= Solution
For a <closed manifold> $M$ with an oriented <Riemannian metric>, define $\delta$ to be the formal $L^2$ <adjoint operator> of the <exterior derivative>, equivalently $\delta=(-1)^{n(q+1)+1}*d*$ on $q$-forms. The <Hodge Laplacian> is $\Delta_q=d\delta+\delta d$. The <Hodge decomposition theorem> states that its space $\mathcal H^q=\ker\Delta_q$ of <harmonic differential forms> is finite-dimensional and smooth, and that
$$
\boxed{\Omega^q(M)=\mathcal H^q\mathbin{\oplus^\perp}d\Omega^{q-1}(M)\mathbin{\oplus^\perp}\delta\Omega^{q+1}(M),\qquad H^q_{\mathrm{dR}}(M)\cong\mathcal H^q.}
$$
The last isomorphism sends a <harmonic differential form> to its <de Rham cohomology> class: every class has exactly one harmonic representative. Out-of-range degree spaces are zero.
We first justify the analytic facts needed in the proof. In local coordinates and a smooth frame for the form bundle, the <principal symbol> of $\Delta_q$ is $g^{ij}\xi_i\xi_j I$. This follows from the exterior/contraction identities for $d$ and $\delta$. Freezing the leading coefficients, the <Fourier transform> estimate for the constant positive quadratic form gives, for each nonnegative integer $s$,
$$
\|u\|_{H^{s+2}}\leq C_s\bigl(\|\Delta_q u\|_{H^s}+\|u\|_{L^2}\bigr).
$$
Here is how to pass to variable coefficients. On a sufficiently small coordinate ball the difference of leading coefficients has arbitrarily small supremum norm. After differentiating $s$ times, its top-order contribution is bounded by $\varepsilon\|u\|_{H^{s+2}}$; derivatives of the coefficients and first-order terms contribute $C_s\|u\|_{H^{s+1}}$. A <partition of unity> gives the same types of terms from commutators with the cutoffs. Absorb the epsilon term and use <Sobolev interpolation> to absorb $\|u\|_{H^{s+1}}$ into a small multiple of $\|u\|_{H^{s+2}}$ plus $C\|u\|_2$. This proves the estimate.
The first-order energy form $Q(u,u)=\|du\|_2^2+\|\delta u\|_2^2$ satisfies $Q(u,u)+c\|u\|_2^2\geq c'\|u\|_{H^1}^2$ for a sufficiently large $c>0$. This follows from the same frozen-symbol argument at first order, or by integrating the principal part and absorbing the lower-order terms. The <Lax-Milgram theorem> thus constructs a bounded inverse $R=(\Delta_q+c)^{-1}:L^2\to H^1$ in the weak sense. To obtain $H^2$ regularity of this weak solution, use a local difference quotient $\delta_hu$ and test the weak equation with $-\delta_{-h}(\chi^2\delta_hu)$. Positivity of $g^{ij}$ bounds $\|\chi\delta_h\nabla u\|_2^2$; derivatives of $\chi$ and difference quotients of the smooth coefficients give terms bounded by $C\|u\|_{H^1}\|\chi\delta_h\nabla u\|_2$ and $C\|u\|_{H^1}^2$. The forcing contributes at most $C\|f\|_2(\|\chi\delta_h\nabla u\|_2+\|u\|_{H^1})$. Absorption gives a bound independent of $h$, so the difference-quotient criterion yields $u\in H^2$. Differentiating the equation and repeating gives the higher estimates and smoothness for smooth forcing. Thus the use of <elliptic regularity> here has been justified rather than assumed.
The inverse $R$ is compact as an $L^2$ operator by the <Rellich-Kondrachov compactness theorem>, and self-adjoint by symmetry of the weak energy form. It is positive and injective with dense range. The <spectral theorem for compact self-adjoint operators> now gives an <orthonormal eigenbasis> of $\Delta_q$, with nonnegative <eigenvalues> tending to infinity. Each <eigenfunction> is smooth by the estimates above. In particular $\mathcal H^q$ is finite-dimensional and there is a positive gap above its zero <eigenvalue>.
Let $H_q$ be the <orthogonal projection> onto $\mathcal H^q$. Define the <Green operator of the Hodge Laplacian> by multiplying a positive-eigenvalue coefficient by $1/\lambda$ and sending the zero coefficients to zero. The gap makes this bounded on $L^2$, and the elliptic estimate shows it maps $H^s$ to $H^{s+2}$; hence it sends smooth forms to smooth forms. It obeys
$$
\Delta_qG_q=G_q\Delta_q=I-H_q.
$$
For a smooth $q$-form $\alpha$, expand the left side to obtain
$$
\alpha=H_q\alpha+d\delta G_q\alpha+\delta dG_q\alpha.
$$
These three summands are mutually orthogonal. The energy identity
$$
\langle\Delta_q h,h\rangle=\|dh\|_2^2+\|\delta h\|_2^2
$$
shows that a <harmonic differential form> has both $dh=0$ and $\delta h=0$, making it orthogonal to the exact and coexact summands by <integration by parts>. Also $\langle d\beta,\delta\gamma\rangle=\langle d^2\beta,\gamma\rangle=0$. This proves the claimed direct sum and its uniqueness.
Finally let $d\alpha=0$ and write $u=G_q\alpha$. We have $d\Delta_qu=\Delta_{q+1}du$ because $d^2=0$. Since $d(\alpha-H_q\alpha)=0$, $du$ is harmonic. It is also exact, so its <orthogonality> to harmonic forms gives $du=0$. The decomposition therefore reduces to $\alpha=H_q\alpha+d\delta u$: every <closed differential form> is cohomologous to a harmonic one. If a harmonic $h$ is exact, $h=d\beta$, then $\|h\|_2^2=\langle\delta h,\beta\rangle=0$. Hence its class is zero only when $h=0$, establishing the isomorphism with <de Rham cohomology>.
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