[proofplan] We verify the three defining properties of an [equivalence relation](/page/Equivalence%20Relation) directly from the path operations. Reflexivity comes from the constant path at a point. Symmetry follows by reversing a path, and transitivity follows by concatenating two paths whose endpoints match. The cited reversal and concatenation results supply the continuity of those two constructed paths. [/proofplan]

[step:Construct constant paths to prove reflexivity] Let $x \in X$. Define the map $\gamma_x : [0,1] \to X$ by $\gamma_x(t)=x$ for every $t \in [0,1]$, where $[0,1]$ has the [subspace topology](/page/Subspace%20Topology) inherited from $\mathbb R$. We verify that $\gamma_x$ is continuous. Let $U \in \tau$ be open in $X$. Then \begin{align*} \gamma_x^{-1}(U)= [0,1] \quad \text{if } x \in U, \qquad \gamma_x^{-1}(U)=\varnothing \quad \text{if } x \notin U. \end{align*} Both $[0,1]$ and $\varnothing$ are open in the subspace topology on $[0,1]$. Hence $\gamma_x$ is continuous. Also $\gamma_x(0)=x$ and $\gamma_x(1)=x$, so $\gamma_x$ is a path from $x$ to $x$. By the definition of $\sim_p$, this proves $x \sim_p x$. Since $x \in X$ was arbitrary, $\sim_p$ is reflexive. [/step]

[step:Reverse paths to prove symmetry] Let $x,y \in X$ and assume $x \sim_p y$. By definition of $\sim_p$, there exists a path $\gamma : [0,1] \to X$ from $x$ to $y$. Since $(X,\tau)$ is a [topological space](/page/Topological%20Space) and $\gamma$ is a path from $x$ to $y$, [citetheorem:9158] applies to the reversed map $\overline{\gamma} : [0,1] \to X$ defined by $\overline{\gamma}(t)=\gamma(1-t)$. It follows that $\overline{\gamma}$ is a path from $y$ to $x$. Therefore $y \sim_p x$ by the definition of $\sim_p$. Thus $\sim_p$ is symmetric. [/step]

[step:Concatenate composable paths to prove transitivity]Let $x,y,z \in X$ and assume $x \sim_p y$ and $y \sim_p z$. By definition of $\sim_p$, there exists a path $\gamma : [0,1] \to X$ from $x$ to $y$, and there exists a path $\eta : [0,1] \to X$ from $y$ to $z$. Since $(X,\tau)$ is a topological space, $\gamma$ is a path from $x$ to $y$, and $\eta$ is a path from $y$ to $z$, the endpoint condition required for concatenation is satisfied. Hence [citetheorem:9159] applies and gives that $\gamma * \eta$ is a path from $x$ to $z$. By the definition of $\sim_p$, this implies $x \sim_p z$. Therefore $\sim_p$ is transitive.[/step]

[guided]We must show that the relation passes through an intermediate point. Start with arbitrary points $x,y,z \in X$ and assume $x \sim_p y$ and $y \sim_p z$. These two assumptions mean, by the definition of path equivalence, that there are continuous maps $\gamma : [0,1] \to X$ and $\eta : [0,1] \to X$ such that $\gamma$ is a path from $x$ to $y$ and $\eta$ is a path from $y$ to $z$. The important compatibility condition is that the endpoint of the first path is the starting point of the second path. Indeed, $\gamma(1)=y$ and $\eta(0)=y$. Thus the hypotheses of [citetheorem:9159] are met: the ambient object is a topological space $(X,\tau)$, $\gamma$ is a path from $x$ to $y$, and $\eta$ is a path from $y$ to $z$. The theorem therefore says that the concatenated path $\gamma * \eta$ is a path from $x$ to $z$. Now return to the definition of $\sim_p$. Since there exists a path in $X$ from $x$ to $z$, namely $\gamma * \eta$, we have $x \sim_p z$. Because $x,y,z \in X$ were arbitrary subject only to $x \sim_p y$ and $y \sim_p z$, the relation $\sim_p$ is transitive.[/guided]

[step:Combine the three properties] We have proved that $\sim_p$ is reflexive, symmetric, and transitive on $X$. Therefore $\sim_p$ is an equivalence relation on $X$. [/step]