# Are contractible open subsets always evenly covered?

Let $$p:E\rightarrow X$$ be a covering map.

Let $$U$$ be a contractible open subset of $$X$$.

Is $$U$$ necessarily an evenly covered open subset, i.e. $$p^{-1}(U)$$ a disjoint union of open subsets of $$E$$ homeomorphic to $$U$$?

I know that if $$X$$ is path connected and locally path connected, then $$U$$ is an evenly covered open set. This can be shown by lifting the inclusion $$U\hookrightarrow X$$ to maps $$U \hookrightarrow E$$.

What if $$X$$ is just a general topological space, not necessarily locally path connected and path connected?

• Generally, covering spaces only behave well under some local conditions on the space, e.g. locally path connected and semilocally simply connected. For instance, if your space is locally path-connected, being semilocally simply connected is necessary and sufficient for the existence of a universal cover. – Joshua Mundinger Nov 17 '18 at 20:17
• @JoshuaMundinger You're completely right, but what you said is somehow off-topic. Under some local conditions, the space does behave well, as I have mentioned in the question. However, I'm looking for a proof or disproof of the statement without those local conditions. ps. I don't think it has something to do with universal cover. – Y. Hu Nov 18 '18 at 13:44

## 1 Answer

The answer is "yes". To see this, let us recall the concept of a pullback. Given a pair of maps $$(p : E \to X,f : Y \to X)$$, a pullback of $$(p,f)$$ is given by a pair of maps $$(p^* : E^* \to Y,f^*: E^* \to E)$$ such that

(1) $$fp^* = pf^*$$

(2) For any pair of maps $$(u : Z \to Y, v : Z \to E)$$ such that $$fu = pv$$ there exists a unique map $$w : Y \to E^*$$ such that $$p^*w = u$$ and $$f^*w = v$$.

This universal property implies that pullbacks are unique up to canonical homeomorphism. Here is an explicit construction: $$E^* = \{ (e,y) \in E \times Y \mid p(e) = f(y) \}, p^*(e,y) = y, f^*(e,y) = e .$$ If $$f$$ is the inclusion map of a subspace $$Y \subset X$$, then we may also take $$E^* = p^{-1}(Y) \subset E, p^*(e) = p(e), f^*(e) = e$$.

It is well-known that pullbacks of covering projections are covering projections. This means that if $$p$$ is a covering projection, then so is $$p^*$$. See Pullback of a covering map.

Lemma : Let $$p : E \to X$$ be a covering projection and $$f_0, f_1 : Y \to X$$ be homotopic maps. Then the pullback covering projections $$p^*_k : E^*_k \to Y$$ along $$f_k$$ are equivalent which means that there exists a homeomorphism $$h : E^*_0 \to E^*_1$$ such that $$p^*_1h = p^*_0$$.

Proof. Let $$H : Y \times I \to X$$ be a homotopy from $$f_0$$ to $$f_1$$ and for $$k = 0,1$$ let $$i_k : Y \times \{ k \} \hookrightarrow Y \times I$$ denote inclusion. Form the pullback $$(p^* : E^* \to Y \times I, H^* : E^* \to E)$$ of $$(p,H)$$ and the pullbacks $$(\pi_k : E^*_k \to Y \times \{ k \}, i^*_k : E^*_k \hookrightarrow E^*)$$ of $$(p^*,i_k)$$, where $$E^*_k = (p^*)^{-1}(Y \times \{ k \})$$. Then we may assume that $$p^*_k = r_k \pi_k : E^*_k \to Y$$, where $$r_k : Y \times \{ k \} \to Y, r_k(y,k) = y$$.

Now the map $$i_0 \pi_0 : E^*_0 \to Y \times I$$ lifts to $$i^*_0 : E^*_0 \to E^*$$. Hence the homotopy $$G : E^*_0 \times I \to Y \times I, G(e,t) = (p^*_0(e),t)$$ which satisfies $$G_0 = i_0 \pi_0$$ lifts to a homotopy $$G' : E^*_0 \times I \to E^*$$. The map $$G'_1 : E^*_0 \to E^*$$ has the property $$G'_1(E^*_0) \subset (p^*)^{-1}(Y \times \{ 1 \}) = i^*_1(E^*_1)$$. Therefore we get a map $$h : E^*_0 \to E^*_1$$ such that $$p^*_1h = p^*_0$$. By lifting $$G' : E^*_1 \times I \to Y \times I, G'(e,t) = (p^*_1(e),1-t)$$ we obtain a map $$h' : E^*_1 \to E^*_0$$ such that $$p^*_0h' = p^*_1$$. It is easy to see that $$h' h = id$$ (this follows from unique path lifting).

Now let $$U \subset X$$ be open and contractible. This means that the inclusion $$i : U \to X$$ is homotopic to a constant map $$c : U \to X, c(y) = x_0$$. Hence the covering projection $$p : p^{-1}(U) \to U$$ is equivalent to the pullback of $$p$$ along $$c$$. But the latter is easily seen to be a trivial covering $$U \times F \to U$$ (recall $$E^* = \{ (e,y) \in E \times U \mid p(e) = c(y) = x_0 \} = \{ (e,y) \in E \times U \mid e \in p^{-1}(x_0)) \} , p^*(e,y) = y$$). This means that $$U$$ is evenly covered.

Note that it suffices to assume that $$i : U \to X$$ is inessential which is weaker than $$U$$ contractible.