Kerodon

Proof. The equivalence $(1) \Leftrightarrow (2)$ follows from Lemma 3.2.4.14 (and Variant 3.2.4.15), the implication $(2) \Rightarrow (3)$ follows from Proposition 1.1.4.12, and the implication $(4) \Rightarrow (2)$ follows from Variant 3.2.4.13. We complete the proof by showing that $(3)$ implies $(4)$. Applying assumption $(3)$ to the inclusion map $\emptyset \subseteq \Delta ^0$, we deduce that there exists a vertex $x \in X$. It will therefore suffice to show that if $A$ is a simplicial set of dimension $< n$, then every pair of morphisms $f_0, f_1: A \rightarrow X$ are homotopic (in particular, $f_0$ is homotopic to the constant map $A \rightarrow \{ x\} $). This follows by applying $(3)$ to the inclusion map $\operatorname{\partial \Delta }^1 \times A \hookrightarrow \Delta ^1 \times A$ (see Proposition 1.1.3.6). $\square$

Proof. Using Proposition 3.1.7.1, we can choose a commutative diagram

where the horizontal maps are inner anodyne, $Y'$ is a Kan complex, and $f'$ is a Kan fibration. Without loss of generality, we may assume that the map $g$ is bijective on vertices (for example, we could take $Y' = \operatorname{Ex}^{\infty }(Y)$; see Proposition 3.3.6.2). It follows from Proposition 3.3.7.1 that for each vertex $y \in Y$, the induced map of Kan complexes $X_{y} \rightarrow X'_{ g(y) }$ is a homotopy equivalence. In particular, $X_{y}$ is $n$-connective if and only if $X'_{ g(y) }$ is $n$-connective (Remark 3.5.1.5). We can therefore replace $f$ by $f'$, and thereby reduce to proving Proposition 3.5.1.22 in the special case where $X$ and $Y$ are Kan complexes.

Without loss of generality, we may assume that $n \geq 0$ (otherwise, the assertion is vacuous). Our proof proceeds by induction on $n$. In the case $n=0$, we must show that $f$ induces a surjection $\pi _0(f): \pi _0(X) \rightarrow \pi _0(Y)$ if and only if every fiber of $f$ is nonempty, which follows from (Corollary 3.2.6.3). Let us therefore assume that $n > 0$ and that $f$ has nonempty fibers. To carry out the inductive step, it will suffice to show that for every vertex $x \in X$ having image $y = f(x)$, the following conditions are equivalent:

$(a)$

The morphism $f$ induces a surjective group homomorphism $\pi _{n}(X,x) \rightarrow \pi _{n}(Y,y)$, and the kernel of the map $\pi _{n-1}(X,x) \rightarrow \pi _{n-1}(Y,y)$ is trivial (by convention, in the case $n = 1$, we define this kernel to be the inverse image of $[y] \in \pi _0(Y)$).

$(b)$

The set $\pi _{n-1}( X_{y}, x )$ consists of a single element.

This follows from Corollary 3.2.6.8 in the case $n > 1$, and from Variant 3.2.6.9 in the case $n =1$. $\square$

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 (3.66) is a homotopy pullback square guarantees that for every vertex $y \in Y$, the induced map of fibers $X_{y} \rightarrow X'_{ g(y) }$ is a homotopy equivalence of Kan complexes (Example 3.4.1.4). The desired result now follows from criterion of Proposition 3.5.1.22 (together with Remark 3.5.1.23). $\square$

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:

$(2')$

For every morphism of Kan complexes $Y' \rightarrow Y$, the projection map $Y' \times _{Y} X \rightarrow Y'$ is $n$-connective.

$(3')$

For every vertex $y \in Y$, the fiber $\{ y\} \times _{Y} X$ is $n$-connective.

The equivalence $(1) \Leftrightarrow (3')$ now follows from Proposition 3.5.1.22, and the equivalence $(1) \Leftrightarrow (2')$ from Corollary 3.5.1.24 $\square$

Proof. Without loss of generality, we may assume that $X$ and $Y$ are Kan complexes. We proceed by induction on $n$. If $n < 0$, then assertions $(1)$ and $(2)$ are vacuous, and $(3)$ reduces to the assertion that if $X$ is nonempty, then $Y$ is also nonempty. When $n=0$, we can restate Proposition 3.5.1.26 as follows:

$(1_0)$

If $Y$ is nonempty and $\pi _0(f): \pi _0(X) \rightarrow \pi _0(Y)$ is surjective, then $X$ is nonempty.

