[proofplan] We use the functoriality of the induced map on homology: a continuous map $f: X \to Y$ induces $f_*: H_n(X) \to H_n(Y)$, and $(f \circ g)_* = f_* \circ g_*$ while $(\operatorname{id}_X)_* = \operatorname{id}_{H_n(X)}$. If $f$ is a homeomorphism with inverse $g$, then both compositions $f_* \circ g_*$ and $g_* \circ f_*$ are identity maps, so $f_*$ is an isomorphism with inverse $g_*$. [/proofplan] [step:Establish functoriality of the induced homomorphism] The assignment $f \mapsto f_*$ satisfies two functorial properties: 1. **Identity.** The identity map $\operatorname{id}_X: X \to X$ induces the identity on chains: $(\operatorname{id}_X)_n(\sigma) = \operatorname{id}_X \circ \sigma = \sigma$ for every generator $\sigma: \Delta^n \to X$. Hence $(\operatorname{id}_X)_n = \operatorname{id}_{C_n(X)}$, and passing to homology, $(\operatorname{id}_X)_* = \operatorname{id}_{H_n(X)}$. 2. **Composition.** For continuous maps $f: X \to Y$ and $g: Y \to Z$, the chain-level map $(g \circ f)_n$ acts on a generator by $(g \circ f)_n(\sigma) = (g \circ f) \circ \sigma = g \circ (f \circ \sigma) = g_n(f_n(\sigma))$. Hence $(g \circ f)_n = g_n \circ f_n$, and passing to homology, $(g \circ f)_* = g_* \circ f_*$. [/step] [step:Apply functoriality to the homeomorphism and its inverse] Let $f: X \to Y$ be a homeomorphism with continuous inverse $g: Y \to X$, so $g \circ f = \operatorname{id}_X$ and $f \circ g = \operatorname{id}_Y$. By functoriality: \begin{align*} g_* \circ f_* = (g \circ f)_* = (\operatorname{id}_X)_* = \operatorname{id}_{H_n(X)}, \end{align*} and \begin{align*} f_* \circ g_* = (f \circ g)_* = (\operatorname{id}_Y)_* = \operatorname{id}_{H_n(Y)}. \end{align*} Therefore $f_*: H_n(X) \to H_n(Y)$ is an isomorphism of abelian groups with inverse $g_*: H_n(Y) \to H_n(X)$. [/step]