Kerodon

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. $\square$

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

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. $\square$

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

admits a solution. For $m > n$, this follows from $(b)$. For $m \leq n$, we can identify (3.69) with a lifting problem

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

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

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). $\square$

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 (3.70) 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 $k \leq n$. $\square$

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 $g$ 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

Since $g'$ is a monomorphism, (3.71) 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 $Y$. $\square$

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

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

admits a solution. Unwinding the definitions, we can rewrite (3.72) as a lifting problem

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). $\square$

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

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

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