Question in Covering space This statement is in Bredon's Topology and Geometry.
I want to show the following statement.
Let $p_i:W_i\to Y, i=1,2$ be covering maps such that $W_1$ is simply connected, and let $w_i\in W_i$ and $y\in Y$ be such that $p_i(w_i)=y$. Then there is a unique map $g:W_1\to W_2$ such that $g(w_1)=w_2$ and $p_2\circ g=p_1$. Morever, $g$ is a covering map.
The book says that the statement follows from this Lemma.
Lemma. Let $W$ be connected. Let $p:X\to Y$ be a covering map and $f:W\to Y$ a map. Let $g_1$ and $g_2$ be maps $W\to X$ both of which are liftings of $f$. If $g_1(w)=g_2(w)$ for some point $w\in W$ then $g_1\equiv g_2$.
The lemma assumed the existence of lift. So I think that lemma is used to show the uniqueness of $g$ but I don't know the existence. I thought about general lifting criterion for the existence of lift but we don't know if $W_1$ is locally path connected. So I don't know how simply connectivity works here. Any help or ideas? Thanks in advance.
 A: The chapter "Covering Spaces" in Bredon's book starts with

The spaces we shall consider in this section will all be Hausdorff, arcwise
connected, and locally arcwise connected.

I think Bredon is wrong when he says that the statement (Corollary 4.5) follows directly from Lemma 4.4. Instead it follows from Corollary 4.2. This Corollary shows that the map $p_1 : (W_1,w_1) \to (Y,y)$ has a unique lift $g : (W_1,w_1) \to (W_2,w_2)$. This means $p_2 \circ g = p_1$.
It remains to show that $g$ is a covering map. Bredon claims it is a simple exercise in the definition of covering maps. I wouldn't say that it is really simple.
We first show that $g$ is surjective. Let $z_2 \in W_2$. Pick any $z_1 \in W_1$. Since $W_2$ is arcwise connected, there exists a path $u : I \to W_2$ such that $u(0) = g(z_1)$ and $u(1) = z_2$. The path $p_2 \circ u : I \to Y$ starts at $p_2(u(0)) = p_2(g(z_1)) = p_1(z_1)$ and therefore has a unique lift $v : I \to W_1$ such that $v(0) =z_1$. Then $g \circ v : I \to W_2$ starts at $g(v(0)) = g(z_1)$. We have $p_2 \circ g \circ v = p_1 \circ v = p_2 \circ u$. Thus $u$ and $g \circ v$ are lifts of $p_2 \circ u$ with both start at $g(z_1)$. Thus $u = g \circ v$ and $z_2 = u(1) = g(v(1)) \in g(W_1)$.
Note that the surjectivity of $g$ implies that for each $M \subset Y$
$$g(p_1^{-1}(M)) = p_2^{-1}(M). \tag{1}$$
We next show that each $z_2 \in W_2$ has an open arcwise connected neigborhood which is evenly covered. Let $V_1$ and $V_2$ be open arcwise connected neigborhoods of $p_2(z_2)$ in $Y$ which are evenly covered by $p_1$ and by $p_2$. There exists an open arcwise connected neigborhood $V$ of $p_2(z_2)$ in $Y$ such that $V \subset V_1 \cap V_2$. The set $V$ is evenly covered by both $p_1$ and $p_2$. Write $p_i^{-1}(V) = \bigcup_{\alpha_i \in A_i} U_{i,\alpha_i}$ with pairwise disjoint open $U_{i,\alpha_i} \subset W_i$ such that the restrictions $p_{i,\alpha_i} : U_{i,\alpha_i} \stackrel{p_i}{\to} V$ are homeomorphisms. Hence all $U_{i,\alpha_i} $ are arcwise connected. Since $g$ is continuous, each $g(U_{1,\alpha_1})$ is arcwise connected and therefore contained in a unique $U_{2,\phi_g(\alpha_1)}$ (it is contained in the disjoint union of arcwise connected open sets, thus only one of these sets can intersect it). Write $g_{\alpha_1} : U_{1,\alpha_1} \stackrel{g}{\to} U_{2,\phi_g(\alpha_1)}$. We have $p_{2,\phi_g(\alpha_1)} \circ g_{\alpha_1} = p_{1,\alpha_1}$, hence $g_{\alpha_1} = p_{2,\phi_g(\alpha_1)}^{-1} \circ p_{1,\alpha_1}$ which shows that $g_{\alpha_1}$ is a homeomorphism. The function $\phi_g : A_1 \to A_2$ must be surjective by $(1)$.
Now $z_2$ is contained in a unique $U_{2,\bar \alpha_2}$. Let $A' = \phi_g^{-1}(\bar \alpha_2)$. Then $g^{-1}(U_{2,\bar \alpha_2}) = \bigcup_{\alpha_1 \in A'} U_{1, \alpha_1}$ which shows that $U_{2,\bar \alpha_2}$ is evenly covered by $g$.
