[proofplan] The agreement set $S = \{y : \tilde f_1(y) = \tilde f_2(y)\}$ is shown to be both open and closed by a local argument based on evenly covered neighbourhoods. For openness, at a point where the two lifts agree we use an evenly covered neighbourhood $U$ of $f(y)$ and restrict to the sheet $\tilde U$ containing the common value; the homeomorphism $p|_{\tilde U}$ then forces the two lifts to agree on a whole neighbourhood. For closedness, at a point where the lifts disagree we select disjoint sheets $U_\beta \ne U_\gamma$ containing $\tilde f_1(y)$ and $\tilde f_2(y)$; the preimages of these sheets under the two lifts give a neighbourhood of $y$ that is disjoint from $S$. Both arguments are local and use only continuity of the lifts, local invertibility of $p$ on each sheet, and the disjointness of distinct sheets. The connectedness corollary is immediate from the fact that a clopen subset of a connected space is either empty or the whole space. [/proofplan] [step:Fix notation and recall the structure of an evenly covered neighbourhood] A [covering map](/page/Covering%20Space) is a continuous surjection $p: \tilde X \to X$ such that every point of $X$ has an open neighbourhood $U$ that is **evenly covered**: this means \begin{align*} p^{-1}(U) = \coprod_{\alpha \in A} U_\alpha, \end{align*} a disjoint union of open subsets $U_\alpha \subseteq \tilde X$ (the *sheets* over $U$), with each restriction \begin{align*} p|_{U_\alpha}: U_\alpha \to U \end{align*} a homeomorphism. A [lift](/page/Lift) of a continuous map $f: Y \to X$ is a continuous map $\tilde f: Y \to \tilde X$ with $p \circ \tilde f = f$. We are given two such lifts $\tilde f_1, \tilde f_2$, and we study the set \begin{align*} S := \{y \in Y : \tilde f_1(y) = \tilde f_2(y)\} \subseteq Y. \end{align*} [/step] [step:Show $S$ is open] Fix $y \in S$; we will produce an open neighbourhood $V \subseteq Y$ of $y$ contained in $S$. Let $\tilde x := \tilde f_1(y) = \tilde f_2(y)$. Then $p(\tilde x) = f(y)$, using that $\tilde f_1$ is a lift (the same value of $f(y)$ is obtained from $\tilde f_2$). By the covering-map property, $f(y) \in X$ has an evenly covered open neighbourhood $U \subseteq X$; write $p^{-1}(U) = \coprod_\alpha U_\alpha$. Since $\tilde x \in p^{-1}(U)$, exactly one index $\beta$ satisfies $\tilde x \in U_\beta$ (uniqueness by disjointness of the sheets). Write \begin{align*} \tilde U := U_\beta, \qquad q := p|_{\tilde U}: \tilde U \to U, \end{align*} so $q$ is a homeomorphism by the definition of an evenly covered neighbourhood. Define \begin{align*} V := \tilde f_1^{-1}(\tilde U) \cap \tilde f_2^{-1}(\tilde U) \subseteq Y. \end{align*} This is open because $\tilde U \subseteq \tilde X$ is open and $\tilde f_1, \tilde f_2$ are continuous, so both preimages are open, and the intersection of two open sets is open. It contains $y$ because $\tilde f_1(y) = \tilde f_2(y) = \tilde x \in \tilde U$. For each $y' \in V$, both $\tilde f_1(y')$ and $\tilde f_2(y')$ lie in $\tilde U$. Applying $p$ and using the lift property $p \circ \tilde f_i = f$, \begin{align*} q(\tilde f_1(y')) = p(\tilde f_1(y')) = f(y') = p(\tilde f_2(y')) = q(\tilde f_2(y')). \end{align*} Since $q$ is a homeomorphism, it is injective, so $\tilde f_1(y') = \tilde f_2(y')$, i.e. $y' \in S$. Thus $V \subseteq S$, and $S$ is open. [guided] The argument for openness has three ingredients: the local structure of the covering map (an evenly covered neighbourhood), the continuity of both lifts (to pull the sheet back to an open set in $Y$), and the injectivity of $p$ on each sheet (to conclude agreement from a common $p$-value). Why do we restrict to a single sheet $\tilde U = U_\beta$ rather than the full preimage $p^{-1}(U)$? Because on the full preimage $p$ is not injective — it could send two different points of $\tilde X$ to the same point of $X$. The definition of "evenly covered" is precisely the requirement that this non-injectivity is resolved by partitioning into sheets, on each of which $p$ is a homeomorphism (hence bijective). Why is the choice of $\beta$ unique? Because the sheets are disjoint: a point $\tilde x \in p^{-1}(U)$ lies in exactly one $U_\alpha$. The cleanest way to record this proof is: we shrink both lifts $\tilde f_1$ and $\tilde f_2$ down to a common open region in $Y$ on which they both factor through a single sheet, and on that region the injective map $q$ sees the same image from both lifts, forcing equality. [/guided] [/step] [step:Show $S$ is closed] We show the complement $Y \setminus S$ is open. Fix $y \notin S$, so $\tilde f_1(y) \ne \tilde f_2(y)$. We produce an open neighbourhood of $y$ disjoint from $S$. Let $U$ be an evenly covered open neighbourhood of $f(y)$ in $X$, with $p^{-1}(U) = \coprod_\alpha U_\alpha$. The lift property gives $\tilde f_1(y), \tilde f_2(y) \in p^{-1}(U)$, so each lies in some sheet: \begin{align*} \tilde f_1(y) \in U_\beta, \qquad \tilde f_2(y) \in U_\gamma. \end{align*} We claim $\beta \ne \gamma$. Indeed if $\beta = \gamma$, then $\tilde f_1(y)$ and $\tilde f_2(y)$ both lie in $U_\beta$; applying the injective homeomorphism $p|_{U_\beta}$ and using $p \circ \tilde f_i = f$, \begin{align*} p|_{U_\beta}(\tilde f_1(y)) = f(y) = p|_{U_\beta}(\tilde f_2(y)), \end{align*} so $\tilde f_1(y) = \tilde f_2(y)$, contradicting $y \notin S$. Thus $\beta \ne \gamma$, and $U_\beta \cap U_\gamma = \varnothing$ by the disjoint-union property. Define \begin{align*} W := \tilde f_1^{-1}(U_\beta) \cap \tilde f_2^{-1}(U_\gamma) \subseteq Y. \end{align*} Both $U_\beta$ and $U_\gamma$ are open in $\tilde X$, and $\tilde f_1, \tilde f_2$ are continuous, so $W$ is open. It contains $y$ by construction. We show $W \cap S = \varnothing$. Suppose for contradiction $y' \in W \cap S$. Then $y' \in S$ gives $\tilde f_1(y') = \tilde f_2(y')$. Call this common point $\tilde z$. From $y' \in W$ we have $\tilde z = \tilde f_1(y') \in U_\beta$ and $\tilde z = \tilde f_2(y') \in U_\gamma$, so $\tilde z \in U_\beta \cap U_\gamma = \varnothing$, a contradiction. Hence $W \cap S = \varnothing$, i.e. $W \subseteq Y \setminus S$. This exhibits an open neighbourhood of $y$ in the complement of $S$, so $Y \setminus S$ is open, and $S$ is closed. [guided] The closedness argument is dual to the openness argument. In the open case we had $y \in S$ and needed to find a neighbourhood in $S$; we did this by picking the single sheet $U_\beta$ containing $\tilde f_1(y) = \tilde f_2(y)$ and restricting both lifts there. In the closed case we have $y \notin S$ and need a neighbourhood in $Y \setminus S$; we pick the *two distinct* sheets $U_\beta$ and $U_\gamma$ containing $\tilde f_1(y)$ and $\tilde f_2(y)$ respectively, and the neighbourhood is the intersection of the two separate preimages. The key structural fact is that distinct sheets are disjoint — this is what makes $U_\beta \cap U_\gamma = \varnothing$ and what rules out any point of $W$ being in $S$. Without the disjointness (i.e. for a general continuous map that is not a covering), this argument breaks down. Why is it important that we proved $\beta \ne \gamma$ before choosing the sheets? Because if we had $\beta = \gamma$ — the two lifts landing in the same sheet at $y$ — then the argument would fail: the preimage intersection $W$ would not be disjoint from $S$, and indeed in that case the openness argument of Step 2 would actually place $y$ in $S$, contradicting our standing assumption that $y \notin S$. The verification of $\beta \ne \gamma$ is what ties the closedness argument to the openness argument and ensures they are consistent. [/guided] [/step] [step:Deduce the connectedness corollary] Steps 2 and 3 show that $S$ is both open and closed in $Y$. If $Y$ is [connected](/page/Connected%20Space), then the only clopen subsets of $Y$ are $\varnothing$ and $Y$ itself (by the definition of connectedness). Therefore either $S = \varnothing$, meaning $\tilde f_1(y) \ne \tilde f_2(y)$ for all $y \in Y$, or $S = Y$, meaning $\tilde f_1 = \tilde f_2$ everywhere on $Y$. This completes the proof. [/step]