have a commutative diagram of Kan complexes
where the vertical maps are Kan fibrations. Using Variant 3.5.9.10, we see that $\delta $ is $(n-k)$-truncated if and only if, for each vertex $y \in Y$, the induced map of fibers
is $(n-k)$-truncated. The desired result now follows from Proposition 3.5.9.15. $\square$
Corollary 3.5.9.17. Let $f: X \rightarrow Y$ be a Kan fibration between Kan complexes and let $n \geq -1$. Then $f$ is $n$-truncated if and only if the relative diagonal $\delta _{X/Y}: X \rightarrow X \times _{Y} X$ is $(n-1)$-truncated.
Proof. Apply Corollary 3.5.9.16 in the case $k = 1$. $\square$
Example 3.5.9.18. Let $X$ be a Kan complex. Then the diagonal map $\delta _{X}: X \hookrightarrow X \times X$ factors as a composition where $u$ is a homotopy equivalence and $q$ is a Kan fibration (Corollary 3.1.3.3). Combining Corollary 3.5.9.17 with Proposition 3.5.9.8, we see that the following conditions are equivalent for every integer $n \geq -1$:
The Kan complex $X$ is $n$-truncated.
The diagonal morphism $\delta _{X}: X \hookrightarrow X \times X$ is $(n-1)$-truncated.
For every pair of vertices $x,y \in X$, the path space
is $(n-1)$-truncated.
\[ X \xrightarrow {u} \operatorname{Fun}( \Delta ^1, X ) \rightarrow {q} \operatorname{Fun}( \operatorname{\partial \Delta }^1, X) \simeq X \times X, \]
\[ \{ x \} \times ^{\mathrm{h}}_{X} \{ y\} = \{ (x,y) \} \times _{ \operatorname{Fun}( \operatorname{\partial \Delta }^1, X) } \operatorname{Fun}(\Delta ^1, X) \]
Variant 3.5.9.19. Let $f: X \rightarrow Y$ be a morphism between Kan complexes, let $n$ be an integer, and let $k$ be a nonnegative integer. If $f$ is $n$-truncated, then the restriction map is $(n-k)$-truncated. The converse holds if $k \geq n+2$.
\[ u: \operatorname{Fun}(\Delta ^ k, X) \rightarrow \operatorname{Fun}(\Delta ^ k, Y) \times _{ \operatorname{Fun}( \operatorname{\partial \Delta }^{k}, Y) } \operatorname{Fun}( \operatorname{\partial \Delta }^{k}, X) \]
Proof. Using Proposition 3.1.7.1, we can factor $f$ as a composition $X \xrightarrow {i} X' \xrightarrow {f'} Y$, where $i$ is anodyne and $f'$ is a Kan fibration. Then $X'$ is a Kan complex (Remark 3.1.1.11), so $i$ is a homotopy equivalence. We then have a commutative diagram
where the vertical maps are homotopy equivalences. It follows that $u$ is $(n-k)$-truncated if and only if $u'$ is $(n-k)$-truncated. We may therefore replace $f$ by $f'$ and thereby reduce to proving Variant 3.5.9.19 in the special case where $f$ is a Kan fibration. In this case, we have a commutative diagram
where the vertical maps are homotopy equivalences (by virtue of the contractibility of $\Delta ^{k}$). It follows that $u$ is $(n-k)$-truncated if and only if $\delta $ is $(n-k)$-truncated, so that Variant 3.5.9.19 is a reformulation of Corollary 3.5.9.16. $\square$
Corollary 3.5.9.20. Let $f: X \rightarrow Y$ be a morphism of Kan complexes and let $n \geq -2$. The following conditions are equivalent:
The morphism $f$ is $n$-truncated.
The restriction map
is a homotopy equivalence.
The diagram of Kan complexes
is a homotopy pullback square.
The diagram of Kan complexes
is a homotopy pullback square.
\[ \theta : \operatorname{Fun}(\Delta ^{n+2}, X) \rightarrow \operatorname{Fun}(\Delta ^{n+2}, Y) \times _{ \operatorname{Fun}( \operatorname{\partial \Delta }^{n+2}, Y) } \operatorname{Fun}( \operatorname{\partial \Delta }^{n+2}, X) \]
\[ \xymatrix@R =50pt@C=50pt{ \operatorname{Fun}(\Delta ^{n+2},X) \ar [d]^{f} \ar [r] & \operatorname{Fun}( \operatorname{\partial \Delta }^{n+2}, X) \ar [d] \\ \operatorname{Fun}(\Delta ^{n+2}, Y) \ar [r] & \operatorname{Fun}( \operatorname{\partial \Delta }^{n+2}, Y) } \]
\[ \xymatrix@R =50pt@C=50pt{ X \ar [d]^{f} \ar [r] & \operatorname{Fun}( \operatorname{\partial \Delta }^{n+2}, X) \ar [d] \\ Y \ar [r] & \operatorname{Fun}( \operatorname{\partial \Delta }^{n+2}, Y) } \]
Proof. The equivalence $(1) \Leftrightarrow (2)$ follows by applying Variant 3.5.9.19 in the special case $k = n+2$. The equivalence $(2) \Leftrightarrow (3)$ follows from Example 3.4.1.3, and the equivalence $(3) \Leftrightarrow (4)$ from Corollary 3.4.1.12. $\square$
Remark 3.5.9.21. In the situation of Corollary 3.5.9.20, suppose that $f$ is a Kan fibration. Then the morphism $\theta $ is also a Kan fibration (Theorem 3.1.3.1). Consequently, $f$ is $n$-truncated if and only if $\theta $ is a trivial Kan fibration (Proposition 3.2.7.2).
