$p: \tilde X \rightarrow X$ covering map. If $p \circ f = p \circ g$, then $f=g$ 
Let $p : \tilde X → X$ be a covering map. Let $Y$ be connected and $y_0 ∈ Y$ . Let $f, g : Y \rightarrow \tilde X$ be continuous maps such that:
(i) $f(y_0) = g(y_0),$ and
(ii) $p ◦ f = p ◦ g.$
Prove that $f = g.$

We suppose that $f(x_0) \neq g(x_0)$ for some point $x_0$ and derive a contradiction.
Here is an immediate conclusion that follows from the above hyphotesis: $f(Y)\cup g(Y) \subseteq \tilde X$ is a connected set, since $f(y_0)=g(y_0)\in f(Y)\cap g(Y)$ and $f(Y), g(Y)$ are connected. I have been trying to figure out a clever way to utilize the hyphotesis that $p: \tilde X \rightarrow X$ is a covering map, however I found nothing useful. For example, since $p\circ f(x_0) = p \circ g(x_0) \in X$, there exists an open set $B$ containing this point such that $p^{-1}(B) = \cup_\alpha V_\alpha$ is a disjoint union such that $p_{|_{V_\alpha}}$ is an homeomorphism between each $V_\alpha$ and $B$. In particular, $f(x_0)$ and $g(x_0)$ would belong to different $V_\alpha$, but I didnt manage to find a way to use that in order to find a separation of $f(Y) \cup g(Y)$. I also know that covering maps behave very well with respect to homotopies, but I don't see how they would be particularly useful in this problem.
Any help or hint on this exercise is very much appreciated!
 A: This is a nice problem. So, define the following set:
$$S = \{y \in Y: f(y) = g(y) \}$$
The goal is to show that $S$ is both open and closed in $Y$. Given that $Y$ is connected and given that $y_0 \in S$, it would follow that $S = Y$ and that would prove the proposition.
So, let $(y_{\delta})_{\delta \in D}$ be a net in $S$ with a limit $y$. Then:
$$\forall \delta \in D: f(y_{\delta}) = g(y_{\delta})$$
By the net characterization of continuity, it follows that $f(y) = g(y)$. That is, $S$ is closed. Next, we show that this is open. Let $y \in S$. Then, $p$ is locally a homeomorphism at $f(y)$. In particular, there is an open nbhd $V$ of $f(y)$ such that $p|_{V}: V \to p(V)$ is a homeomorphism.
Observe that $f^{-1}(V)$ is an open nbhd of $y$. We'll show that $f^{-1}(V) \subseteq S$. Indeed, let $x \in f^{-1}(V)$. Then, $f(x) \in V$. Observe that $p(f(x)) = p(g(x))$. Since $p$ is injective under this restriction, it now follows that $f(x) = g(x)$. But this is exactly what we wanted to show so we are done.
Edit:
The argument above only works if $Y$ is assumed to be Hausdorff. There's actually a similar result in Raghavan Narasimhan's Complex Analysis In One Variable, Chapter 2, where he does assume Hausdorffness.
