Let ($\tilde{X}$,$p$) be a Covering Space of $X$, $Y$ a connected and locally arcwise-connected Space, $y_{0}\in Y$, $\tilde{x}_{0}\in\tilde{X}$, and $x_{0}=p(\tilde{x}_{0})$. Given a map $\phi : (Y,y_{0})\rightarrow (X,x_{0})$, there exists a lifting $\tilde{\phi} : (Y,y_{0})\rightarrow (\tilde{X},\tilde{x}_{0})$ if and only if $\phi_{*} (\pi_{1}(Y,y_{0}))\subset p_{*}(\pi_{1}(\tilde{X},\tilde{x}_{0}))$.
I have a problem to prove that this condiction is sufficient. In particular, we define $\tilde{\phi}(y)$ as the final point of a lifting of $\phi\circ f$, where $f$ if a path in $Y$ from $y_{0}$ to $y$. I understood that this definition is well defined, but I have some problems to prove that $\tilde{\phi}$ is continuous. The proof says:
let $U$ an arbitrary neighborhood of $\tilde{\phi}(y)$; choose an elementary neighborhood $U'$ of $p\circ\tilde{\phi}(y)=\phi(y)$ such that $U'\subset p(U)$. Let $W$ be the arc component of $p^{-1}(U')$ which contains $\tilde{\phi}(y)$, and let $U''$ be an elementary neighborhood of $\phi(y)$ such that $U''\subset p(U\cap W)$. It's clear that the arc component of $p^{-1}(U'')$ which contains $\tilde{\phi}(y)$ is contained in $U$. Because $\phi$ is continuous, we can choose $V$ such that $\phi(V)\subset U''$. We can also choose $V$ arcwise connected for the hypotesis. So $\tilde{\phi}(V)\subset U$ and $\tilde{\phi}$ is continuous.
I don't get why $\tilde{\phi}(V)\subset U$. Can someone help me? Thanks before!
Best Answer
If one only reads the part of the proof presented in your question, it is indeed not very clear. The essential point is the construction of $\tilde \phi$. Usually (and I guess also in your textbook) for each $y \in Y$ a path $u$ is chosen in $Y$ such that $u(0) = y_0, u(1) = y$. The path $\phi \circ u : I \to X$ starts at $x_0 = \phi(y_0)$ and ends at $\phi(y)$. This path has unique lift $l_u : I \to \tilde X$ such that $l_u(0) = \tilde x_0$. Now define $\tilde \phi(y) = l_u(1) \in p^{-1}(\phi(y))$. The condition $\phi_{*} (\pi_{1}(Y,y_{0}))\subset p_{*}(\pi_{1}(\tilde{X},\tilde{x}_{0}))$ is then used to show that this does not depend on the choice of $u$. This produces a function $\tilde \phi : Y \to \tilde X$ such that $p \circ \tilde \phi = \phi$.
Why is it continuous? The above construction shows the following:
We have $\tilde \phi (y) = l_u(1)$ for some path $u$ from $y_0$ to $y$. If $y' \in V$, we may choose a path $v$ in $V$ from $y$ to $y'$. Then $u * v$ is a path from $y_0$ to $y'$ ($*$ denotes composition of paths). The path $\phi \circ v$ is a path in $U''$ which starts at $\phi(y)$ and ends at $\phi(y')$. It has a unique lift $l_v : I \to \tilde X$ such that $l_v(0) = \tilde \phi(y)$. The set $l_v(I)$ is an arc connected subset of $p^{−1}(U'')$, hence it must be contained in the arc component $V$ of $p^{−1}(U'')$ which contains $\tilde \phi(y)$. We thus get $l_v(I) \subset V \subset U$. But $l_u * l_v$ is a lift of $(\phi \circ u) * (\phi \circ v) = \phi \circ (u * v)$ such that $l(0) = \tilde x_0$. Thus $l_u * l_v = l_{u*v}$ and therefore $$\tilde \phi(y') = l_{u*v}(1) = l_v(1) \in U .$$