[proofplan] We prove that the reversed parametrisation satisfies the two requirements in the definition of a path: continuity and the correct endpoint values. The key auxiliary map is the interval reversal map $r : [0,1] \to [0,1]$, $r(t)=1-t$, which is continuous. Since $\overline{\gamma}=\gamma \circ r$, continuity follows from continuity of composition, and the endpoint identities follow by direct evaluation. [/proofplan] [step:Express the reversed path as a composition with the interval reversal map] Define the interval reversal map \begin{align*} r : [0,1] \to [0,1],\qquad t \mapsto 1-t. \end{align*} For every $t \in [0,1]$, the number $1-t$ belongs to $[0,1]$, so $r$ is well-defined. By the definition of $\overline{\gamma}$, for every $t \in [0,1]$ we have \begin{align*} \overline{\gamma}(t)=\gamma(1-t)=(\gamma \circ r)(t). \end{align*} Thus $\overline{\gamma}=\gamma \circ r$ as maps from $[0,1]$ to $X$. [/step] [step:Prove that the reversed parametrisation is continuous] The map $r : [0,1] \to [0,1]$ is continuous because it is the restriction to $[0,1]$ of the continuous affine map $\mathbb{R}\to\mathbb{R}$, $s\mapsto 1-s$. Since $\gamma : [0,1] \to X$ is a path from $x$ to $y$, $\gamma$ is continuous. Therefore the composition $\gamma \circ r : [0,1]\to X$ is continuous. Since $\overline{\gamma}=\gamma \circ r$, the map $\overline{\gamma}$ is continuous. [guided] We need to verify the continuity part of the definition of a path. The reversed map is not a new kind of object; it is the original path $\gamma$ precomposed with a continuous reparametrisation of the interval. Define \begin{align*} r : [0,1] \to [0,1],\qquad t \mapsto 1-t. \end{align*} This map is well-defined because if $t \in [0,1]$, then $0 \leq 1-t \leq 1$, so $1-t \in [0,1]$. The map $r$ is continuous with respect to the [subspace topology](/page/Subspace%20Topology) on $[0,1]$ because it is the restriction of the continuous affine map $\mathbb{R} \to \mathbb{R}$ given by $s \mapsto 1-s$. Now use the hypothesis that $\gamma$ is a path from $x$ to $y$. By definition, this means in particular that \begin{align*} \gamma : [0,1] \to X \end{align*} is continuous. The codomain of $r$ is $[0,1]$, which is the domain of $\gamma$, so the composition \begin{align*} \gamma \circ r : [0,1] \to X \end{align*} is defined. By continuity of composition, $\gamma \circ r$ is continuous. Finally, for each $t \in [0,1]$, \begin{align*} (\gamma \circ r)(t)=\gamma(r(t))=\gamma(1-t)=\overline{\gamma}(t). \end{align*} Hence $\overline{\gamma}=\gamma \circ r$, and therefore $\overline{\gamma}$ is continuous. [/guided] [/step] [step:Evaluate the endpoints of the reversed path] Since $\gamma$ is a path from $x$ to $y$, its endpoint values are \begin{align*} \gamma(0)=x \end{align*} and \begin{align*} \gamma(1)=y. \end{align*} Using the definition of $\overline{\gamma}$, we compute \begin{align*} \overline{\gamma}(0)=\gamma(1-0)=\gamma(1)=y. \end{align*} Similarly, \begin{align*} \overline{\gamma}(1)=\gamma(1-1)=\gamma(0)=x. \end{align*} Thus $\overline{\gamma}$ starts at $y$ and ends at $x$. [/step] [step:Conclude that the reversed map is a path from $y$ to $x$] We have shown that $\overline{\gamma} : [0,1]\to X$ is continuous, that $\overline{\gamma}(0)=y$, and that $\overline{\gamma}(1)=x$. By the definition of a path, $\overline{\gamma}$ is a path from $y$ to $x$. [/step]