# $A \xrightarrow{\alpha, \beta} E \xrightarrow{p} B$, where $p$ is a covering map and $A$ is connected.

Problem: Let $\alpha, \beta$ be continuous maps from a connected space $A$ to a space $E$, and $p:E \rightarrow B$ a covering map. If $a \in A$ with $\alpha(a) = \beta(a)$ and $p \circ \alpha = p \circ \beta$, then $\alpha = \beta$.

I feel like I'm very close. For each $t \in A$, I let $W^t$ be an evenly covered neighborhood of $p \circ \alpha(t) = p \circ \beta(t)$, and write $p^{-1}W^t$ as a disjoint union of open sets $$\bigcup\limits_i V_i^t$$ where $p|V_i^t: V_i \rightarrow W$ is a homeomorphism. What I'd like to do is use the connectedness of $A$ to ensure that ensure that $\alpha(t), \beta(t)$ lie in the same slice $V_i^t$; it would follow that $\alpha(t) = \beta(t)$ by the injectivity of $p|V_i$. Is this trivial? Could someone please give me a hint?

Now you need to prove that the set where $\alpha$ and $\beta$ agree on is both open and closed.
First assume $\alpha(t)=\beta(t)$ and then assume $\alpha(t)\ne \beta(t)$.
Let $S = \{t \in A : \alpha(t) = \beta(t)\}$. We want to show that $S$ is the entire space $A$. Since $A$ is connected and $S$ is nonempty, it is enough to show that $S$ and $A \setminus S$ are open.
To show $S$ is open, let $t_0 \in S$. Let $W$ be an evenly covered neighborhood of $p(\alpha(t_0)) = p(\beta(t_0))$. Write $p^{-1}W$ as a disjoint union of slices $\bigcup\limits_i V_i$. Let $\alpha(t_0) = \beta(t_0)$ be in the slice $V$. Then $$\alpha^{-1} V \cap \beta^{-1}V$$ is a neighborhood of $t_0$, and this neighborhood is contained in $S$: if $t$ is in this neighborhood, the fact that $\alpha(t), \beta(t) \in V$ and $p_{|V}$ is injective means that $p(\alpha(t)) = p(\beta(t))$ implies $\alpha(t) = \beta(t)$. Thus $S$ is open.
To show $A \setminus S$ is open, let $t_0 \in A \setminus S$. Again let $W$ be an evenly covered neighborhood of $p(\alpha(t_0)) = p(\beta(t_0))$, with the $V_i$ as before. Since $\alpha(t_0) \neq \beta(t_0)$, these elements must be in distinct slices $V_1$ and $V_2$. Then $$\alpha^{-1}V_1 \cap \beta^{-1} V_2$$ is a neighborhood of $t_0$ which is contained in $A \setminus S$: if $t$ is in this neighborhood, the fact that $\alpha(t) \in V_1$ and $\beta(t) \in V_2$ with $V_1, V_2$ disjoint means that $\alpha(t) \neq \beta(t)$. Thus $A \setminus S$ is open.