1. **State the problem:** We want to prove that the mapping $\pi : x \to x/\sim$ is onto. 2. **Understanding the mapping:** The function $\pi$ maps an element $x$ from a set $X$ to its equivalence class $x/\sim$ under an equivalence relation $\sim$. 3. **Definition of onto (surjective) mapping:** A function $f : A \to B$ is onto if for every element $b \in B$, there exists at least one element $a \in A$ such that $f(a) = b$. 4. **Apply the definition to $\pi$:** For every equivalence class $C \in X/\sim$, we need to find an $x \in X$ such that $\pi(x) = C$. 5. **Proof:** By definition, each equivalence class $C$ is a subset of $X$ containing elements equivalent to each other. Choose any element $x \in C$. Then $\pi(x) = x/\sim = C$. 6. **Conclusion:** Since for every equivalence class $C$ there exists an $x$ with $\pi(x) = C$, the mapping $\pi$ is onto. **Final answer:** The mapping $\pi : x \to x/\sim$ is onto because every equivalence class has a preimage in $X$.