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3.1.4 Covering Maps

Let $X$ and $S$ be topological spaces. Recall that a continuous function $f: X \rightarrow S$ is a covering map if every point $s \in S$ has an open neighborhood $U \subseteq S$ for which the inverse image $f^{-1}(U)$ is homeomorphic to a disjoint union of copies of $U$. This definition has a counterpart in the setting of simplicial sets:

Definition 3.1.4.1. Let $f: X \rightarrow S$ be a morphism of simplicial sets. We say that $f$ is a covering map if, for every pair of integers $0 \leq i \leq n$ with $n > 0$, every lifting problem

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

has a unique solution.

Remark 3.1.4.2. Let $f: X \rightarrow S$ be a morphism of simplicial sets. Then $f$ is a covering map if and only if the opposite morphism $f^{\operatorname{op}}: X^{\operatorname{op}} \rightarrow S^{\operatorname{op}}$ is a covering map.

Remark 3.1.4.3. Let $f: X \rightarrow S$ be a morphism of simplicial sets, and let $\delta : X \rightarrow X \times _{S} X$ be the relative diagonal of $f$. Then $f$ is a covering map if and only if both $f$ and $\delta $ are Kan fibrations. In particular, every covering map is a Kan fibration.

Remark 3.1.4.4. Suppose we are given a pullback diagram of simplicial sets

\[ \xymatrix@R =50pt@C=50pt{ X' \ar [r] \ar [d]^-{f'} & X \ar [d]^-{f} \\ S' \ar [r] & S. } \]

If $f$ is a covering map, then $f'$ is also a covering map.

Remark 3.1.4.5. Let $f: X \rightarrow Y$ and $g: Y \rightarrow Z$ be morphisms of simplicial sets. Suppose that $g$ is a covering map. Then $f$ is a covering map if and only if $g \circ f$ is a covering map. In particular, the collection of covering maps is closed under composition.

Remark 3.1.4.6. Let $f: X_{} \rightarrow S_{}$ be a morphism of simplicial sets. The following conditions are equivalent:

$(a)$

The morphism $f$ is a covering map (Definition 3.1.4.1).

$(b)$

For every square diagram of simplicial sets

\[ \xymatrix@R =50pt@C=50pt{ A_{} \ar [d]^{i} \ar [r] & X_{} \ar [d]^{f} \\ B_{} \ar [r] \ar@ {-->}[ur] & S_{} } \]

where $i$ is anodyne, there exists a unique dotted arrow rendering the diagram commutative.

This follows by combining Remarks 3.1.2.7 and 3.1.4.3.

Proposition 3.1.4.7. Let $f: X_{} \rightarrow S_{}$ be a covering map of simplicial sets, and let $i: A_{} \hookrightarrow B_{}$ be any monomorphism of simplicial sets. Then the induced map

\[ \operatorname{Fun}( B_{}, X_{} ) \rightarrow \operatorname{Fun}( B_{}, S_{} ) \times _{ \operatorname{Fun}( A_{}, S_{} )} \operatorname{Fun}( A_{}, X_{} ) \]

is a covering map.

Proof. By virtue of Remark 3.1.4.6, it will suffice to show that if $i': A' \hookrightarrow B'$ is an anodyne morphism of simplicial sets, then every lifting problem of the form

\[ \xymatrix@C =100pt{ A'_{} \ar [d]^{i'} \ar [r] & \operatorname{Fun}( B_{}, X_{} ) \ar [d] \\ B'_{} \ar@ {-->}[ur] \ar [r] & \operatorname{Fun}( B_{}, S_{} ) \times _{ \operatorname{Fun}( A_{}, S_{} )} \operatorname{Fun}( A_{}, X_{} ) } \]

admits a unique solution. Equivalently, we must show that every lifting problem

\[ \xymatrix@C =100pt{ (A_{} \times B'_{} ) \coprod _{ A_{} \times A'_{} } ( B_{} \times A'_{} ) \ar [r] \ar [d] & X_{} \ar [d]^{f} \\ B_{} \times B'_{} \ar [r] \ar@ {-->}[ur] & S_{} } \]

admits a unique solution. This follows from Remark 3.1.4.6, since the left vertical map is anodyne (Proposition 3.1.2.8) and $f$ is a covering map. $\square$

Corollary 3.1.4.8. Let $f: X_{} \rightarrow S_{}$ be a covering map of simplicial sets. Then, for every simplicial set $B_{}$, composition with $f$ induces a covering map $\operatorname{Fun}( B_{}, X_{} ) \rightarrow \operatorname{Fun}( B_{}, S_{} )$.

