# Local homeomorphism + curve lifting property, gives a covering map

The Theorem 4.19. is from Otto Forster's Lectures on Riemann Surfaces.

Theorem 4.19. Suppose $$X$$ is a manifold, $$Y$$ is a Hausdorff space and $$p:Y\rightarrow X$$ is a local homeomorphism with the curve lifting property. Then $$p$$ is a covering map.

Proof. Suppose $$x_0\in X$$ is an arbitrary point and $$y_j$$, $$j\in J$$, are the preimages of $$x_0$$ with respect to $$p$$. Take $$U$$ to be an open neighborhood of $$x_0$$ which is homeomorphic to a ball and $$f:U\rightarrow X$$ be the canonical injection. Then for each $$j\in J$$ there is a lifting $$g_j:U\rightarrow Y$$ such that $$g_j(x_0)=y_j$$. Let $$V_j=g_j(U)$$. Now one can easily convince oneself that $$p^{-1}(U)=\bigcup_{j\in J}V_j,$$ that the $$V_j$$ are pairwise disjoint open sets and that every mapping $$p|V_j\rightarrow U$$ is a homeomorphism.

I have a few difficulties in understanding the proof, first of all how the choice of $$U$$ has been made? Secondly, however, it is chosen how does the conclusion of the proof follow easily? Am I missing something trivial? The suggestions and comments will be appreciated.

• @freakish, That's all fine that we can lift the canonical injection in the proof. The condition of being Manifold is also not very important for me. The ambiguous part is how the sets $V_j '$s are becoming open and disjoint, and why the restricted map $p|V_j\rightarrow U$ becomes a homeomorphism? I suppose there is more to the selection of open nbhd U. – Dèö Feb 22 at 12:42
• Yeah, that's quite a good amount of arguments. So that's where we require X to be a topological manifold. Thanks, @freakish, let me also think about disjointness of $V_j$'s – Dèö Feb 22 at 15:06
• Yes, that completes the proof. :-) – Dèö Feb 22 at 18:44

$$X$$ is a manifold so every point has a neighbourhood $$U$$ homeomorphic to an Euclidean ball. Since $$U$$ is a ball then it is path connected, locally path connected and contractible. Meaning every map starting from it can be lifted. The "curve" lifting term is probably misleading. It should apply to wider range of maps (at least with balls as a domain, not sure if it is equivalent to lifting just paths), see here.

Now since $$X$$ is a manifold and $$p$$ is a local homeomorphism then $$Y$$ is a manifold. Additionally each $$g_j$$ is injective and thus the invariance of domain applies proving that each $$g_j$$ is open. In particular $$V_j$$ is open and $$g_j$$ is a homeomorphism onto image. Since $$p|V_j$$ is its partial inverse then $$p|V_j$$ is a homeomorphism onto image as well.

The invariance of domain may be an overkill but it's the simpliest way IMO to show that each $$g_j$$ is open.

Finally we will show that $$V_j$$ are pairwise disjoint. Assume that $$V_j\cap V_k\neq\emptyset$$. Then $$g_j(z)=g_k(z')$$ for some $$z,z'\in U$$ but

$$z=f(z)=p(g_j(z))=p(g_k(z'))=f(z')=z'$$

so both $$g_j$$ and $$g_k$$ are also liftings with respect to the fixed point $$z=z'$$. Since they share value at the fixed point then they have to be equal by the uniqueness of the lifting. It follows that $$V_j=V_k$$.

Forster is in fact a little short, but he is right. A detailed proof is somewhat lengthy.

Let us start with some definitions. A path in a space $$X$$ is a map $$u : I = [0,1] \to X$$. This is what Forster calls a curve, but that's a non-standard notation. A curve in our sense is a map $$c : J \to X$$ defined on an arbitrary closed interval $$J$$.

Lemma. Let $$p : Y \to X$$ have the path lifting property. If $$p$$ is a local homeomorphism, then it has the unique path lifting property (the definition of this property should be obvious).

Proof. Let $$u : I \to X$$ be a path and $$u', u'' : I \to Y$$ be lifts such that $$u'(0) = u''(0)$$. Define $$R = \sup \{r \in I \mid u'(t) = u''(t) \text{ for all } t \le r \}$$. Then $$u'(R) = u''(R) = y$$. Assume $$R < 1$$. Choose an open neighborhood $$V$$ of $$y$$ which is mapped by $$p$$ homeomorphically onto $$p(V)$$. For some $$R'$$ between $$R$$ and $$1$$ we have $$u'([R,R')), u''([R,R')) \subset V$$. For $$t \in [R,R')$$ we have $$p(u'(t)) = u(t)= p(u''(t))$$ and we infer $$u'(t) = u''(t)$$. This contradicts the definition of $$R$$. Thus $$R = 1$$, i.e. $$u' = u''$$.

Remark. If $$p : Y \to X$$ has the unique path lifting property, then it has the unique curve lifting property. This means that if $$c : J \to X$$ is a curve, $$t \in J$$ and $$y \in p^{-1}(c(t))$$, then there exists a unique lift $$c' : I \to Y$$ of $$c$$ such that $$c'(t) = y$$.

