Definition 3.5.1.1. Let $X$ be a Kan complex and let $n$ be a nonnegative integer. We say that $X$ is *$n$-connective* if it is nonempty and, for every vertex $x \in X$ and every integer $0 \leq m < n$, the set $\pi _{m}(X,x)$ consists of a single element.

### 3.5.1 Connectivity

Recall that a simplicial set $X$ is *connected* if the set of path components $\pi _0(X)$ has exactly one element (Corollary 1.2.1.15). When $X$ is a Kan complex, we can use the homotopy groups introduced in ยง3.2 to formulate a hierarchy of stronger connectivity conditions.

Remark 3.5.1.2. It will sometimes be useful to extend Definition 3.5.1.1 to the case where $n$ is an arbitrary integer. By convention, if $n < 0$, then *every* Kan complex $X$ is $n$-connective.

Warning 3.5.1.3. The terminology of Definition 3.5.1.1 is not standard. Many authors refer to a Kan complex $X$ as *$n$-connected* if it is $(n+1)$-connective in the sense of Definition 3.5.1.1.

Remark 3.5.1.4. Let $X$ be a Kan complex. It follows from Example 3.2.2.18 that the isomorphism class of the homotopy group $\pi _{m}(X,x)$ depends only on the connected component $[x] \in \pi _0(X)$. Consequently, if $n > 0$, then $X$ is $n$-connective if and only if it is connected and the homotopy groups $\pi _{m}(X,x)$ are trivial for $0 < m < n$ for *some* choice of vertex $x \in X$.

Remark 3.5.1.5 (Homotopy Invariance). Let $X$ and $Y$ be Kan complexes which are homotopy equivalent. Then $X$ is $n$-connective if and only if $Y$ is $n$-connective. See Remark 3.2.2.17.

Variant 3.5.1.6. Let $X$ be a simplicial set and let $n$ be an integer. Using Corollary 3.1.7.2, we can choose an anodyne map $X \hookrightarrow Q$, where $Q$ is a Kan complex. We will say that $X$ is *$n$-connective* if the Kan complex $Q$ is $n$-connective, in the sense of Definition 3.5.1.1. By virtue of Remark 3.5.1.5 (and Warning 3.1.7.3), this condition is independent of the choice of $Q$.

Example 3.5.1.7. A simplicial set $X$ is $0$-connective if and only if it is nonempty.

Example 3.5.1.8. A simplicial set $X$ is $1$-connective if and only if it is connected (see Corollary 1.2.1.15).

Example 3.5.1.9. A Kan complex $X$ is $2$-connective if and only if it is *simply connected*: that is, $X$ is connected and the fundamental group $\pi _{1}(X,x)$ vanishes (by virtue of Remark 3.5.1.4, this condition does not depend on the choice of base point $x \in X$).

Example 3.5.1.10. Let $X$ be a Kan complex which has only a single $k$-simplex for $0 \leq k \leq n$ (that is, the $n$-skeleton $\operatorname{sk}_{n}(X)$ is isomorphic to $\Delta ^0$). Then $X$ is $(n+1)$-connective. For a partial converse, see Proposition 3.5.2.9.

Remark 3.5.1.11. Let $X$ be a simplicial set. Then $X$ is weakly contractible if and only if it is $n$-connective for every integer $n$. To prove this, we can use Corollary 3.1.7.2 to reduce to the case where $X$ is a Kan complex, in which case it is a reformulation of Proposition 3.5.1.12.

Definition 3.5.1.1 admits a number of alternative formulations.

Proposition 3.5.1.12. Let $X$ be a Kan complex and let $n$ be a nonnegative integer. The following conditions are equivalent:

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

- $(2)$
For every integer $0 \leq m \leq n$, every morphism $\operatorname{\partial \Delta }^{m} \rightarrow X$ can be extended to an $m$-simplex of $X$.

- $(3)$
Let $B$ be a simplicial set of dimension $\leq n$ and let $A \subseteq B$ be a simplicial subset. Then every morphism $f_0: A \rightarrow X$ admits an extension $f: B \rightarrow X$.

- $(4)$
Let $A$ be a simplicial set of dimension $< n$. Then every morphism $f: A \rightarrow X$ is nullhomotopic.

