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Here is the key step you need to finish the proof: We are supposing $X$ is locally path-connected and semilocally simply connected, $\pi:P(X,x_0)\to \widetilde{X}$ is the quotient map identifying path-homotopy classes of paths, and $p:\widetilde{X}\to X$ is the induced surjection. Also, $ev:P(X,x_0)\to X$, $ev(\alpha)=\alpha(1)$ is endpoint evaluation.

Lemma: For any $\alpha\in P(X,x_0)$ there exists an open neighborhood $V$ of $\alpha(1)$ in $X$ and an open neighborhood $\mathcal{V}$ of $\alpha$ in $P(X,x_0)$ such that

1. $ev_1(\mathcal{V})=V$,
2. $\pi(\mathcal{V}\cap ev^{-1}(\alpha(1))=\{[\alpha]\}$, that is, every path $\beta\in\mathcal{V}$ that ends at $\alpha(1)$ is path-homotopic to $\alpha$.

Proof. Pick an open cover $\mathscr{U}$ of $\alpha([0,1])$ such that each $U\in\mathscr{U}$ is path connected and every loop in $U$ is null-homotopic in $X$. Using the Lebesgue number lemma, find $0=t_0<t_1<t_2<\cdots <t_n=1$ such that $\alpha([t_{i-1},t_i])\subseteq U_i$ for some $U_i\in\mathscr{U}$. In your notation, this means $\alpha\in\bigcap_{i=1}^{n}\left[[t_{i-1},t_i]\right],U_i]$. This open neighborhood of $\alpha$ is not quite good enough because the intersections $U_{i}\cap U_{i+1}$ need not be path-connected. So for each $i\in\{1,2,\dots ,n-1\}$, let $W_i$ be a path-connected neighborhood of $\alpha(t_i)$ in $U_i\cap U_{i+1}$. Now set $$\mathcal{V}=\bigcap_{i=1}^{n}\left[[t_{i-1},t_i]\right],U_i]\cap \bigcap_{i=1}^{n-1}\left[\{t_i\},W_i\right]$$ and $V=U_n$. Since $V$ is path-connected, it's clear that $ev_1(\mathcal{V})=V$. So it suffices to check that 2. holds.

Let $\beta\in \mathcal{V}$ with $\alpha(1)=\beta(1)$. It helps to draw a picture, here but basically, you want to pick paths $\gamma_i:[0,1]\to W_i$ from $\alpha(t_i)$ to $\beta(t_i)$ for each $i\in\{1,2,\dots,n-1\}$. For notational convenience, let $\gamma_0$ and $\gamma_{n}$ be the constant paths at $\alpha(0)$ and $\alpha(1)$ respectively. Now $\alpha_i$ and $\gamma_{i-1}\cdot \beta_i\cdot\gamma_{i-1}^{-}$ are paths in $U_i$ that start and end at the same point. Therefore, $\alpha_i\simeq \gamma_{i-1}\cdot \beta_i\cdot\gamma_{i}^{-}$ in $X$. By adjoining all of these path-homotopies together, you can construct a homotopy $$\alpha\simeq\prod_{i=1}^{n}\alpha_i\simeq \prod_{i=1}^{n}\gamma_{i-1}\cdot \beta_i\cdot\gamma_{i}^{-}\simeq \beta.$$Hence, if $\beta\in\mathcal{V}$ with $\beta(1)=\alpha(1)$, then $[\alpha]=[\beta]$. $\square$

Equipped with this lemma, you can finish your proof. Indeed, this is the key "technical" part of proving $\pi$ is a covering map when $\widetilde{X}$ has the natural quotient topology.

There are some interesting applications of the quotient topology on $\widetilde{X}$ when $X$ is not necessarily locally path connected or semilocally simply connected. In such cases, the quotient topology is not always equivalent to the "standard" or "whisker" topology.

Here is the key step you need to finish the proof: We are supposing $X$ is locally path-connected and semilocally simply connected, $\pi:P(X,x_0)\to \widetilde{X}$ is the quotient map identifying path-homotopy classes of paths, and $p:\widetilde{X}\to X$ is the induced surjection. Also, $ev:P(X,x_0)\to X$, $ev(\alpha)=\alpha(1)$ is endpoint evaluation.

Lemma: For any $\alpha\in P(X,x_0)$ there exists an open neighborhood $V$ of $\alpha(1)$ in $X$ and an open neighborhood $\mathcal{V}$ of $\alpha$ in $P(X,x_0)$ such that

1. $ev(\mathcal{V})=V$,
2. $\pi(\mathcal{V}\cap ev^{-1}(\alpha(1))=\{[\alpha]\}$, that is, every path $\beta\in\mathcal{V}$ that ends at $\alpha(1)$ is path-homotopic to $\alpha$.

