Definition 3.2.7.1. Let $f: X \rightarrow Y$ be a morphism of Kan complexes. We say that $f$ is $0$-connective if the underlying map of connected components $\pi _0(X) \rightarrow \pi _0(Y)$ is surjective. For $n > 0$, we say that $f$ is $n$-connective if it is $0$-connective and, for every vertex $x \in X$ 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 $m = n$.
$\Newextarrow{\xhookrightarrow}{10,10}{0x21AA}$
3.2.7 Connectivity of Morphisms
We now introduce a relative version of Definition 3.2.4.5.
Remark 3.2.7.2. 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.
\[ \xymatrix { X' \ar [d]^{f'} \ar [r] & X \ar [d]^{f} \\ Y' \ar [r] & Y } \]
Variant 3.2.7.3. 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 3.2.7.1. It follows from Remark 3.2.7.2 that this condition does not depend on the choice of the diagram (3.31).
3.31
\begin{equation} \begin{gathered}\label{equation:n-connective-for-simplicial-set} \xymatrix { X \ar [r] \ar [d]^{f} & X' \ar [d]^{f'} \\ Y \ar [r] & Y', } \end{gathered} \end{equation}
Example 3.2.7.4. Let $X$ be a simplicial set and let $n$ be a nonnegative integer. Then $X$ is $n$-connective (in the sense of Variant 3.2.4.9) if and only if the projection map $X \rightarrow \Delta ^0$ is $n$-connective (in the sense of Variant 3.2.7.3).
Remark 3.2.7.5 (Homotopy Invariance). 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 3.2.7.6 (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 nonnegative integer $m \leq n$.
Proposition 3.2.7.7. Let $f: X \rightarrow Y$ be a morphism of simplicial sets and let $n \geq 0$ be an integer. Then:
If the $Y$ is $n$-connective and $f$ is $n$-connective, then $X$ is $n$-connective.
If $X$ is $n$-connective and $Y$ is $(n+1)$-connective, then $f$ is $n$-connective.
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 $n$. When $n=0$, we can restate Proposition 3.2.7.7 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.
Proposition 3.2.7.8. Let $f: X \rightarrow Y$ be a Kan fibration between Kan complexes and let $n \geq 0$ be a nonnegative integer. Then the morphism $f$ is $n$-connective (in the sense of Definition 3.2.7.1) 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 3.2.4.5).
Proof. We proceed 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 is 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.
The equivalence of $(a)$ and $(b)$ now follows from the exact sequence
\[ \pi _{n}(X,x) \rightarrow \pi _{n}(Y,y) \xrightarrow {\partial } \pi _{n-1}(X_ y,x) \rightarrow \pi _{n-1}(X,x) \rightarrow \pi _{n-1}(Y,y) \]
of Theorem 3.2.6.1. $\square$
Remark 3.2.7.9. In the statement of Proposition 3.2.7.8, it is not necessary to assume that $X$ and $Y$ are Kan complexes. See Variant 3.3.7.5.
Corollary 3.2.7.10 (Transitivity). Let $f: X \rightarrow Y$ and $g: Y \rightarrow Z$ be morphisms of simplicial sets and let $n \geq 0$ be an integer. Then:
If $f$ and $g$ are $n$-connective, then the composition $(g \circ f): X \rightarrow Z$ is $n$-connective.
If $g \circ f$ is $n$-connective and $g$ is $(n+1)$-connective, then $f$ is $n$-connective.
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 3.2.7.8, we can further reduce to the case $Z = \Delta ^0$. In this case, Corollary 3.2.7.10 is a restatement of Proposition 3.2.7.7. $\square$
Corollary 3.2.7.11. Let $f: X \rightarrow Y$ be a Kan fibration between Kan complexes and let $n$ be a positive integer. Then $f$ is $n$-connective if and only if it satisfies the following pair of conditions:
The map of connected components $\pi _0(f): \pi _0(X) \rightarrow \pi _0(Y)$ is surjective.
