### 3.5.2 Connectivity as a Lifting Property

Suppose we are given a commutative diagram of simplicial sets

3.67
\begin{equation} \begin{gathered}\label{equation:connectivity-diagram} \xymatrix@R =50pt@C=50pt{ X \ar [d]^{f} \ar [r] & X' \ar [d]^{f'} \\ Y \ar [r] & Y'. } \end{gathered} \end{equation}

If (3.67) is a homotopy pullback square and $f'$ is $n$-connective, then Corollary 3.5.1.24 guarantees that $f$ is also $n$-connective. In this section, we will prove a dual result: if (3.67) is a homotopy pushout square and $f$ is $n$-connective, then $f'$ is also $n$-connective (Corollary 3.5.2.7). We will deduce this from the following relative version of Proposition 3.5.1.12:

Proposition 3.5.2.1. Let $f: X \rightarrow Y$ be a Kan fibration of simplicial sets and let $n$ be an integer. The following conditions are equivalent:

- $(1)$
The morphism $f$ is $n$-connective.

- $(2)$
For every vertex $y \in Y$, the Kan complex $X_{y} = \{ y\} \times _{Y} X$ is $n$-connective.

- $(3)$
For every simplicial set $B$ of dimension $\leq n$ and every simplicial subset $A \subseteq B$, every lifting problem

\[ \xymatrix@R =50pt@C=50pt{ A \ar [r] \ar [d] & X \ar [d]^{f} \\ B \ar [r] \ar@ {-->}[ur] & Y } \]

admits a solution.

- $(4)$
For every integer $0 \leq m \leq n$, every lifting problem

\[ \xymatrix@R =50pt@C=50pt{ \operatorname{\partial \Delta }^{m} \ar [r] \ar [d] & X \ar [d]^{f} \\ \Delta ^{m} \ar [r]^-{\sigma } \ar@ {-->}[ur] & Y } \]

admits a solution.

**Proof.**
The equivalence $(1) \Leftrightarrow (2)$ is Proposition 3.5.1.22, the implication $(3) \Rightarrow (4)$ is immediate, and the converse follows from Proposition 1.1.4.12. Note that any morphism $\sigma : \Delta ^{m} \rightarrow Y$ is homotopic to a constant map. Using the homotopy extension lifting property (Remark 3.1.5.3), we see that $(4)$ is equivalent to the following *a priori* weaker assertion:

- $(4')$
For every integer $0 \leq m \leq n$ and every vertex $y \in Y$, every lifting problem

\[ \xymatrix@R =50pt@C=50pt{ \operatorname{\partial \Delta }^{m} \ar [r] \ar [d] & X_{y} \ar [d]^{f_ y} \\ \Delta ^{m} \ar [r]^-{\sigma } \ar@ {-->}[ur] & \{ y\} } \]

admits a solution.

The equivalence of $(2)$ and $(4')$ now follows from Proposition 3.5.1.12.
$\square$

Corollary 3.5.2.2. Let $n$ be an integer and let $f: X \rightarrow Y$ be a morphism of simplicial sets which is bijective on $k$-simplices for $k < n$ and surjective for $k = n$. Then $f$ is $n$-connective.

**Proof.**
For $n \leq 0$, this follows immediately from Example 3.5.1.17. We will therefore assume that $n > 0$. Our assumptions on $f$ guarantee that for $0 < m \leq n$, every lifting problem

\[ \xymatrix@R =50pt@C=50pt{ \Lambda ^{m}_{i} \ar [r] \ar [d] & X \ar [d]^{f} \\ \Delta ^{m} \ar [r] \ar@ {-->}[ur] & Y } \]

admits a solution. Using Variant 3.1.7.12, we can factor $f$ as a composition $X \xrightarrow {i} X' \xrightarrow {f'} Y$, where $i$ is an anodyne morphism which is bijective on $k$-simplices for $k < n$, and $f'$ is a Kan fibration. Since the collection of $n$-connective morphisms is closed under composition, it will suffice to show that $f'$ is $n$-connective. By virtue of Proposition 3.5.1.22, this is equivalent to the assertion that for each vertex $y \in Y$, the fiber $X'_{y} = \{ y\} \times _{Y} X$ is an $n$-connective Kan complex. This follows from Example 3.5.1.10.
$\square$

Example 3.5.2.3. Let $X$ be a simplicial set, let $n$ be an integer, and let $\operatorname{sk}_{n}(X)$ denote the $n$-skeleton of $X$. Then the inclusion map $i: \operatorname{sk}_{n}(X) \hookrightarrow X$ is bijective on $m$-simplices for $m \leq n$. Applying Corollary 3.5.2.2, we conclude that $i$ is $n$-connective.

