[proofplan] We prove the three defining properties of an [equivalence relation](/page/Equivalence%20Relation) directly from the definition of homeomorphism. Reflexivity is witnessed by the identity map on each [topological space](/page/Topological%20Space). Symmetry follows because the inverse of a homeomorphism is again a homeomorphism, and transitivity follows by composing the witnessing homeomorphisms. [/proofplan] [step:Use the identity map to prove reflexivity] Let $(X,\tau_X) \in \mathcal{C}$ be arbitrary. Define the identity map $\operatorname{id}_X:X \to X$ by $\operatorname{id}_X(x)=x$ for every $x \in X$. The map $\operatorname{id}_X$ is bijective, and its inverse is itself. To verify continuity, let $U \subset X$ satisfy $U \in \tau_X$. Then $\operatorname{id}_X^{-1}(U)=U \in \tau_X$. Thus $\operatorname{id}_X$ is continuous. Since $\operatorname{id}_X^{-1}=\operatorname{id}_X$, its inverse is continuous as well. Therefore $\operatorname{id}_X$ is a homeomorphism from $(X,\tau_X)$ to itself, so $(X,\tau_X)\sim (X,\tau_X)$. [guided] We need to show that every element of $\mathcal{C}$ is related to itself. Fix an arbitrary topological space $(X,\tau_X)\in \mathcal{C}$. The natural candidate witness is the identity map \begin{align*} \operatorname{id}_X:X \to X. \end{align*} It is defined by $\operatorname{id}_X(x)=x$ for every $x\in X$. To prove that this map is a homeomorphism, we check the definition. First, $\operatorname{id}_X$ is bijective because every point $x\in X$ has exactly one preimage under $\operatorname{id}_X$, namely $x$ itself. Its inverse map is again $\operatorname{id}_X$. Next, we verify continuity using the open-set definition. Let $U\subset X$ be open in $(X,\tau_X)$, meaning $U\in \tau_X$. The preimage of $U$ under the identity map is \begin{align*} \operatorname{id}_X^{-1}(U)=U. \end{align*} Since $U\in\tau_X$, this preimage is open. Hence $\operatorname{id}_X$ is continuous. Because the inverse of $\operatorname{id}_X$ is also $\operatorname{id}_X$, the same argument proves that the inverse is continuous. Therefore $\operatorname{id}_X$ is a homeomorphism from $(X,\tau_X)$ to itself, so $(X,\tau_X)\sim (X,\tau_X)$. [/guided] [/step] [step:Invert a witnessing homeomorphism to prove symmetry] Let $(X,\tau_X),(Y,\tau_Y)\in\mathcal{C}$ and suppose $(X,\tau_X)\sim (Y,\tau_Y)$. By definition of $\sim$, there exists a homeomorphism \begin{align*} f:X \to Y. \end{align*} Since $f$ is a homeomorphism, $f$ is bijective, $f$ is continuous, and the inverse map $f^{-1}:Y \to X$ is continuous. The map $f^{-1}$ is bijective because it is the inverse of a bijection. Its inverse is $f$, which is continuous. Hence $f^{-1}:Y\to X$ is a bijective continuous map with continuous inverse, so $f^{-1}$ is a homeomorphism from $(Y,\tau_Y)$ to $(X,\tau_X)$. Therefore $(Y,\tau_Y)\sim (X,\tau_X)$. [/step] [step:Compose witnessing homeomorphisms to prove transitivity] Let $(X,\tau_X),(Y,\tau_Y),(Z,\tau_Z)\in\mathcal{C}$ and suppose \begin{align*} (X,\tau_X)\sim (Y,\tau_Y) \end{align*} and \begin{align*} (Y,\tau_Y)\sim (Z,\tau_Z). \end{align*} By definition of $\sim$, there exist homeomorphisms \begin{align*} f:X \to Y \end{align*} and \begin{align*} g:Y \to Z. \end{align*} Define $h:X \to Z$ by $h=g\circ f$. Because $f$ and $g$ are bijections, $h=g\circ f$ is a bijection. Since $f$ and $g$ are continuous, their composition $h$ is continuous. The inverse of $h$ is \begin{align*} h^{-1}=f^{-1}\circ g^{-1}. \end{align*} Since $g^{-1}:Z\to Y$ and $f^{-1}:Y\to X$ are continuous, their composition $f^{-1}\circ g^{-1}:Z\to X$ is continuous. Thus $h:X\to Z$ is a bijective continuous map with continuous inverse, so $h$ is a homeomorphism from $(X,\tau_X)$ to $(Z,\tau_Z)$. Therefore $(X,\tau_X)\sim (Z,\tau_Z)$. [/step] [step:Conclude that the homeomorphism relation is an equivalence relation] We have proved that $\sim$ is reflexive, symmetric, and transitive on the arbitrary class $\mathcal{C}$ of topological spaces. Therefore the relation of being homeomorphic is an equivalence relation on $\mathcal{C}$. [/step]