Section 3.1 Morphisms and Deck Transformations

Definition 3.1.

Let \(p_1 \colon \tilde{X}_1 \to X\) and \(p_2 \colon \tilde{X}_2 \to X\) be coverings of \(X\text{.}\) A morphism of coverings from \(p_1\) to \(p_2\) is a continuous map \(f \colon \tilde{X}_1 \to \tilde{X}_2\) with \(p_2 \circ f = p_1\text{,}\) that is, a map making

commute. Coverings of \(X\) and morphisms between them form a category \(\operatorname{Cov}(X)\text{.}\)

It turns out that \(f\) itself is actually a cover, at least for nice enough spaces.

Definition 3.2.

A space \(X\) is locally path-connected if for every \(x \in X\) and every neighbourhood \(U\) of \(x\) there is an open path-connected set \(V\) with \(x \in V \subseteq U\text{.}\) Equivalently, every point has a neighbourhood basis of path-connected open sets.

A space can be path-connected without being locally path-connected (for example, the comb space), or locally path-connected without being path-connected (for example \((0,1) \cup (2,3)\)). However a connected, locally path-connected space is path-connected.

Proposition 3.3.

Let \(X\) be locally connected, let \(p \colon Y \to X\) and \(q \colon Z \to X\) be coverings, and let \(f \colon Y \to Z\) be a morphism of coverings, so that \(q \circ f=p\text{.}\) Then \(f\) is a covering map.

Proof.

Let \(z\in Z\) and put \(x=q(z)\text{.}\) Choose an open neighbourhood of \(x\) that is evenly covered by both \(p\) and \(q\text{,}\) and then, using local connectedness, shrink it to a connected open neighbourhood \(U\) of \(x\text{.}\) The restrictions of both coverings over \(U\) are trivial, so there are discrete spaces \(D\) and \(E\) and identifications

\begin{equation*} p^{-1}(U)\cong U\times D, \qquad q^{-1}(U)\cong U\times E \end{equation*}

under which \(p\) and \(q\) are the projections onto \(U\text{.}\) Since \(q\circ f=p\text{,}\) the restriction of \(f\) has the form

\begin{equation*} f|_{p^{-1}(U)}\colon U\times D\longrightarrow U\times E, \qquad (u,d)\longmapsto \bigl(u,\varphi(u,d)\bigr) \end{equation*}

for a continuous map \(\varphi\colon U\times D\to E\text{.}\)

For each \(d\in D\text{,}\) the map \(u\mapsto\varphi(u,d)\) is constant because \(U\) is connected and \(E\) is discrete. Thus \(\varphi(u,d)=\Phi(d)\) for some map \(\Phi\colon D\to E\text{,}\) and hence

\begin{equation*} f(u,d)=\bigl(u,\Phi(d)\bigr). \end{equation*}

For every \(e\in E\text{,}\) the sheet \(U\times\{e\}\) of \(q^{-1}(U)\) satisfies

\begin{equation*} f^{-1}\bigl(U\times\{e\}\bigr) =\coprod_{d\in\Phi^{-1}(e)}U\times\{d\}, \end{equation*}

and \(f\) maps each summand \(U\times\{d\}\) homeomorphically onto \(U\times\{e\}\text{.}\) Therefore every sheet of \(q^{-1}(U)\) is evenly covered by \(f\text{.}\) Such sheets form a neighbourhood cover of \(Z\text{,}\) so \(f\) is a covering map.

Definition 3.4.

An isomorphism of coverings is a morphism admitting an inverse morphism. An automorphism of a covering \(p \colon \tilde{X} \to X\) in \(\operatorname{Cov}(X)\) is called a deck transformation. The group of deck transformations is written

\begin{equation*} \operatorname{Deck}(p) := \operatorname{Aut}_{\operatorname{Cov}(X)}(p). \end{equation*}

Exercise 3.5.

Show that a morphism of coverings is an isomorphism if and only if it is a homeomorphism.

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