Corollary 3.5.2.4. Let $f: X \rightarrow Z$ be a Kan fibration of simplicial sets and let $n$ be an integer. The following conditions are equivalent:

- $(1)$
The morphism $f$ is $n$-connective.

- $(2)$
The morphism $f$ factors as a composition $X \xrightarrow {f'} Y \xrightarrow {f''} Z$, where $f'$ is a monomorphism which is bijective on $k$-simplices for $k \leq n$ and $f''$ is a trivial Kan fibration.

- $(3)$
The morphism $f$ factors as a composition $X \xrightarrow {f'} Y \xrightarrow {f''} Z$ where $f'$ is bijective on $k$-simplices for $k < n$ and surjective for $k = n$, and $f''$ is $n$-connective.

**Proof.**
The implication $(2) \Rightarrow (3)$ is immediate and the implication $(3) \Rightarrow (1)$ follows from Corollary 3.5.2.2 (since the collection of $n$-connective morphisms is closed under composition; see Corollary 3.5.1.28). We will complete the proof by showing that $(1)$ implies $(2)$. Using a variant of Exercise 3.1.7.11, we can choose a factorization of $f$ as a composition $X \xrightarrow {f'} Y \xrightarrow {f''} Z$ with the following properties;

- $(a)$
For every integer $m > n$, every lifting problem

\[ \xymatrix@R =50pt@C=50pt{ \operatorname{\partial \Delta }^{m} \ar [r] \ar [d] & Y \ar [d]^{f''} \\ \Delta ^{m} \ar [r] \ar@ {-->}[ur] & Z } \]

admits a solution.

- $(b)$
The morphism $f'$ can be realized as a transfinite pushout of inclusion maps $\operatorname{\partial \Delta }^ m \hookrightarrow \Delta ^ m$ for $m > n$.

It follows immediately from $(b)$ that the morphism $f'$ is bijective on $k$-simplices for $0 \leq k \leq n$. We will complete the proof by showing that, if $f$ is $n$-connective, then $f''$ is a trivial Kan fibration: that is, every lifting problem

3.68
\begin{equation} \begin{gathered}\label{equation:connective-Kan-complex-factorization} \xymatrix@R =50pt@C=50pt{ \operatorname{\partial \Delta }^{m} \ar [r] \ar [d] & Y \ar [d]^{f''} \\ \Delta ^{m} \ar [r] \ar@ {-->}[ur] & Z } \end{gathered} \end{equation}

admits a solution. For $m > n$, this follows from $(b)$. For $m \leq n$, we can identify (3.68) with a lifting problem

\[ \xymatrix@R =50pt@C=50pt{ \operatorname{\partial \Delta }^{m} \ar [r] \ar [d] & X \ar [d]^{f} \\ \Delta ^{m} \ar [r] \ar@ {-->}[ur] & Z, } \]

which admits a solution by virtue of our assumption that $f$ is an $n$-connective Kan fibration (Proposition 3.5.2.1).
$\square$

Corollary 3.5.2.5. Let $X$ be a Kan complex and let $n$ be an integer. The following conditions are equivalent:

- $(1)$
The Kan complex $X$ is $n$-connective.

- $(2)$
There exists a monomorphism of Kan complexes $f: X \hookrightarrow Y$ where $Y$ is contractible and $f$ is bijective on $k$-simplices for $0 \leq k \leq n$.

- $(3)$
There exists a morphism of simplicial sets $f: X \rightarrow Y$ where $Y$ is $n$-connective, $f$ is bijective on $k$-simplices for $k < n$, and $f$ is surjective on $n$-simplices.

**Proof.**
Apply Corollary 3.5.2.4 in the special case $Z = \Delta ^0$ (together with Example 3.5.1.18).
$\square$

Corollary 3.5.2.6. For every nonnegative integer $n$, the simplicial set $\operatorname{\partial \Delta }^{n}$ is $(n-1)$-connective.

**Proof.**
Since the inclusion map $\operatorname{\partial \Delta }^{n} \hookrightarrow \Delta ^{n}$ is bijective on $k$-simplices for $k < n$, it will suffice to show that the standard simplex $\Delta ^{n}$ is $(n-1)$-connective (Corollary 3.5.2.5). This is clear, since $\Delta ^{n}$ is contractible (Example 3.2.4.2).
$\square$

Corollary 3.5.2.7. Let $n$ be an integer, and suppose we are given a homotopy pushout square of simplicial sets

3.69
\begin{equation} \begin{gathered}\label{equation:pushout-of-connective} \xymatrix@R =50pt@C=50pt{ X \ar [r] \ar [d]^{f} & X' \ar [d]^{f'} \\ Y \ar [r] & Y'. } \end{gathered} \end{equation}

If $f$ is $n$-connective, then $f'$ is also $n$-connective.

