[proofplan] We use the homotopy invariance of singular homology: homotopic maps induce equal maps on homology. If $f: X \to Y$ is a homotopy equivalence with homotopy inverse $g$, then $g \circ f \simeq \operatorname{id}_X$ implies $g_* \circ f_* = \operatorname{id}_{H_n(X)}$, and $f \circ g \simeq \operatorname{id}_Y$ implies $f_* \circ g_* = \operatorname{id}_{H_n(Y)}$. This shows $f_*$ is an isomorphism. The cohomological statement follows by applying the same argument to the induced maps on cohomology, using that $f^*$ reverses composition. [/proofplan] [step:Apply homotopy invariance to the compositions $g \circ f$ and $f \circ g$] Let $g: Y \to X$ be a homotopy inverse of $f$, so $g \circ f \simeq \operatorname{id}_X$ and $f \circ g \simeq \operatorname{id}_Y$. By functoriality of the induced map on homology, $(g \circ f)_* = g_* \circ f_*$ and $(f \circ g)_* = f_* \circ g_*$. By homotopy invariance (homotopic maps induce equal maps on homology): \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 with inverse $g_*$. [/step] [step:Deduce the cohomological statement] For cohomology, the induced maps reverse composition: $(g \circ f)^* = f^* \circ g^*$ and $(f \circ g)^* = g^* \circ f^*$. By homotopy invariance applied to cohomology: \begin{align*} f^* \circ g^* = (g \circ f)^* = (\operatorname{id}_X)^* = \operatorname{id}_{H^n(X)}, \end{align*} and \begin{align*} g^* \circ f^* = (f \circ g)^* = (\operatorname{id}_Y)^* = \operatorname{id}_{H^n(Y)}. \end{align*} Therefore $f^*: H^n(Y) \to H^n(X)$ is an isomorphism with inverse $g^*$. [/step]