Proposition 3.1.4.9. Let $f: X \rightarrow S$ be a covering map of topological spaces. Then the induced map $\operatorname{Sing}_{\bullet }(f): \operatorname{Sing}_{\bullet }(X) \rightarrow \operatorname{Sing}_{\bullet }(S)$ is a covering map of simplicial sets (in the sense of Definition 3.1.4.1).

Proof. Let $\delta : X \rightarrow X \times _{S} X$ be the relative diagonal of $f$. We first claim $\delta $ exhibits $X$ as a summand of $X \times _{S} X$ in the category of topological spaces (that is, it is a homeomorphism of $X$ onto a closed and open subset of the fiber product $X \times _{S} X$). To verify this, we can work locally on $S$ and thereby reduce to the case where $X$ is a product of $S$ with a discrete topological space, in which case the result is clear. It follows that the induced map of singular simplicial sets

\[ \operatorname{Sing}_{\bullet }(\delta ): \operatorname{Sing}_{\bullet }(X) \hookrightarrow \operatorname{Sing}_{\bullet }(X \times _{S} X) \simeq \operatorname{Sing}_{\bullet }(X) \times _{ \operatorname{Sing}_{\bullet }(S)} \operatorname{Sing}_{\bullet }(X) \]

is also the inclusion of a summand (Remark 1.1.7.3), and is therefore a Kan fibration by virtue of Example 3.1.1.4. Consequently, to show that $\operatorname{Sing}_{\bullet }(f)$ is a covering map, it will suffice to show that it is a Kan fibration (Remark 3.1.4.3). This is a special case of Corollary 3.5.6.11, since $f: X \rightarrow S$ exhibits $X$ as a fiber bundle over $S$ (with discrete fibers). $\square$

Warning 3.1.4.10. The converse of Proposition 3.1.4.9 is false. For example, let $f: X \rightarrow S$ be a continuous function between topological spaces where $S = \ast $ consists of a single point. In this case, the function $f$ is a covering map if and only if the topology on $X$ is discrete. However, the induced map of simplicial sts $\operatorname{Sing}_{\bullet }(f): \operatorname{Sing}_{\bullet }(X) \rightarrow \operatorname{Sing}_{\bullet }(S)$ is a covering map if and only if the simplicial set $\operatorname{Sing}_{\bullet }(X)$ is discrete: that is, if and only if every continuous function $[0,1] \rightarrow X$ is constant (Example 3.1.4.13). Many non-discrete topological spaces satisfy this weaker condition (for example, we could take $X$ to be the Cantor set).

Remark 3.1.4.11. Let $f: X \rightarrow S$ be a morphism of simplicial sets. Then $f$ is a covering map (in the sense of Definition 3.1.4.1) if and only if the induced map of geometric realizations $| X | \rightarrow |S|$ is a covering map of topological spaces (see Proposition ).

Covering maps of simplicial sets have a very simple local structure:

Proposition 3.1.4.12. Let $f: X_{\bullet } \rightarrow S_{\bullet }$ be a morphism of simplicial sets. The following conditions are equivalent:

$(1)$

The morphism $f$ is a covering map.

$(2)$

For every map of standard simplices $u: \Delta ^ m \rightarrow \Delta ^ n$, composition with $u$ induces a bijection $X_{n} \rightarrow X_{m} \times _{ S_{m} } S_{n}$.

$(3)$

For every $n$-simplex $\sigma : \Delta ^ n \rightarrow S_{\bullet }$, the projection map $\Delta ^{n} \times _{S_{\bullet }} X_{\bullet } \rightarrow \Delta ^ n$ restricts to an isomorphism on each connected component of $\Delta ^ n \times _{S_{\bullet }} X_{\bullet }$.