**Proof.**
Using Proposition 3.1.7.1, we can factor $f$ as a composition $X \xrightarrow {i} \overline{X} \xrightarrow {\overline{f}} Y$, where $i$ is anodyne and $\overline{f}$ is a Kan fibration. Replacing $X$ by $\overline{X}$ and $X'$ by the pushout $X' \coprod _{X} \overline{X}$, we are reduced to proving Corollary 3.5.2.7 in the special case where $f$ is a Kan fibration. In this case, we can use Corollary 3.5.2.4 to factor $f$ as a composition $X \xrightarrow { \widetilde{f} } \widetilde{Y} \xrightarrow {q} Y$, where $q$ is a trivial Kan fibration and $\widetilde{f}$ is a monomorphism which is bijective on $k$-simplices for $k \leq n$. In this case, our assumption that (3.69) is a homotopy pushout square guarantees that the induced map $X' \coprod _{X} \widetilde{Y} \rightarrow Y'$ is a weak homotopy equivalence (Proposition 3.4.2.11). Consequently, to show that $f'$ is $n$-connective, it will suffice to show that the inclusion map $j: X' \hookrightarrow X' \coprod _{X} \widetilde{Y}$ is $n$-connective (Corollary 3.5.1.28). This is a special case of Corollary 3.5.2.2, since $j$ is bijective on $k$-simplices for $k \leq n$.
$\square$

Definition 3.5.2.8. Let $X$ be a simplicial set and let $n$ be a nonnegative integer. We say that $X$ is *$n$-reduced* if, for every nonnegative integer $m \leq n$, the $n$-skeleton $\operatorname{sk}_{n}(X)$ is isomorphic to the standard $0$-simplex $\Delta ^0$: that is, $X$ has a single $m$-simplex for every integer $0 \leq m \leq n$.

Proposition 3.5.2.9. Let $X$ be a simplicial set and let $n \geq 0$ be an integer. Then $X$ is $(n+1)$-connective if and only if there exists a weak homotopy equivalence $f: X \rightarrow Y$, where $Y$ is $n$-reduced.

**Proof.**
Assume first that there exists a weak homotopy equivalence $f: X \rightarrow Y$, where $Y$ is $n$-reduced. Choose a vertex $y \in Y$. Our assumption that $Y$ is $n$-reduced guarantees that the inclusion map $i: \{ y\} \hookrightarrow Y$ is bijective on $m$-simplices for $m \leq n$. Applying Corollary 3.5.2.2, we deduce that $i$ is $n$-connective. It follows that $Y$ is $(n+1)$-connective (Corollary 3.5.1.27). Since $f$ is a weak homotopy equivalence, the simplicial set $X$ is also $(n+1)$-connective.

We now prove the converse. Assume that $X$ is $(n+1)$-connective. In particular, $X$ is nonempty; we can therefore choose a vertex $x \in X$. Using Proposition 3.1.7.1, we can factor the inclusion map $\{ x\} \hookrightarrow X$ as a composition $\{ x \} \xrightarrow {j} E \xrightarrow {g} X$, where $j$ is anodyne and $g$ is a Kan fibration. Since the simplicial set $E$ is weakly contractible, our hypothesis that $X$ is $(n+1)$-connective guarantees that $f$ is $n$-connective (Proposition 3.5.1.26). Applying Corollary 3.5.2.4, we can factor $g$ as a composition $E \xrightarrow {g'} \widetilde{X} \xrightarrow {g''} X$, where $g'$ is a monomorphism which is bijective on $m$-simplices for $m \leq n$ and $g''$ is a trivial Kan fibration. Let $s$ be a section of $g''$ and let $Y = \widetilde{X} / E$ be the simplicial set obtained from $\widetilde{X}$ by collapsing the image of $g'$, so that we have a pushout square

3.70
\begin{equation} \begin{gathered}\label{equation:connective-vs-reduced} \xymatrix@R =50pt@C=50pt{ E \ar [r]^-{g'} \ar [d] & \widetilde{X} \ar [d]^{q} \\ \Delta ^0 \ar [r] & Y. } \end{gathered} \end{equation}

Since $g'$ is a monomorphism, (3.70) is a homotopy pushout square (Example 3.4.2.12). Since $E$ is weakly contractible, it follows that $q$ is a weak homotopy equivalence (Proposition 3.4.2.10). It follows that the composite map $X \xrightarrow {s} \widetilde{X} \xrightarrow {q} Y$ is a weak homotopy equivalence from $X$ to an $n$-reduced simplicial set $Y$.
$\square$

We now record a few other consequences of Proposition 3.5.2.1.

