I am reading Algebraic topology by W. Massey and I have a problem with the proof of property 5.1:
Let $(\tilde{X},p)$ be a covering space of $X$, $Y$ a connected and arcwise connected space, $\tilde{x}_0 \in \tilde{X}$, $y_0 \in Y$ and $x_0=p(\tilde{x}_0) \in X$. Given a map $\varphi : (Y,y_0) \to (X,x_0)$, there exists a lifting $\tilde{\varphi} : (Y,y_0) \to (\tilde{X},\tilde{x}_0)$ iff $\varphi_* \pi_1(Y,y_0) \subset p_* \pi_1 (\tilde{X},\tilde{x}_0)$.
To construct $\tilde{\varphi}$, let $y \in Y$ and $f : I \to Y$ a path from $y_0$ to $y$. So $\varphi p : I \to X$ is a path from $x_0$ to $\varphi(y)$. There exists only one path $g : I \to \tilde{X}$ such that $g(0)=\tilde{x}_0$ and $pg= \varphi p$. Let define $\tilde{\varphi}(y)= g(1)$.
We can show that $\tilde{\varphi}$ is well defined (that is, $\tilde{\varphi}$ doesn't depend on the choice of $f$) and it is obvious that $p \tilde{\varphi} = \varphi$.
To prove the continuity of $\tilde{\varphi}$, let $y \in Y$ and $U \subset \tilde{X}$ a open neighborhood of $\tilde{\varphi}(y)$. We can suppose $U$ arc connected. Take $U'$ an elementary neighborhood of $p \tilde{\varphi}(y)= \varphi(y)$ such that $U' \subset p(U)$ and $V \subset Y$ arc connected such that $\varphi(V) \subset U'$ ($\varphi$ is continuous).
Finally, the arc component of $p^{-1}(U')$ containing $\tilde{\varphi}(y)$ is included in $U$.
But how do you prove that for all $y' \in V$, $\tilde{\varphi}(y')$ is in the same arc component of $p^{-1}(U')$ than $\tilde{\varphi}(y)$?