[proofplan] Both the upper and lower bounds follow from the [Lipschitz Bound on Hausdorff Measure](/theorems/3056). The upper bound applies the Lipschitz bound to $f$ (Lipschitz constant $C$). The lower bound applies the same theorem to $f^{-1}: f(A) \to A$ (Lipschitz constant $1/c$) and rearranges. The dimension equality then follows because the two-sided bound ensures $\mathcal{H}^s(E) = 0$ if and only if $\mathcal{H}^s(f(E)) = 0$. [/proofplan] [step:Establish the upper bound $\mathcal{H}^s(f(E)) \leq C^s \mathcal{H}^s(E)$] Since $f: A \to \mathbb{R}^n$ satisfies $|f(x) - f(y)| \leq C|x - y|$ for all $x, y \in A$, the map $f$ is Lipschitz with constant $C$. Applying the [Lipschitz Bound on Hausdorff Measure](/theorems/3056) to $f$ with exponent $s$ and the subset $E \subset A$: \begin{align*} \mathcal{H}^s(f(E)) \leq C^s \, \mathcal{H}^s(E). \end{align*} [/step] [step:Establish the lower bound $c^s \mathcal{H}^s(E) \leq \mathcal{H}^s(f(E))$] Since $f$ is bi-Lipschitz with lower constant $c > 0$, the map $f$ is injective and the inverse $f^{-1}: f(A) \to A$ satisfies \begin{align*} |f^{-1}(u) - f^{-1}(v)| \leq \frac{1}{c} |u - v| \end{align*} for all $u, v \in f(A)$. Indeed, setting $u = f(x)$ and $v = f(y)$ in the lower Lipschitz bound $c|x - y| \leq |f(x) - f(y)|$ and rearranging gives $|x - y| \leq \frac{1}{c}|f(x) - f(y)|$, which is the stated Lipschitz bound on $f^{-1}$. Applying the [Lipschitz Bound on Hausdorff Measure](/theorems/3056) to $f^{-1}: f(A) \to A$ with Lipschitz constant $1/c$, exponent $s$, and the subset $f(E) \subset f(A)$: \begin{align*} \mathcal{H}^s(f^{-1}(f(E))) \leq \left(\frac{1}{c}\right)^s \mathcal{H}^s(f(E)). \end{align*} Since $f$ is injective, $f^{-1}(f(E)) = E$, so \begin{align*} \mathcal{H}^s(E) \leq \frac{1}{c^s} \mathcal{H}^s(f(E)). \end{align*} Multiplying both sides by $c^s$ yields $c^s \mathcal{H}^s(E) \leq \mathcal{H}^s(f(E))$. [/step] [step:Combine the bounds to show $\dim_{\mathcal{H}}(f(E)) = \dim_{\mathcal{H}}(E)$] From the previous two steps, for every $s \geq 0$: \begin{align*} c^s \, \mathcal{H}^s(E) \leq \mathcal{H}^s(f(E)) \leq C^s \, \mathcal{H}^s(E). \end{align*} Since $c > 0$ and $C < \infty$, the constants $c^s$ and $C^s$ are positive and finite for every $s \geq 0$. Therefore $\mathcal{H}^s(E) = 0$ if and only if $\mathcal{H}^s(f(E)) = 0$. By definition, the Hausdorff dimension is the critical exponent \begin{align*} \dim_{\mathcal{H}}(E) = \inf\{s \geq 0 : \mathcal{H}^s(E) = 0\}, \end{align*} and similarly for $f(E)$. Since the set $\{s \geq 0 : \mathcal{H}^s(E) = 0\}$ equals $\{s \geq 0 : \mathcal{H}^s(f(E)) = 0\}$, the infima agree: \begin{align*} \dim_{\mathcal{H}}(f(E)) = \dim_{\mathcal{H}}(E). \end{align*} [/step]