Proof. Pick an open cover $\mathscr{U}$ of $\alpha([0,1])$ such that each $U\in\mathscr{U}$ is path connected and every loop in $U$ is null-homotopic in $X$. Using the Lebesgue number lemma, find $0=t_0<t_1<t_2<\cdots <t_n=1$ such that $\alpha([t_{i-1},t_i])\subseteq U_i$ for some $U_i\in\mathscr{U}$. In your notation, this means $\alpha\in\bigcap_{i=1}^{n}\left[[t_{i-1},t_i],U_i\right]$. This open neighborhood of $\alpha$ is not quite good enough because the intersections $U_{i}\cap U_{i+1}$ need not be path-connected. So for each $i\in\{1,2,\dots ,n-1\}$, let $W_i$ be a path-connected neighborhood of $\alpha(t_i)$ in $U_i\cap U_{i+1}$. Now set $$\mathcal{V}=\bigcap_{i=1}^{n}\left[[t_{i-1},t_i],U_i\right]\cap \bigcap_{i=1}^{n-1}\left[\{t_i\},W_i\right]$$ and $V=U_n$. Since $V$ is path-connected, it's clear that $ev(\mathcal{V})=V$. So it suffices to check that 2. holds.

Let $\beta\in \mathcal{V}$ with $\alpha(1)=\beta(1)$. It helps to draw a picture, here but basically, you want to pick paths $\gamma_i:[0,1]\to W_i$ from $\alpha(t_i)$ to $\beta(t_i)$ for each $i\in\{1,2,\dots,n-1\}$. For notational convenience, let $\gamma_0$ and $\gamma_{n}$ be the constant paths at $\alpha(0)$ and $\alpha(1)$ respectively. Now $\alpha_i$ and $\gamma_{i-1}\cdot \beta_i\cdot\gamma_{i-1}^{-}$ are paths in $U_i$ that start and end at the same point. Therefore, $\alpha_i\simeq \gamma_{i-1}\cdot \beta_i\cdot\gamma_{i}^{-}$ in $X$. By adjoining all of these path-homotopies together, you can construct a homotopy $$\alpha\simeq\prod_{i=1}^{n}\alpha_i\simeq \prod_{i=1}^{n}\gamma_{i-1}\cdot \beta_i\cdot\gamma_{i}^{-}\simeq \beta.$$Hence, if $\beta\in\mathcal{V}$ with $\beta(1)=\alpha(1)$, then $[\alpha]=[\beta]$. $\square$

Equipped with this lemma, you can finish your proof. Indeed, this is the key "technical" part of proving $\pi$ is a covering map when $\widetilde{X}$ has the natural quotient topology.

There are some interesting applications of the quotient topology on $\widetilde{X}$ when $X$ is not necessarily locally path connected or semilocally simply connected. In such cases, the quotient topology is not always equivalent to the "standard" or "whisker" topology.

Here is the key step you need to finish the proof: We are supposing $X$ is locally path-connected and semilocally simply connected, $\pi:P(X,x_0)\to \widetilde{X}$ is the quotient map identifying path-homotopy classes of paths, and $p:\widetilde{X}\to X$ is the induced surjection. Also, $ev:P(X,x_0)\to X$, $ev(\alpha)=\alpha(1)$ is endpoint evaluation.

Lemma: For any $\alpha\in P(X,x_0)$ there exists an open neighborhood $V$ of $\alpha(1)$ in $X$ and an open neighborhood $\mathcal{V}$ of $\alpha$ in $P(X,x_0)$ such that

1. $ev_1(\mathcal{V})=V$,
2. $\pi(\mathcal{V}\cap ev^{-1}(\alpha(1))=\{[\alpha]\}$, that is, every path $\beta\in\mathcal{V}$ that ends at $\alpha(1)$ is path-homotopic to $\alpha$.