**Proof.**
The equivalence $(1) \Leftrightarrow (2)$ follows from Lemma 3.2.4.13 (and Variant 3.2.4.14), the implication $(2) \Rightarrow (3)$ follows from Proposition 1.1.4.12, and the implication $(4) \Rightarrow (2)$ follows from Variant 3.2.4.12. 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).
$\square$

We now introduce a relative version of Definition 3.5.1.1.

Definition 3.5.1.13. Let $f: X \rightarrow Y$ be a morphism of Kan complexes. We say that $f$ is *$n$-connective* if it satisfies the following conditions:

If $n \geq 0$, then the underlying map of connected components $\pi _0(X) \rightarrow \pi _0(Y)$ is surjective.

If $n > 0$ and $x \in X$ is a vertex having image $y = f(x)$, the induced map $\pi _{m}(X,x) \rightarrow \pi _{m}(Y,y)$ is a bijection when $m < n$ and a surjection when $m = n$.

Remark 3.5.1.14. Suppose we are given a diagram of Kan complexes

which commutes up to homotopy. If the horizontal maps are homotopy equivalences, then $f$ is $n$-connective if and only if $f'$ is $n$-connective.

Variant 3.5.1.15. Let $f: X \rightarrow Y$ be a morphism of simplicial sets and let $n$ be a nonnegative integer. Using Proposition 3.1.7.1, we can choose a commutative diagram

where $X'$ and $Y'$ are Kan complexes and the horizontal maps are weak homotopy equivalences. We will say that $f$ is *$n$-connective* if the morphism of Kan complexes $f'$ is $n$-connective, in the sense of Definition 3.5.1.13. It follows from Remark 3.5.1.14 that this condition does not depend on the choice of the diagram (3.65).

Example 3.5.1.16. For $n < 0$, every morphism of simplicial sets $f: X \rightarrow Y$ is $n$-connective.

Example 3.5.1.17. A morphism of simplicial sets $f: X \rightarrow Y$ is $0$-connective if and only if the induced map $\pi _0(f): \pi _0(X) \rightarrow \pi _0(Y)$ is surjective.

Example 3.5.1.18. Let $X$ be a simplicial set and let $n$ be an integer. Then $X$ is $n$-connective (in the sense of Variant 3.5.1.6) if and only if the projection map $X \rightarrow \Delta ^0$ is $n$-connective (in the sense of Variant 3.5.1.15).

Remark 3.5.1.19. Let $f: X \rightarrow Y$ be a morphism of simplicial sets. Then $f$ is a weak homotopy equivalence if and only if it is $n$-connective for every integer $n$. To see this, we can assume without loss of generality that $X$ and $Y$ are Kan complexes, in which case it is a restatement of Theorem 3.2.7.1.

Remark 3.5.1.20. Let $f,f': X \rightarrow Y$ be morphisms of simplicial sets which are homotopic. Then $f$ is $n$-connective if and only if $f'$ is $n$-connective.

Remark 3.5.1.21 (Monotonicity). Let $n$ be a nonnegative integer and let $f: X \rightarrow Y$ be an $n$-connective morphism of simplicial sets. Then $f$ is also $m$-connective for every integer $m \leq n$.

Proposition 3.5.1.22. Let $f: X \rightarrow Y$ be a Kan fibration of simplicial sets and let $n$ be an integer. Then $f$ is $n$-connective (in the sense of Variant 3.5.1.15) if and only if, for every vertex $y \in Y$, the Kan complex $X_{y} = \{ y\} \times _{Y} X$ is $n$-connective (in the sense of Definition 3.5.1.1).

**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 $n =1$. $\square$

Remark 3.5.1.23. In the situation of Proposition 3.5.1.22, it is not necessary to verify that $X_{y}$ is $n$-connective for *every* vertex $y \in Y$; it is enough to check this condition at one vertex in each connected component of $Y$ (see Remark 3.3.7.3). In particular, if $Y$ is connected, then it is enough to check that $X_{y}$ is $n$-connective for *any* vertex $y \in Y$.

Corollary 3.5.1.24. Let $n$ be an integer and suppose we are given a homotopy pullback square of simplicial sets

If the morphism $f'$ is $n$-connective (in the sense of Variant 3.5.1.6), then $f$ is also $n$-connective. Moreover, the converse holds if $g$ is surjective on connected components.

**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).
$\square$

Corollary 3.5.1.25. Let $f: X \rightarrow Y$ be a morphism of Kan complexes and let $n$ be an integer. The following conditions are equivalent:

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

- $(2)$
For every morphism of Kan complexes $Y' \rightarrow Y$, the projection map $Y' \times _{Y}^{\mathrm{h}} X \rightarrow Y'$ is $n$-connective.