Proof. Assume first that $(1)$ is satisfied; we will prove $(2)$. Let $u: \Delta ^ m \rightarrow \Delta ^ n$ be a morphism of simplicial sets. Choose a vertex $v: \Delta ^0 \rightarrow \Delta ^ m$. It follows from Example 3.1.2.5 that $v$ and $u \circ v$ are anodyne morphisms of simplicial sets. Invoking Remark 3.1.4.6, we conclude that the right square and outer rectangle in the diagram

\[ \xymatrix@R =50pt@C=50pt{ X_{n} \ar [d] \ar [r]^-{\circ u} & X_{m} \ar [r]^-{ \circ v} & X_{0} \ar [d] \\ S_{n} \ar [r]^-{ \circ u} & S_{m} \ar [r]^-{\circ v} & S_0 } \]

are pullback diagrams. It follows that the right square is a pullback diagram as well.

We next show that $(2)$ implies $(3)$. Fix a map $\sigma : \Delta ^ n \rightarrow S_{\bullet }$, and let $T = X_{n} \times _{S_{n}} \{ \sigma \} $ denote the collection of all $n$-simplices $\tau $ of $X_{\bullet }$ satisfying $f( \tau ) = \sigma $. To prove $(3)$, it will suffice to show that the tautological map

\[ g: \coprod _{\tau \in T} \Delta ^ n \rightarrow \Delta ^ n \times _{ S_{\bullet } } X_{\bullet } \]

is an isomorphism of simplicial sets. Equivalently, we must show that for every map of simplices $u: \Delta ^ m \rightarrow \Delta ^ n$, the induced map $T \rightarrow X_{m} \times _{S_{m}} \{ \sigma \circ u \} $ is bijective, which follows immediately from $(2)$.

We now complete the proof by showing that $(3)$ implies $(1)$. Assume that $(3)$ is satisfied. We wish to show that, for every pair of integers $0 \leq i \leq n$ with $n \geq 1$, every lifting problem

\[ \xymatrix@R =50pt@C=50pt{ \Lambda ^{n}_{i} \ar [d]^{^{(}->} \ar [r] & X_{\bullet } \ar [d]^{f} \\ \Delta ^ n \ar [r] \ar@ {-->}[ur] & S_{\bullet } } \]

admits a unique solution. To prove this, we are free to replace $f$ by the projection map $\Delta ^ n \times _{S_{\bullet }} X_{\bullet } \rightarrow \Delta ^ n$, and thereby reduce to the case where $S_{\bullet }$ is a standard simplex. In this case, assumption $(3)$ guarantees that each connected component of $X_{\bullet }$ is isomorphic to $S_{\bullet }$. The desired result now follows from the observation that the simplicial sets $\Lambda ^{n}_{i}$ and $\Delta ^ n$ are connected. $\square$

Example 3.1.4.13. Let $X$ be a simplicial set. Then the unique morphism $f: X \rightarrow \Delta ^{0}$ is a covering map of simplicial sets if and only if $X$ is discrete (see Definition 1.1.4.9).

Corollary 3.1.4.14. Let $f: X \rightarrow S$ be a monomorphism of simplicial sets. The following conditions are equivalent:

$(1)$

The morphism $f$ exhibits $X$ as a summand of $S$ (Definition 1.1.6.1).

$(2)$

The morphism $f$ is a covering map.

$(3)$

The morphism $f$ is a Kan fibration.

Proof. The implication $(1) \Rightarrow (2)$ and $(2) \Rightarrow (3)$ are immediate. Moreover, if $f$ is a monomorphism, then the relative diagonal $\delta : X \rightarrow X \times _{S} X$ is an isomorphism, so the implication $(3) \Rightarrow (2)$ follows from Remark 3.1.4.3. We will complete the proof by showing that $(2) \Rightarrow (1)$. Let $u: \Delta ^{m} \rightarrow \Delta ^{n}$ be a morphism of standard simplices and let $\sigma : \Delta ^{n} \rightarrow S$ be a simplex of $S$; we wish to show that $\sigma $ factors through $f$ if and only if $\sigma \circ u$ factors through $f$. This follows immediately from the criterion of Proposition 3.1.4.12. $\square$