Let $M$ be $m$-dimensional compact Riemannian manifold.
We choose a local coordinate system $\{(U_i,\phi_i)\}$ such that $U_i$ is diffeomorphic to $\mathbb{R}^m$ and $\overline{U_i}$ is compact and choose a partition of unity $\{\rho_i\}$ subordinate to $\{U_i\}$.
Then we define $$H^s(M):=\{u\in D'(M)|(\phi_i^{-1})^*(u|_{U_i})\in H^s_{loc}(\phi_i(U_i))\}$$ and semilinear pairing$\langle\cdot,\cdot\rangle:H^s(M)\times H^{-s}(M)\to\mathbb{C}$ $$\langle u,v\rangle:=\sum\langle(\phi_i^{-1})^*\rho_i u,(\phi_i^{-1})^*\rho_i u\rangle_s$$ where $\langle\cdot,\cdot\rangle_s$ is a pairing of $H^s(\mathbb{R}^m)$ and $H^{-s}(\mathbb{R}^m)$.
I am trying to proof that this pairing identifies $H^{-s}(M)$ as the dual space of $H^s(M)$.
Here is a sketch up to the middle of the proof.
Let $v$ be in $H^{-s}(M)$.Then $\langle\cdot,v\rangle:H^s(M)\to\mathbb{C}$ is in $(H^s(M))^*$. So We define $\Phi:H^{-s}(M)\to (H^s(M))^*,\Phi(v)=\langle\cdot,v\rangle$.
To prove this proposition, it is sufficient to show that $\Phi$ is bijective.
Let $f$ be in $(H^s(M))^*$ and $u$ be in $H^s(M)$ and we define $g\in H^s(\mathbb{R}^m)$ as $g(u')=f((\phi_i)^*u')$.
Then $f(\rho_iu)=g((\phi_i^{-1})^*(\rho_iu))$ so $f(u)=\sum f(\rho_iu)=\sum g((\phi^{-1})^*(\rho_i u))$.
Since pairing $\langle\cdot,\cdot\rangle_s$ identifies $H^{-s}(\mathbb{R}^m)$ as $(H^s(\mathbb{R}^m))^*$,there uniquely exists $v'\in H^{-s}(\mathbb{R}^m)$ such that forall $i,f(\rho_iu)=\langle(\phi_i^{-1})^*(\rho_iu),v'\rangle_s$.
Therefore $f(u)=\sum\langle(\phi_i^{-1})^*(\rho_iu),v'\rangle_s$.
I don't know what to do after. To get $v\in H^{-s}(M)$ such that $f(u)=\langle u,v\rangle$,what I need to do?