$(2_0)$

If $X$ is nonempty and $Y$ is connected, then the map $\pi _0(f): \pi _0(X) \rightarrow \pi _0(Y)$ is surjective.

$(3_0)$

If $\pi _0(f): \pi _0(X) \rightarrow \pi _0(Y)$ is surjective and $X$ is connected, then $Y$ is connected.

Assume that $n > 0$, and let $x \in X$ be a vertex having image $y = f(x)$. The inductive step is a consequence of the following observations:

$(1_ n)$

If the morphism $\pi _{n-1}(f): \pi _{n-1}(X,x) \rightarrow \pi _{n-1}(Y,y)$ is bijective and $\pi _{n-1}(Y,y)$ is a singleton, then $\pi _{n-1}(X,x)$ is also a singleton.

$(2_ n)$

If the sets $\pi _{n-1}(X,x)$ and $\pi _{n}(Y,y)$ are singletons, then $\pi _{n-1}(f)$ is injective and $\pi _{n}(f)$ is surjective.

$(3_ n)$

If $\pi _{n}(f)$ is surjective and $\pi _{n}(X,x)$ is a singleton, then $\pi _{n}(Y,y)$ is a singleton.

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 3.5.1.22, we can further reduce to the case $Z = \Delta ^0$. In this case, Corollary 3.5.1.28 is a restatement of Proposition 3.5.1.26. $\square$

Proof. Without loss of generality, we may assume that condition $(a)$ is satisfied. Since $f$ is a Kan fibration, every vertex $y \in Y$ has the form $f(x)$ for some vertex $x \in X$ (Corollary 3.2.6.3). It follows that every fiber of $f$ can also be viewed as a fiber of the map $q: X \times _{Y} X \rightarrow X$ given by projection onto the first factor. Using the criterion of Proposition 3.5.1.22, we see that $f$ is $n$-connective if and only if $q$ is $n$-connective. The desired result now follows by applying Corollary 3.5.1.28 to the morphisms $X \xrightarrow {\delta _{X/Y}} X \times _{Y} X \xrightarrow {q} X$, since the composite map $q \circ \delta _{X/Y} = \operatorname{id}_{X}$ is $n$-connective. $\square$

Proof. Using Proposition 3.1.7.1, we can factor $f$ as a composition $X \xrightarrow {i} \overline{X} \xrightarrow {\overline{f}} Y$, where $\overline{f}$ is a Kan fibration and $i$ is a homotopy equivalence. Replacing $\overline{X}$ by a full simplicial subset if necessary, we may further assume that $i$ is surjective on vertices. It follows from Remark 3.5.1.14 (and Proposition 3.4.0.9) that conditions $(1)$, $(2)$, or $(3)$ is satisfied by $f$ if and only if it is satisfied by $\overline{f}$. Consequently, we may replace $f$ by $\overline{f}$ and thereby reduce to proving Variant 3.5.1.30 in the special case where $f$ is a Kan fibration. In this case, we have a commutative diagram

where the horizontal maps are homotopy equivalences (Proposition 3.4.0.7), so the equivalence of $(1)$ and $(2)$ follows from Corollary 3.5.1.29 (together with Remark 3.5.1.14). Since $\theta $ is a Kan fibration (Theorem 3.1.3.1), the equivalence of $(2)$ and $(3)$ follows from Proposition 3.5.1.22. $\square$

Proof. Combine Remark 3.5.1.19 with Corollary 3.5.1.29. $\square$

Proof. Using Corollary 3.1.7.2 and Proposition 3.1.6.23, we can reduce to the situation where $X$ is a Kan complex. In this case, the follows by applying Corollary 3.5.1.31 in the special case $Y = \Delta ^0$. $\square$

Proof. Combine Remark 3.5.1.19 with Corollary 3.5.1.32. $\square$