Proof. Pick an open cover $\mathscr{U}$ of $\alpha([0,1])$ such that each $U\in\mathscr{U}$ is path connected and every loop in $U$ is null-homotopic in $X$. Using the Lebesgue number lemma, find $0=t_0<t_1<t_2<\cdots <t_n=1$ such that $\alpha([t_{i-1},t_i])\subseteq U_i$ for some $U_i\in\mathscr{U}$. In your notation, this means $\alpha\in\bigcap_{i=1}^{n}\left[[t_{i-1},t_i]\right],U_i]$. This open neighborhood of $\alpha$ is not quite good enough because the intersections $U_{i}\cap U_{i+1}$ need not be path-connected. So for each $i\in\{1,2,\dots ,n-1\}$, let $W_i$ be a path-connected neighborhood of $\alpha(t_i)$ in $U_i\cap U_{i+1}$. Now set $$\mathcal{V}=\bigcap_{i=1}^{n}\left[[t_{i-1},t_i]\right],U_i]\cap \bigcap_{i=1}^{n-1}\left[\{t_i\},W_i\right]$$ and $V=U_n$. Since $V$ is path-connected, it's clear that $ev_1(\mathcal{V})=V$. So it suffices to check that 2. holds.

Let $\beta\in \mathcal{V}$ with $\alpha(1)=\beta(1)$. It helps to draw a picture, here but basically, you want to pick paths $\gamma_i:[0,1]\to W_i$ from $\alpha(t_i)$ to $\beta(t_i)$ for each $i\in\{1,2,\dots,n-1\}$. For notational convenience, let $\gamma_0$ and $\gamma_{n}$ be the constant paths at $\alpha(0)$ and $\alpha(0)$ respectively. Now $\alpha_i$ and $\gamma_{i-1}\cdot \beta_i\cdot\gamma_{i-1}^{-}$ are paths in $U_i$ that start and end at the same point. Therefore, $\alpha_i\simeq \gamma_{i-1}\cdot \beta_i\cdot\gamma_{i}^{-}$ in $X$. By adjoining all of these path-homotopies together, you can construct a homotopy $$\alpha\simeq\prod_{i=1}^{n}\alpha_i\simeq \prod_{i=1}^{n}\gamma_{i-1}\cdot \beta_i\cdot\gamma_{i}^{-}\simeq \beta.$$Hence, if $\beta\in\mathcal{V}$ with $\beta(1)=\alpha(1)$, then $[\alpha]=[\beta]$. $\square$

Equipped with this lemma, you can finish your proof. Indeed, this is the key "technical" part of proving $\pi$ is a covering map when $\widetilde{X}$ has the natural quotient topology.

There are some interesting applications of the quotient topology on $\widetilde{X}$ when $X$ is not necessarily locally path connected or semilocally simply connected. In such cases, the quotient topology is not always equivalent to the "standard" or "whisker" topology.

Here is the key step you need to finish the proof: We are supposing $X$ is locally path-connected and semilocally simply connected, $\pi:P(X,x_0)\to \widetilde{X}$ is the quotient map identifying path-homotopy classes of paths, and $p:\widetilde{X}\to X$ is the induced surjection. Also, $ev:P(X,x_0)\to X$, $ev(\alpha)=\alpha(1)$ is endpoint evaluation.

Lemma: For any $\alpha\in P(X,x_0)$ there exists an open neighborhood $V$ of $\alpha(1)$ in $X$ and an open neighborhood $\mathcal{V}$ of $\alpha$ in $P(X,x_0)$ such that

1. $ev_1(\mathcal{V})=V$,
2. $\pi(\mathcal{V}\cap ev^{-1}(\alpha(1))=\{[\alpha]\}$, that is, every path $\beta\in\mathcal{V}$ that ends at $\alpha(1)$ is path-homotopic to $\alpha$.

Proof. Pick an open cover $\mathscr{U}$ of $\alpha([0,1])$ such that each $U\in\mathscr{U}$ is path connected and every loop in $U$ is null-homotopic in $X$. Using the Lebesgue number lemma, find $0=t_0<t_1<t_2<\cdots <t_n=1$ such that $\alpha([t_{i-1},t_i])\subseteq U_i$ for some $U_i\in\mathscr{U}$. In your notation, this means $\alpha\in\bigcap_{i=1}^{n}\left[[t_{i-1},t_i]\right],U_i]$. This open neighborhood of $\alpha$ is not quite good enough because the intersections $U_{i}\cap U_{i+1}$ need not be path-connected. So for each $i\in\{1,2,\dots ,n-1\}$, let $W_i$ be a path-connected neighborhood of $\alpha(t_i)$ in $U_i\cap U_{i+1}$. Now set $$\mathcal{V}=\bigcap_{i=1}^{n}\left[[t_{i-1},t_i]\right],U_i]\cap \bigcap_{i=1}^{n-1}\left[\{t_i\},W_i\right]$$ and $V=U_n$. Since $V$ is path-connected, it's clear that $ev_1(\mathcal{V})=V$. So it suffices to check that 2. holds.