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, it follows that 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 3.2.7.8, we see that $f$ is $n$-connective if and only if $q$ is $n$-connective. The desired result now follows by applying Corollary 3.2.7.10 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. $\square$
Corollary 3.2.7.12. Let $X$ be a Kan complex and let $n$ be a positive integer. Then $X$ is $n$-connective if and only if it is nonempty and the diagonal map $X \rightarrow X \times X$ is $(n-1)$-connective.
Proof. Apply Corollary 3.2.7.11 in the special case $Y = \Delta ^0$ (see Example 3.2.7.4). $\square$
In the case of a Kan fibration, Definition 3.2.7.1 can be reformulated as a lifting property.
Proposition 3.2.7.13. Let $f: X \rightarrow Y$ be a Kan fibration of simplicial sets and let $n \geq 0$ be an integer. The following conditions are equivalent:
For every vertex $y \in Y$, the Kan complex $X_{y} = \{ y\} \times _{Y} X$ is $n$-connective.
For every simplicial set $B$ of dimension $\leq n$ and every simplicial subset $A \subseteq B$, every lifting problem
admits a solution.
For every integer $0 \leq m \leq n$, every lifting problem
admits a solution.
\[ \xymatrix { A \ar [r] \ar [d] & X \ar [d]^{f} \\ B \ar [r] \ar@ {-->}[ur] & Y } \]
\[ \xymatrix { \operatorname{\partial \Delta }^{m} \ar [r] \ar [d] & X \ar [d]^{f} \\ \Delta ^{m} \ar [r]^{\sigma } \ar@ {-->}[ur] & Y } \]
Proof. The implication $(2) \Rightarrow (3)$ is immediate, and the converse follows from Proposition 1.1.3.13. 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 $(3)$ is equivalent to the following a priori weaker assertion:
- $(3')$
For every integer $0 \leq m \leq n$ and every vertex $y \in Y$, every lifting problem
\[ \xymatrix { \operatorname{\partial \Delta }^{m} \ar [r] \ar [d] & X_{y} \ar [d]^{f_ y} \\ \Delta ^{m} \ar [r]^{\sigma } \ar@ {-->}[ur] & \{ y\} } \]admits a solution.
The equivalence of $(1)$ and $(3')$ now follows from Proposition 3.2.4.18. $\square$
Corollary 3.2.7.14. Let $m$ and $n$ be nonnegative integers, and let $f: X \rightarrow Y$ be a Kan fibration between Kan complexes which is $(m+n)$-connective. Let $B$ be a simplicial set of dimension $m$, and let $A \subseteq B$ be a simplicial subset. Then the restriction map is $n$-connective.
\[ f': \operatorname{Fun}( B, X ) \rightarrow \operatorname{Fun}( A, X ) \times _{ \operatorname{Fun}( A, Y) } \operatorname{Fun}( B, Y ) \]
Proof. Our hypothesis that $f$ is a Kan fibration guarantees that $f'$ is also a Kan fibration (Theorem 3.1.3.1). By virtue of Proposition 3.2.7.13, 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
3.32
\begin{equation} \begin{gathered}\label{equation:exponentiation-for-connectivity} \xymatrix { A' \ar [r] \ar [d] & \operatorname{Fun}( B, X ) \ar [d]^{f'} \\ 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 (3.32) as a lifting problem
\[ \xymatrix { (A \times B') \coprod _{ (A \times A' ) } (B \times A') \ar [r] \ar [d] & X \ar [d]^{f} \\ B \times B' \ar [r] \ar@ {-->}[ur] & Y. } \]
Since the simplicial set $B \times B'$ has dimension $\leq m+n$ (Proposition 1.1.3.11), the existence of a solution follows from our assumption that $f$ is $(m+n)$-connective (Proposition 3.2.7.13). $\square$
Corollary 3.2.7.15. Let $m$ and $n$ be nonnegative 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 Corollary 3.2.7.14 in the special case $Y = \Delta ^0$. $\square$
Corollary 3.2.7.16. Let $m$ and $n$ be nonnegative 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. 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 Corollary 3.2.7.14 in the special case $A = \emptyset $. $\square$
Corollary 3.2.7.17. Let $m$ and $n$ be nonnegative 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 3.2.7.15 in the special case $A = \emptyset $ (or Corollary 3.2.7.16 in the special case $Y = \Delta ^0$). $\square$