- $(3)$
For every vertex $y \in Y$, the homotopy fiber $\{ y\} \times ^{\mathrm{h}}_{Y} X$ is $n$-connective

**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 $\square$

Proposition 3.5.1.26. Let $f: X \rightarrow Y$ be a morphism of simplicial sets and let $n$ be an integer. Then:

- $(1)$
If the $Y$ is $n$-connective and $f$ is $n$-connective, then $X$ is $n$-connective.

- $(2)$
If $X$ is $n$-connective and $Y$ is $(n+1)$-connective, then $f$ is $n$-connective.

- $(3)$
If $f$ is $n$-connective and $X$ is $(n+1)$-connective, then $Y$ is $(n+1)$-connective.

**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.

Corollary 3.5.1.27. Let $Y$ be a simplicial set and let $n \geq -1$ be an integer. The following conditions are equivalent:

- $(1)$
The simplicial set $Y$ is $n$-connective.

- $(2)$
The simplicial set $Y$ is nonempty and, for every vertex $y \in Y$, the inclusion map $\{ y\} \hookrightarrow Y$ is $(n-1)$-connective.

- $(3)$
There exists a vertex $y \in Y$ for which the inclusion map $\{ y\} \hookrightarrow Y$ is $(n-1)$-connective.

Corollary 3.5.1.28 (Transitivity). Let $f: X \rightarrow Y$ and $g: Y \rightarrow Z$ be morphisms of simplicial sets and let $n$ be an integer. Then:

- $(1)$
If $f$ and $g$ are $n$-connective, then the composition $(g \circ f): X \rightarrow Z$ is $n$-connective.

- $(2)$
If $g \circ f$ is $n$-connective and $g$ is $(n+1)$-connective, then $f$ is $n$-connective.

- $(3)$
If $f$ is $n$-connective and $(g \circ f)$ is $(n+1)$-connective, then $g$ is $(n+1)$-connective.

**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.
$\square$

Corollary 3.5.1.29. Let $f: X \rightarrow Y$ be a Kan fibration of simplicial sets and let $n$ be a nonnegative integer. Then $f$ is $n$-connective if and only if it satisfies the following pair of conditions:

- $(a)$
The map of connected components $\pi _0(f): \pi _0(X) \rightarrow \pi _0(Y)$ is surjective.

- $(b)$
The diagonal map $\delta _{X/Y}: X \rightarrow X \times _{Y} X$ is $(n-1)$-connective.

**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.
$\square$

Variant 3.5.1.30. Let $f: X \rightarrow Y$ be a morphism of Kan complexes which is surjective on connected components and let $n \geq 0$ be an integer. The following conditions are equivalent:

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

- $(2)$
The induced map

\[ \theta : \operatorname{Fun}( \Delta ^1, X ) = X \times ^{\mathrm{h}}_{X} X \rightarrow X \times ^{\mathrm{h}}_{Y} X \]is $(n-1)$-connective.

- $(3)$
For every pair of vertices $x,x' \in X$, the map of path spaces

\[ \{ x\} \times ^{\mathrm{h}}_{X} \{ x'\} \rightarrow \{ f(x) \} \times ^{\mathrm{h}}_{Y} \{ f(x') \} \]is $(n-1)$-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 $\overline{f}$ 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. $\square$

Corollary 3.5.1.31. Let $f: X \rightarrow Y$ be a Kan fibration of simplicial sets. Then $f$ is a weak homotopy equivalence if and only if the relative diagonal $\delta _{X/Y}: X \rightarrow X \times _{Y} X$ is a weak homotopy equivalence and the map of connected components $\pi _0(X) \rightarrow \pi _0(Y)$ is a surjection.

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

Corollary 3.5.1.32. Let $n$ be a nonnegative integer. Then a simplicial set $X$ is $n$-connective if and only if it is nonempty and the diagonal map $\delta _{X}: X \rightarrow X \times X$ is $(n-1)$-connective.

**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 $Y = \Delta ^0$.
$\square$

Corollary 3.5.1.33. A simplicial set $X$ is weakly contractible if and only if it is nonempty and the diagonal map $\delta _{X}: X \hookrightarrow X \times X$ is a weak homotopy equivalence.

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