Let $\beta\in \mathcal{V}$ with $\alpha(1)=\beta(1)$. It helps to draw a picture, here but basically, you want to pick paths $\gamma_i:[0,1]\to W_i$ from $\alpha(t_i)$ to $\beta(t_i)$ for each $i\in\{1,2,\dots,n-1\}$. For notational convenience, let $\gamma_0$ and $\gamma_{n}$ be the constant paths at $\alpha(0)$ and $\alpha(1)$ respectively. Now $\alpha_i$ and $\gamma_{i-1}\cdot \beta_i\cdot\gamma_{i-1}^{-}$ are paths in $U_i$ that start and end at the same point. Therefore, $\alpha_i\simeq \gamma_{i-1}\cdot \beta_i\cdot\gamma_{i}^{-}$ in $X$. By adjoining all of these path-homotopies together, you can construct a homotopy $$\alpha\simeq\prod_{i=1}^{n}\alpha_i\simeq \prod_{i=1}^{n}\gamma_{i-1}\cdot \beta_i\cdot\gamma_{i}^{-}\simeq \beta.$$Hence, if $\beta\in\mathcal{V}$ with $\beta(1)=\alpha(1)$, then $[\alpha]=[\beta]$. $\square$

Equipped with this lemma, you can finish your proof. Indeed, this is the key "technical" part of proving $\pi$ is a covering map when $\widetilde{X}$ has the natural quotient topology.

There are some interesting applications of the quotient topology on $\widetilde{X}$ when $X$ is not necessarily locally path connected or semilocally simply connected. In such cases, the quotient topology is not always equivalent to the "standard" or "whisker" topology.

Here is the key step you need to finish the proof: We are supposing $X$ is locally path-connected and semilocally simply connected, $\pi:P(X,x_0)\to \widetilde{X}$ is the quotient map identifying path-homotopy classes of paths, and $p:\widetilde{X}\to X$ is the induced surjection. Also, $ev:P(X,x_0)\to X$, $ev(\alpha)=\alpha(1)$ is endpoint evaluation.

Lemma: For any $\alpha\in P(X,x_0)$ there exists an open neighborhood $V$ of $\alpha(1)$ in $X$ and an open neighborhood $\mathcal{V}$ of $\alpha$ in $P(X,x_0)$ such that

1. $ev_1(\mathcal{V})=V$,
2. $\pi(\mathcal{V}\cap ev^{-1}(\alpha(1))=\{[\alpha]\}$, that is, every path $\beta\in\mathcal{V}$ that ends at $\alpha(1)$ is path-homotopic to $\alpha$.

Proof sketch. Pick an open cover $\mathscr{U}$ of $\alpha([0,1])$ such that each $U\in\mathscr{U}$ is path connected and every loop in $U$ is null-homotopic in $X$. Using the Lebesgue number lemma, find $0=t_0<t_1<t_2<\cdots <t_n=1$ such that $\alpha([t_{i-1},t_i])\subseteq U_i$ for some $U_i\in\mathscr{U}$. In your notation, this means $\alpha\in\bigcap_{i=1}^{n}\left[[t_{i-1},t_i]\right],U_i]$. This open neighborhood of $\alpha$ is not quite good enough because the intersections $U_{i}\cap U_{i+1}$ need not be path-connected. So for each $i\in\{1,2,\dots ,n-1\}$, let $W_i$ be a path-connected neighborhood of $\alpha(t_i)$ in $U_i\cap U_{i+1}$. Now set $$\mathcal{V}=\bigcap_{i=1}^{n}\left[[t_{i-1},t_i]\right],U_i]\cap \bigcap_{i=1}^{n-1}\left[\{t_i\},W_i\right]$$ and $V=U_n$. Since $V$ is path-connected, it's clear that $ev_1(\mathcal{V})=V$. So it suffices to check that 2. holds.

Let $\beta\in \mathcal{V}$ with $\alpha(1)=\beta(1)$. It helps to draw a picture, here but basically, you want to pick paths $\gamma_i:[0,1]\to W_i$ from $\alpha(t_i)$ to $\beta(t_i)$ for each $i\in\{1,2,\dots,n-1\}$. For notational convenience, let $\gamma_0$ and $\gamma_{n}$ be the constant paths at $\alpha(0)$ and $\alpha(0)$ respectively. Now $\alpha_i$ and $\gamma_{i-1}\cdot \beta_i\cdot\gamma_{i-1}^{-}$ are paths in $U_i$ that start and end at the same point. Therefore, $\alpha_i\simeq \gamma_{i-1}\cdot \beta_i\cdot\gamma_{i}^{-}$ in $X$. By adjoining all of these path-homotopies together, you can construct a homotopy $$\alpha\simeq\prod_{i=1}^{n}\alpha_i\simeq \prod_{i=1}^{n}\gamma_{i-1}\cdot \beta_i\cdot\gamma_{i}^{-}\simeq \beta.$$Hence, if $\beta\in\mathcal{V}$ with $\beta(1)=\alpha(1)$, then $[\alpha]=[\beta]$. $\square$

Equipped with this lemma, you can finish your proof. Indeed, this is the key "technical" part of proving $\pi$ is a covering map when $\widetilde{X}$ has the natural quotient topology.