In what follows let $$p : Y \to X$$ be a local homeomorphism with the path lifting property.

By a curve over $$W \subset X$$ we mean a curve $$v : J \to Y$$ such that $$p(v(J)) \subset W$$.

A sheet over an open $$W \subset X$$ is an open subset $$V \subset Y$$ mapped by $$p$$ bijectively onto $$W$$. Observe that the restriction $$p_V : V \to W = p(V)$$ is automatically a homeomorphism because local homeomophisms are open maps. For an open $$W' \subset W$$ we define $$V \mid_{W'} = V \cap p^{-1}(W')$$.

Claim 0. For any open $$W' \subset W$$ the set $$V \mid_{W'}$$ is a sheet over $$W'$$.

Proof. Obvious.

Claim 1. Let $$V$$ be a sheet over $$W$$. Then any curve $$v$$ over $$W$$ which meets $$V$$ (i.e. such that $$v(J) \cap V \ne \emptyset$$) is contained in $$V$$ (i.e. $$v(J) \subset V$$).

Proof. Let $$v$$ be a curve over $$W$$ such that $$v(t) \in V$$ for some $$t \in J$$. Let $$u' = p_V^{-1} p v$$. Then $$u'(t) = v(t)$$, thus by unique curve lifting $$u' = v$$ and $$v(J) = u'(J) \subset V$$.

Claim 2. Let $$V,V'$$ be sheets over a path connected $$W$$. Then either $$V = V'$$ or $$V \cap V' = \emptyset$$.

Proof. Let $$V \cap V' \ne \emptyset$$. Consider $$y \in V$$ and $$y' \in V'$$. Since both of these sets are path connected, there is a path $$v$$ in $$V \cup V'$$ from $$y$$ to $$y'$$. It is a path over $$W$$. By Claim 1 $$y' \in V$$ and $$y \in V'$$.

Claim 3. Let $$V, V'$$ be sheets over open $$W, W'$$. If $$D = W \cap W'$$ is nonempty and path connected and $$V \mid_D \cap V' \mid_D = V \cap V' \cap p^{-1}(D) \ne \emptyset$$, then $$V \mid_D = V' \mid_D$$.

Proof. Apply Claim 0 and Claim 2.

Claim 4. Let $$V_\alpha$$ be sheets over open $$W_\alpha$$. For all $$\alpha, \beta$$ such that $$D_{\alpha,\beta} = W_\alpha \cap W_\beta \ne \emptyset$$ assume that $$D_{\alpha,\beta}$$ is path connected and $$V_\alpha \mid_{D_{\alpha,\beta}} \cap V_\beta \mid_{D_{\alpha,\beta}} \ne \emptyset$$. Then $$V = \bigcup_\alpha V_\alpha$$ is a sheet over $$W = \bigcup_\alpha W_\alpha$$.

Proof. $$V,W$$ are open and $$p$$ maps $$V$$ onto $$W$$. It remains to show that $$p$$ is injective on $$V$$. So let $$y \in V_\alpha, y' \in V_\beta$$ such that $$p(y) = p(y')$$. This implies $$D_{\alpha,\beta} \ne \emptyset$$ and $$y,y' \in p^{-1}(D_{\alpha,\beta})$$. Claim 3 yields $$V_\alpha \mid_{D_{\alpha,\beta}} = V_\beta \mid_{D_{\alpha,\beta}} = V'$$. But $$y,y' \in V'$$ and $$V'$$ is mapped by $$p$$ bijectively onto $$D_{\alpha,\beta}$$. Hence $$y = y'$$.

Claim 5. Let $$x_0 \in X$$ and $$W$$ be an open path connected neighborhood of $$x_0$$. For any $$y_j \in p^{-1}(x_0)$$ let $$V_j$$ be a sheet over $$W$$ containing $$y_j$$. Then $$W$$ is evenly covered by $$\{ V_j \}$$.

Proof. If $$V_j \cap V_k \ne \emptyset$$, then Claim 2 shows $$V_j = V_k$$ which implies $$y_j \in V_k$$. But then $$p_{V_k}(y_j) = p_{V_k}(y_k)$$ which implies $$y_j = y_k$$, i.e. $$j = k$$. Hence the $$V_j$$ are pairwise disjoint open sets which are mapped by $$p$$ homeomorphically onto $$W$$. It remains to show $$\bigcup_j V_j = p^{-1}(W)$$. Let $$y \in p^{-1}(W)$$. Choose a path $$u$$ in $$W$$ from $$p(y)$$ to $$x_0$$ and let $$v$$ be the unique lift such that $$v(0) = y$$. $$v$$ is a path over $$W$$ and we have $$v(1) = y_j$$ for some $$j$$. By Claim 1 we get $$y \in V_j$$.