Proposition 3.5.2.11. Let $m$ and $n$ be integers and let $f: X \rightarrow Y$ be an $(m+n)$-connective morphism of Kan complexes. Let $B$ be a simplicial set of dimension $\leq m$, and let $A \subseteq B$ be a simplicial subset. Then the restriction map

\[ u: \operatorname{Fun}( B, X ) \rightarrow \operatorname{Fun}( A, X ) \times _{ \operatorname{Fun}( A, Y) } \operatorname{Fun}( B, Y ) \]

is $n$-connective.

**Proof.**
Without loss of generality, we may assume that $m \geq 0$ (otherwise, our hypothesis guarantees that $A$ and $B$ are empty, so that $u$ is an isomorphism) and that $n \geq 0$ (otherwise, the conclusion that $u$ is $n$-connective is vacuous). Using Proposition 3.1.7.1, we can factor $f$ as a composition $X \xrightarrow {j} X' \xrightarrow {f'} Y$, where $j$ is anodyne and $f'$ is a Kan fibration. Since $Y$ is a Kan complex, the simplicial set $X'$ is a Kan complex (Remark 3.1.1.11), so $j$ is a homotopy equivalence of Kan complexes. We then have a commutative diagram

\[ \xymatrix@R =50pt@C=50pt{ \operatorname{Fun}(B,X) \ar [d]^{j \circ } \ar [r]^-{u} & \operatorname{Fun}( A, X ) \times _{ \operatorname{Fun}( A, Y) } \operatorname{Fun}( B, Y ) \ar [d] \\ \operatorname{Fun}(B, X' ) \ar [r]^-{u'} & \operatorname{Fun}( A, X ) \times _{ \operatorname{Fun}( A, Y) } \operatorname{Fun}( B, Y ) } \]

where the vertical maps are homotopy equivalences (see Proposition 3.4.0.2). Consequently, to show that $u$ is $n$-connective, it will suffice to show that $u'$ is $n$-connective. We may therefore replace $f$ by $f'$, and thereby reduce to proving Proposition 3.5.2.11 in the special case where $f$ is a Kan fibration. In this case, $u$ is also a Kan fibration (Theorem 3.1.3.1). By virtue of Proposition 3.5.2.1, it will suffice to show that if $B'$ is a simplicial set of dimension $\leq n$ and $A' \subseteq B'$ is a simplicial subset, then every lifting problem

3.71
\begin{equation} \begin{gathered}\label{equation:exponentiation-for-connectivity} \xymatrix@R =50pt@C=50pt{ A' \ar [r] \ar [d] & \operatorname{Fun}( B, X ) \ar [d]^{u} \\ B' \ar [r] \ar@ {-->}[ur] & \operatorname{Fun}( A, X ) \times _{ \operatorname{Fun}( A, Y) } \operatorname{Fun}( B, Y ) } \end{gathered} \end{equation}

admits a solution. Unwinding the definitions, we can rewrite (3.71) as a lifting problem

\[ \xymatrix@R =50pt@C=50pt{ (A \times B') \coprod _{ (A \times A' ) } (B \times A') \ar [r] \ar [d] & X \ar [d]^{f} \\ B \times B' \ar [r] \ar@ {-->}[ur] & Y. } \]

Since the simplicial set $B \times B'$ has dimension $\leq m+n$ (Proposition 1.1.3.6), the existence of a solution follows from our assumption that $f$ is $(m+n)$-connective (Proposition 3.5.2.1).
$\square$

Corollary 3.5.2.12. Let $m$ and $n$ be integers, let $B$ be a simplicial set of dimension $\leq m$, and let $X$ be a Kan complex which is $(m+n)$-connective. Then, for every simplicial subset $A \subseteq B$, the restriction map $\operatorname{Fun}(B,X) \rightarrow \operatorname{Fun}(A,X)$ is $n$-connective.

**Proof.**
Apply Proposition 3.5.2.11 in the special case $Y = \Delta ^0$.
$\square$

Corollary 3.5.2.13. Let $m$ and $n$ be integers, let $B$ be a simplicial set of dimension $\leq m$, and let $f: X \rightarrow Y$ be a morphism of Kan complexes which is $(m+n)$-connective. Then the induced map $\operatorname{Fun}(B, X) \rightarrow \operatorname{Fun}(B,Y)$ is $n$-connective.

**Proof.**
Applying Proposition 3.5.2.11 in the special case $A = \emptyset $.
$\square$

Corollary 3.5.2.14. Let $m$ and $n$ be integers, let $X$ be a Kan complex which is $(m+n)$-connective, and let $B$ be a simplicial set of dimension $\leq m$. Then the Kan complex $\operatorname{Fun}(B,X)$ is $n$-connective.

**Proof.**
Apply Corollary 3.5.2.12 in the special case $A = \emptyset $ (or Corollary 3.5.2.13 in the special case $Y = \Delta ^0$).
$\square$