There are some interesting applications of the quotient topology on $\widetilde{X}$ when $X$ is not necessarily locally path connected or semilocally simply connected. In such cases, the quotient topology is not always equivalent to the "standard" or "whisker" topology.

Here is the key step you need to finish the proof: We are supposing $X$ is locally path-connected and semilocally simply connected, $\pi:P(X,x_0)\to \widetilde{X}$ is the quotient map identifying path-homotopy classes of paths, and $p:\widetilde{X}\to X$ is the induced surjection. Also, $ev:P(X,x_0)\to X$, $ev(\alpha)=\alpha(1)$ is endpoint evaluation.

Lemma: For any $\alpha\in P(X,x_0)$ there exists an open neighborhood $V$ of $\alpha(1)$ in $X$ and an open neighborhood $\mathcal{V}$ of $\alpha$ in $P(X,x_0)$ such that

1. $ev_1(\mathcal{V})=V$,
2. $\pi(\mathcal{V}\cap ev^{-1}(\alpha(1))=\{[\alpha]\}$, that is, every path $\beta\in\mathcal{V}$ that ends at $\alpha(1)$ is path-homotopic to $\alpha$.

Proof. Pick an open cover $\mathscr{U}$ of $\alpha([0,1])$ such that each $U\in\mathscr{U}$ is path connected and every loop in $U$ is null-homotopic in $X$. Using the Lebesgue number lemma, find $0=t_0<t_1<t_2<\cdots <t_n=1$ such that $\alpha([t_{i-1},t_i])\subseteq U_i$ for some $U_i\in\mathscr{U}$. In your notation, this means $\alpha\in\bigcap_{i=1}^{n}\left[[t_{i-1},t_i]\right],U_i]$. This open neighborhood of $\alpha$ is not quite good enough because the intersections $U_{i}\cap U_{i+1}$ need not be path-connected. So for each $i\in\{1,2,\dots ,n-1\}$, let $W_i$ be a path-connected neighborhood of $\alpha(t_i)$ in $U_i\cap U_{i+1}$. Now set $$\mathcal{V}=\bigcap_{i=1}^{n}\left[[t_{i-1},t_i]\right],U_i]\cap \bigcap_{i=1}^{n-1}\left[\{t_i\},W_i\right]$$ and $V=U_n$. Since $V$ is path-connected, it's clear that $ev_1(\mathcal{V})=V$. So it suffices to check that 2. holds.

Let $\beta\in \mathcal{V}$ with $\alpha(1)=\beta(1)$. It helps to draw a picture, here but basically, you want to pick paths $\gamma_i:[0,1]\to W_i$ from $\alpha(t_i)$ to $\beta(t_i)$ for each $i\in\{1,2,\dots,n-1\}$. For notational convenience, let $\gamma_0$ and $\gamma_{n}$ be the constant paths at $\alpha(0)$ and $\alpha(0)$ respectively. Now $\alpha_i$ and $\gamma_{i-1}\cdot \beta_i\cdot\gamma_{i-1}^{-}$ are paths in $U_i$ that start and end at the same point. Therefore, $\alpha_i\simeq \gamma_{i-1}\cdot \beta_i\cdot\gamma_{i}^{-}$ in $X$. By adjoining all of these path-homotopies together, you can construct a homotopy $$\alpha\simeq\prod_{i=1}^{n}\alpha_i\simeq \prod_{i=1}^{n}\gamma_{i-1}\cdot \beta_i\cdot\gamma_{i}^{-}\simeq \beta.$$Hence, if $\beta\in\mathcal{V}$ with $\beta(1)=\alpha(1)$, then $[\alpha]=[\beta]$. $\square$

Equipped with this lemma, you can finish your proof. Indeed, this is the key "technical" part of proving $\pi$ is a covering map when $\widetilde{X}$ has the natural quotient topology.

There are some interesting applications of the quotient topology on $\widetilde{X}$ when $X$ is not necessarily locally path connected or semilocally simply connected. In such cases, the quotient topology is not always equivalent to the "standard" or "whisker" topology.