How does Hatcher conclude that the set of points where $\tilde f_1$ and $\tilde f_2$ agree is both open and closed?

Question

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?

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Answer 1

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.

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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 $(e_1, e_2) \in \Delta_E$.

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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.