How does Hatcher conclude that the set of points where $\tilde f_1$ and $\tilde f_2$ agree is both open and closed? The last line of the proof he notes that the set of points where $\tilde f_1$ and $\tilde f_2$ agree is both open and closed.
How does this follow from the argument?




 A: The comment section is too short to post this, hence this non-answer.
Let $p \colon E \twoheadrightarrow B$ be a covering map, we want to show that
$\Delta_E$ is both open and closed in the fibred product
$$E \times_p E = \{(e_1, e_2) : p(e_1) = p(e_2)\}.$$
Here follows the proof in lenghty details.

Open:
Let $e \in E$, we want to show that there exists an open subset
$A \subseteq E \times_p E$ such that $(e, e) \in A \subseteq \Delta_E$.
Let $U$ be an evenly covered open neighbourhood of $p(e)$, and let
$V$ be the sheet (one of the opens in which the preimage of $U$ decomposes)
containing $e$. Then the restriction
$$p_| \colon V \to U$$
is a homeomorphism, and in particular it is injective.
The subset
$$V \times_p V = (V \times V) \cap (E \times_p E) =
\{(e_1, e_2) \in V \times V : p(e_1) = p(e_2)\}$$
is an open neighbourhood of $(e, e)$. We want to show that it is contained
in $\Delta_E$.
Indeed, let $(e_1, e_2) \in V \times_p V$. Then $p(e_1) = p(e_2)$, and by injectivity $e_1 = e_2$, that is $(e_1, e_2) \in \Delta_E$.

Closed:
We prove that the complement $\Delta_E^c = E \times_p E \setminus \Delta_E$
is open. To do this, we adopt the same idea as before,
except that this time we use two (disjoint) sheets.
Let $(e_1, e_2) \in \Delta_E^c$, meaning $e_1 \neq e_2$
and $p(e_1) = p(e_2)$.
Let $U$ be an evenly covered open neighbourhood of $p(e_i)$ and let
$V_i$ be the sheet containing $e_i$. The open subsets $V_1$ and $V_2$ are disjoint by definition of covering map.
The subset
$$V_1 \times_p V_2 = (V_1 \times V_2) \cap (E \times_p E)$$
is an open neighbourhood of $(e_1, e_2)$. To show that $V_1 \times_p V_2$
is contained in $\Delta_E^c$, it is enough to use the fact that
$V_1$ and $V_2$ are disjoint.