Corollary 3.5.9.22. Let $X$ be a Kan complex and let $n \geq -2$ be an integer. Then $X$ is $n$-truncated if and only if the diagonal map $X \rightarrow \operatorname{Fun}( \operatorname{\partial \Delta }^{n+2}, X)$ is a homotopy equivalence.
Corollary 3.5.9.23. Let $f: X \rightarrow Y$ be a Kan fibration between Kan complexes and let $n$ be an integer. The following conditions are equivalent:
The morphism $f$ is $n$-truncated.
For every nonnegative integer $m \geq n+2$, the induced map
is a trivial Kan fibration.
For every nonnegative integer $m \geq n+2$, every lifting problem
has a solution.
For every simplicial set $B$ and every simplicial subset $A \subseteq B$ which contains the $(n+1)$-skeleton $\operatorname{sk}_{n+1}(B)$, every lifting problem
admits a solution.
\[ \theta : \operatorname{Fun}( \Delta ^{m}, X) \rightarrow \operatorname{Fun}( \operatorname{\partial \Delta }^{m}, X) \times _{ \operatorname{Fun}( \operatorname{\partial \Delta }^{m}, Y) } \operatorname{Fun}( \Delta ^{m}, Y) \]
\[ \xymatrix@R =50pt@C=50pt{ \operatorname{\partial \Delta }^{m} \ar [r] \ar [d] & X \ar [d]^{f} \\ \Delta ^{m} \ar@ {-->}[ur] \ar [r] & Y } \]
\[ \xymatrix@R =50pt@C=50pt{ A \ar [r] \ar [d] & X \ar [d]^{f} \\ B \ar@ {-->}[ur] \ar [r] & Y } \]
Proof. If $f$ is $n$-truncated, then it is also $n'$-truncated for every integer $n' \geq n$ (Remark 3.5.9.6). Consequently, the implication $(1) \Rightarrow (2)$ follows from Remark 3.5.9.21. The implication $(2) \Rightarrow (3)$ is immediate from the definitions, and the implication $(3) \Rightarrow (1)$ follows from Proposition 3.5.9.8. The equivalence $(3) \Leftrightarrow (4)$ follows from Proposition 1.1.4.12. $\square$
Proposition 3.5.9.24. Let $f: X \rightarrow Y$ be an $n$-truncated Kan fibration between Kan complexes, let $k$ be an integer, and let $j: A \rightarrow B$ be a morphism of simplicial sets which is $(k-1)$-connective. Then the induced map is $(n-k)$-truncated.
\[ \theta : \operatorname{Fun}(B, X ) \rightarrow \operatorname{Fun}(A, X) \times _{ \operatorname{Fun}(A,Y) } \operatorname{Fun}(B, Y) \]
Proof. Using Proposition 3.1.7.1, we can factor $j$ as a composition $A \xrightarrow {i} A' \xrightarrow {j} B$, where $i$ is anodyne and $j$ is a Kan fibration. In this case, $\theta $ factors as a composition
where $\rho $ is a trivial Kan fibration (Theorem 3.1.3.5). It will therefore suffice to show that $\theta '$ is $(n-k)$-truncated. Using Corollary 3.5.2.4 (or Exercise 3.1.7.11, in the case $k = 0$), we can factor $j'$ as a composition $A' \xrightarrow {\widetilde{j}} \widetilde{B} \xrightarrow {q} B$, where $\widetilde{j}$ is a monomorphism which is bijective on simplices of dimension $\leq k-1$ and $q$ is a trivial Kan fibration. In this case, we have a commutative diagram
where the vertical maps are homotopy equivalences. Consequently, to prove that $\theta $ is $(n-k)$-truncated, it will suffice to show that $\widetilde{\theta }$ is $(n-k)$-truncated. We may therefore replace $j$ by $\widetilde{j}$ and thereby reduce to proving Proposition 3.5.9.24 in the special case where $j$ is a monomorphism which is bijective on simplices of dimension $\leq k-1$.
