[proofplan] Each of the three properties (compactness, connectedness, path-connectedness) is preserved under continuous surjections, and the quotient projection $\pi: X \to X/{\sim}$ is a continuous surjection. We apply the [Continuous Image of a Compact Space is Compact](/theorems/305) for compactness, the [Continuous Image of a Connected Space is Connected](/theorems/296) for connectedness, and the [Continuous Image of a Path-Connected Space](/theorems/1056) for path-connectedness. [/proofplan] [step:Verify that the quotient map $\pi: X \to X/{\sim}$ is a continuous surjection] The quotient map $\pi: X \to X/{\sim}$ sends each point $x$ to its equivalence class $[x]$. By definition of the quotient topology, a subset $V \subset X/{\sim}$ is open if and only if $\pi^{-1}(V)$ is open in $X$. In particular, if $V$ is open in $X/{\sim}$, then $\pi^{-1}(V)$ is open in $X$, so $\pi$ is continuous. The map $\pi$ is surjective: every equivalence class $[x] \in X/{\sim}$ is the image of its representative $x$. Since $\pi$ is a continuous surjection and $\pi(X) = X/{\sim}$, each of the three properties follows from preservation under continuous surjections. [/step] [step:Compactness: apply the continuous image theorem] Suppose $X$ is compact. By the [Continuous Image of a Compact Space is Compact](/theorems/305), the image of a compact space under a continuous map is compact. Since $\pi: X \to X/{\sim}$ is continuous and surjective, $X/{\sim} = \pi(X)$ is compact. [guided] The hypotheses of theorem 305 require a continuous map from a compact topological space. The map $\pi$ is continuous (by definition of the quotient topology) and $X$ is compact by hypothesis. The conclusion gives that $\pi(X) = X/{\sim}$ is compact. Note that the quotient map property (the "if and only if" characterisation of open sets) is not needed — only the continuity direction. [/guided] [/step] [step:Connectedness: apply the continuous image theorem] Suppose $X$ is connected. By the [Continuous Image of a Connected Space is Connected](/theorems/296), the image of a connected space under a continuous map is connected. Since $\pi: X \to X/{\sim}$ is continuous and surjective, $X/{\sim} = \pi(X)$ is connected. [guided] Theorem 296 requires a continuous map from a connected topological space. The continuity of $\pi$ and the connectedness of $X$ are both given, so $X/{\sim} = \pi(X)$ is connected. Concretely, if $X/{\sim} = A \cup B$ with $A, B$ disjoint and open in $X/{\sim}$, then $X = \pi^{-1}(A) \cup \pi^{-1}(B)$ with $\pi^{-1}(A), \pi^{-1}(B)$ disjoint and open (by continuity of $\pi$). Since $X$ is connected, one of $\pi^{-1}(A), \pi^{-1}(B)$ is empty, so one of $A, B$ is empty (as $\pi$ is surjective). [/guided] [/step] [step:Path-connectedness: apply the continuous image theorem] Suppose $X$ is path-connected. By the [Continuous Image of a Path-Connected Space](/theorems/1056), the image of a path-connected space under a continuous surjection is path-connected. Since $\pi: X \to X/{\sim}$ is continuous and surjective, $X/{\sim} = \pi(X)$ is path-connected. [guided] Let $[x], [y] \in X/{\sim}$. Since $X$ is path-connected, there exists a continuous path $\gamma: [0,1] \to X$ with $\gamma(0) = x$ and $\gamma(1) = y$. The composition $\pi \circ \gamma: [0,1] \to X/{\sim}$ is continuous (as the composition of continuous maps) and satisfies $(\pi \circ \gamma)(0) = [x]$ and $(\pi \circ \gamma)(1) = [y]$. Hence $[x]$ and $[y]$ are connected by a continuous path in $X/{\sim}$. [/guided] [/step]