Existence and uniqueness of homotopy lifts through a covering map

Given: The objects, hypotheses, and choice principles stated above.

Let $p:E\to B$ be a covering and $f:Y\to B$ continuous. A lift of $f$ through $p$ is a continuous map $\tilde{f}:Y\to E$ with $p\circ \tilde{f}=f$. This includes lifts of paths $I\to B$ and of homotopies $Y\times I\to B$; an initial lift prescribes the restriction at time $0$ (def-homotopy-relative-and-path-homotopy, def-path-connected). (Lifts of maps, paths, and homotopies through a covering map).

Let $p:E\to B$ be a covering, let $\alpha :I\to B$ be a path, and let $e_0\in E$ satisfy $p(e_0)=\alpha (0)$. There is a unique path $\tilde{\alpha }:I\to E$ with $\tilde{\alpha }(0)=e_0$ and $p\circ \tilde{\alpha }=\alpha$. (Existence and uniqueness of path lifts through a covering map).

The product set. Let $I$ be a set and let $X_i$ be a set for each $i\in I$. The product is ${\prod }_{i\in I}{X}_i:=\{x:x\text{ is a function with domain }I\text{ and }x(i)\in X_i\text{ for every }i\in I\},$ and we write $x_i:=x(i)$, the $i$-th coordinate of $x$. Two elements of the product are equal exactly when they agree at every index, functions being equal when they have the same domain and the same values. For $j\in I$ the $j$-th projection is ${\pi }_j:{\prod }_{i\in I}{X}_i\to X_j,{\pi }_j(x):=x_j\mathrm{.}$. The product topology ${\mathcal{T}}^{\mathrm{\Pi }}$ on ${\prod }_i{X}_i$ is the initial topology of the projections: the topology generated by the subbasis $\{{\pi }_i^{-1}[U]:i\in I,U\in {\mathcal{T}}_i\}$. Finite intersections of subbasic sets form a basis for it, and they are exactly the boxes ${\prod }_{i\in I}{U}_i$ with every $U_i$ open in $X_i$ and $U_i=X_i$ for all but finitely many $i$. (The product set ${\prod }_{i\in I}{X}_i$ of functions choosing a point in each factor, the projections, the box topology, and the product topology as the initial topology of the projections; the empty product is a one-point space).

Let $X$, $Y$ and $Z$ be topological spaces, with subspaces carrying the subspace topology (def-subspace-topology-top). Then: 1. Composites. If $f:X\to Y$ and $g:Y\to Z$ are continuous (def-continuous-map-top) then $g\circ f:X\to Z$ is continuous. 2. Open cover. Let $f:X\to Y$ be a function and let $\{U_i:i\in I\}$ be a family of open subsets of $X$ with ${\bigcup }_{i\in I}{U}_i=X$. If $f{\mathrm{\mid }}_{U_i}:U_i\to Y$ is continuous for every $i\in I$, then $f$ is continuous. 3. Finite closed cover. Let $f:X\to Y$ be a function, let $n\ge 1$ and let $F_1,\dots ,F_n$ be closed subsets of $X$ with $F_1\cup \cdots \cup F_n=X$. If $f{\mathrm{\mid }}_{F_k}:F_k\to Y$ is continuous for every $k$, then $f$ is continuous. The converses of claims 2 and 3 hold with no hypothesis on the cover at all: every restriction of a continuous map to a subspace is continuous (def-subspace-topology-top). The finiteness in claim 3 is not removable; see the remarks. (Continuity may be checked on any open cover, and on any finite closed cover; composites of continuous maps are continuous).

Let $(X,\mathcal{T})$ be a topological space (def-topological-space). An open cover of $(X,\mathcal{T})$ is a family $\mathcal{U}\subseteq \mathcal{T}$ of open sets with $X=\bigcup \mathcal{U}$; a subcover of $\mathcal{U}$ is a subfamily that is itself an open cover; and $(X,\mathcal{T})$ is compact when every open cover of it has a finite subcover. (Open cover, subcover, and compact topological space; a compact subset is a subspace that is compact in its own right).