If $j$ is a monomorphism, then $\theta $ is a Kan fibration (Theorem 3.1.3.1). Consequently, to show that $\theta $ is $(n-k)$-connective, it will suffice to show that every lifting problem
has a solution, provided that $m \geq n-k+2$ (Corollary 3.5.9.23) Unwinding the definitions, we can rewrite this as a lifting problem
Since $f$ is a Kan fibration, $\theta '$ is also a Kan fibration (Theorem 3.1.3.1), and our assumption that $f$ is $n$-truncated guarantees that $\theta '$ is $(n-m)$-truncated (Variant 3.5.9.19). In particular, $\theta '$ is $(k-2)$-truncated (Remark 3.5.9.6), so the existence of the desired solution follows from Corollary 3.5.9.23. $\square$
Corollary 3.5.9.25. Let $X$ be an $n$-truncated Kan complex, let $k$ be an integer, and let $j: A \rightarrow B$ be a $(k-1)$-connective morphism of simplicial sets. Then the induced map $\operatorname{Fun}(B,X) \rightarrow \operatorname{Fun}(A,X)$ is $(n-k)$-connective.
Proof. Apply Proposition 3.5.9.24 in the special case $Y = \Delta ^0$ (see Example 3.5.9.4). $\square$
Corollary 3.5.9.26. Let $f: X \rightarrow Y$ be an $n$-truncated morphism between Kan complexes. Then, for every simplicial set $B$, the induced map $\operatorname{Fun}(B,X) \rightarrow \operatorname{Fun}(B,Y)$ is $n$-truncated.
Proof. Using Proposition 3.1.7.1, we can reduce to the case where $f$ is a Kan fibration. In this case, the desired result follows by applying Proposition 3.5.9.24 in the special case $A = \emptyset $ (and the integer $k$ is equal to zero). $\square$
Corollary 3.5.9.27. Let $X$ be an $n$-truncated Kan complex. Then, for any simplicial set $B$, the Kan complex $\operatorname{Fun}(B,X)$ is also $n$-truncated.
Proof. Apply Corollary 3.5.9.25 in the special case $A = \emptyset $ (or Corollary 3.5.9.26 in the special case $Y = \Delta ^0$). $\square$
Corollary 3.5.9.28. Let $n \geq -2$ be an integer and let $f: X \rightarrow Y$ be a morphism of Kan complexes. The following conditions are equivalent:
The morphism $f$ is $n$-truncated.
For every $(n+1)$-connective morphism of simplicial sets $A \rightarrow B$, the diagram of Kan complexes
is a homotopy pullback square.
The diagram
is a homotopy pullback square.
\[ \xymatrix@R =50pt@C=50pt{ \operatorname{Fun}(B, X) \ar [r] \ar [d] & \operatorname{Fun}(A, X) \ar [d] \\ \operatorname{Fun}( B, Y) \ar [r] & \operatorname{Fun}(A, Y) } \]
\[ \xymatrix@R =50pt@C=50pt{ X \ar [r] \ar [d] & \operatorname{Fun}( \operatorname{\partial \Delta }^{n+2}, X ) \ar [d] \\ Y \ar [r] & \operatorname{Fun}( \operatorname{\partial \Delta }^{n+2}, Y) } \]
Proof. The equivalence $(1) \Leftrightarrow (3)$ follows from Corollary 3.5.9.20, and the implication $(2) \Rightarrow (3)$ from Corollary 3.5.2.6. It will therefore suffice to show that $(1)$ implies $(2)$. Using Proposition 3.1.7.1, we can reduce to the case where $f$ is a Kan fibration. In this case, the map $\operatorname{Fun}(A, X) \rightarrow \operatorname{Fun}(A,Y)$ is also a Kan fibration (Corollary 3.1.3.2), so condition $(2)$ is equivalent to the requirement that the map $\theta : \operatorname{Fun}(B, X) \rightarrow \operatorname{Fun}(A, X) \times _{ \operatorname{Fun}(A,Y) } \operatorname{Fun}(B, Y)$ is a homotopy equivalence (Example 3.4.1.3). This follows from Proposition 3.5.9.24 (and Example 3.5.9.2). $\square$
Corollary 3.5.9.29. Let $X$ be a Kan complex and let $n \geq -2$ be an integer. The following conditions are equivalent:
The Kan complex $X$ is $n$-truncated.
For $m \geq n+2$, every morphism $f: \operatorname{\partial \Delta }^{m} \rightarrow X$ is nullhomotopic.
If $A$ is an $(n+1)$-connective simplicial set, then every morphism $f: A \rightarrow X$ is nullhomotopic.
If $A$ is an $(n+1)$-connective simplicial set, then the diagonal map $X \rightarrow \operatorname{Fun}(A,X)$ is a homotopy equivalence.
For every $(n+1)$-connective morphism of simplicial sets $A \rightarrow B$, the induced map $\operatorname{Fun}(B,X) \rightarrow \operatorname{Fun}(A,X)$ is a homotopy equivalence.
Proof. The equivalence of $(1) \Leftrightarrow (2)$ follows from Variant 3.2.4.13. The implication $(3) \Rightarrow (2)$ follows from Corollary 3.5.2.6 and the implications $(5) \Rightarrow (4) \Rightarrow (3)$ are immediate (see Example 3.5.1.18). To complete the proof, it will suffice to show that $(1)$ implies $(5)$. This follows by applying Corollary 3.5.9.28 in the special case $Y = \Delta ^0$. $\square$