Claim 6. Let $$U \subset X$$ be an open set and $$h : U \to B_r(0)$$ be a homeomorphism to an open ball such that $$h(x_0) = 0$$. Then for each $$y_j \in p^{-1}(x_0)$$ there exists a sheet over $$U$$ containing $$y_j$$. Note that then Claim 5 applies to show that $$U$$ is evenly covered.

Proof. Let us consider the map $$q : Z = p^{-1}(U) \stackrel{p}{\rightarrow} U \stackrel{h}{\rightarrow} B_r(0)$$. We are done if we can show for each $$y_j$$ there exists a sheet of $$q$$ over $$B_r(0)$$ containing $$y_j$$.

Clearly $$q$$ is local homeomorphism with the path lifting property.

Because $$q$$ is a local homeomorphism, for each $$x \in B_r(0)$$ and each $$y \in q^{-1}(x)$$ there exists an open ball $$B_{r_x}(x) \subset B_r(0)$$ and sheet $$V_x$$ over $$B_{r_x}(x)$$ containing $$y$$.

Let $$R = \sup \{ s \le r \mid \text{There exists a sheet } V_s^j \text{ over } B_s(0) \text{ containing } y_j \}$$. Obviously $$0 < R \le r$$. There exists an increasing sequence $$(s_n)$$ of positive real numbers converging to $$R$$ such that there exists a sheet $$V_{s_n}^j$$ over $$B_{s_n}(0)$$ containing $$y_j$$. By Claim 3 $$V_R = \bigcup_{n=1}^\infty V_{s_n}$$ is a sheet over $$B_R(0) = \bigcup_{n=1}^\infty B_{s_n}(0)$$ which contains $$y_j$$.

We show $$R = r$$; this completes the proof of Claim 6.

Assume $$R < r$$. Let $$S_R$$ denote the boundary sphere of $$B_R(0)$$. For each $$x \in S_R$$ let $$u_x$$ be the radial path from $$0$$ to $$x$$. Let $$u^j_x$$ be the lift to $$Z$$ such that $$u^j_x(0) = y_j$$. Next choose an open ball $$B_{r_x}(x) \subset B_r(0)$$ and a sheet $$V^j_x$$ over $$B_{r_x}(x)$$ containing $$u^j_x(1)$$. Below we show that $$\mathfrak{V} = \{ V_R \} \cup \{V^j_x \mid x \in S_R \}$$ satisfies the assumptions of Claim 4. This proves that $$V' = V_R \cup \bigcup_{x \in S_R} V^j_x$$ is a sheet over $$W' = B_R(0) \cup \bigcup_{x \in S_R} B_{r_x}(x)$$. But $$W'$$ is an open neighborhood of the closed ball with radius $$R$$ and thus contains some $$B_{R'}(0)$$ with $$R' > R$$. Hence $$V' \mid_{B_{R'}(0)}$$ is a sheet over $$B_{R'}(0)$$ which contradicts the definition of $$R$$.

$$\mathfrak{V}$$ satisfies the assumptions of Claim 4:

For any $$x$$ we have $$D_x = B_R(0) \cap B_{r_x}(x) \ne \emptyset$$. This intersection is convex, hence path connected. Let us show that $$V_R \mid_{D_x} \cap V^j_x \mid_{D_x} \ne \emptyset$$. For some $$a < 1$$ we have $$u_x([a,1]) \subset B_{r_x}(x)$$. Clearly $$u^j_x(a) \in q^{-1}(D_x)$$. Let $$v^j_x$$ be the restriction of $$u^j_x$$ to $$[a,1]$$. This is a curve over $$B_{r_x}(x)$$ which meets $$V^j_x$$ in $$u^j_x(1)$$, hence by Claim 1 $$u^j_x(a) = v^j_x(a) \in V^j_x$$. Similarly the restriction $$w^j_x$$ of $$u^j_x$$ to $$[0,a]$$ is a curve over $$B_R(0)$$ which meets $$V_R$$ in $$y_j$$, hence by Claim 1 $$u^j_x(a) = w^j_x(a) \in V_R$$. Thus $$V_R \mid_{D_x} \cap V^j_x \mid_{D_x} \ne \emptyset$$.

Let $$x,x'$$ such that $$D_{x,x'} = B_{r_x}(x) \cap B_{r_{x'}}(x') \ne \emptyset$$. The set $$D_{x,x'}$$ is convex, hence path connected, and contains some point $$m$$ on the open line segment between $$x$$ and $$x'$$. Hence $$m \in B_R(0)$$ and $$D = B_R(0) \cap B_{r_x}(x) \cap B_{r_{x'}}(x') \ne \emptyset$$. From the previous part and Claim 3 we conclude $$V_R \mid_{D_x} = V^j_x \mid_{D_x}$$ and hence $$V^R \mid_D = (V_R \mid_{D_x}) \mid_D = (V^j_x \mid_{D_x}) \mid_D = V^j_x \mid_D$$. Similarly $$V_R \mid D = V^j_{x'} \mid_D$$. This shows that $$V^j_x \mid_{D{x,x'}} \cap V^j_{x'} \mid_{D{x,x'}} \ne \emptyset$$.