Lifting of maps to a covering space 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 $\tilde{\varphi}(y)$?
 A: Let $f$ be a path from $y_0$ to $y$, let $y' \in V$ and let $f'$ be a path in $V$ (who is arc connected) from $y$ to $y'$. Then by concatenating $f$ and $f'$, you obtain a path $f \oplus f'$ from $y_0$ to $y'$, from which you can compute $\tilde \varphi(y')$ : $\varphi \circ (f \oplus f')$ is the concatenation of a path in $X$ from $x_0$ to $\varphi(y)$ and a path in $U'$ from $\varphi(y)$ to $\varphi(y')$. The lifting of the first part is a path in $\tilde X$ from $\tilde x_0$ to $\tilde \varphi(y)$, and the lifting of the second second part is a path in $p^{-1}(U')$ from $\tilde \varphi(y)$ to $\tilde \varphi(y')$.
Therefore, $\tilde \varphi(y)$ and $\tilde \varphi(y')$ are in the same arc component of $p^{-1}(U')$.
Unfortunately you removed a step from the proof so you can't conclude that $\tilde \varphi(y')$ is in $U$ just from that. Let $W$ be the arc component of $\tilde \varphi(y)$ in $p^{-1}(U')$. It is not necessarily contained in $U$ so we have to do the extra work. 
Let $U''$ be an elementary neighborhood of $\varphi(y)$ such that $U'' \subset p(W \cap U)$, then choose $V$ such that $V$ is arc connected and $\varphi(V) \subset U''$ instead. Then $\tilde \varphi(y)$ and $\tilde \varphi(y')$ are in the same arc component of $p^{-1}(U'')$. Since $U'' \subset p(W) \subset U'$, they are also in the same arc component of $p^{-1}(U')$, thus $\tilde \varphi(y') \in W$.
Since $W$ is arc connected and $p(W) \subset U'$ is an elementary neighborhood, $p|_W$ is injective (otherwise there would be a nontrivial loop in $U'$). Since $\varphi(y') \in p(W \cap U)$ and $\tilde \varphi(y') \in W$, $p|_W(\tilde \varphi(y')) = \varphi(y') \in p|_W(W \cap U)$, thus $\tilde \varphi(y') \in W \cap U$ . And so we have shown that $\tilde \varphi(y') \in U$ for all $y' \in V$
