= Solution
Fix the convention $\iota_{X_f}\omega=-df$ for a <Hamiltonian vector field>. Nondegeneracy of the <symplectic form> gives one unique smooth $X_f$ for each smooth real-valued function $f$. Define the <Poisson bracket> by
$$
\{f,g\}=\omega(X_f,X_g)=X_f(g).
$$
In coordinates with $\omega=\sum dq^i\wedge dp_i$, this is $\sum(f_{q^i}g_{p_i}-f_{p_i}g_{q^i})$. Thus the sign convention is explicit and agrees with the usual coordinate <Poisson bracket>. By <Cartan's magic formula>, $\mathcal L_{X_f}\omega=d\iota_{X_f}\omega+\iota_{X_f}d\omega=0$. The contraction–<Lie derivative> commutator identity now gives
$$
\iota_{[X_f,X_g]}\omega=\mathcal L_{X_f}(\iota_{X_g}\omega)-\iota_{X_g}(\mathcal L_{X_f}\omega)=-d(X_f g)=-d\{f,g\}.
$$
Hence \b[the <Hamiltonian Lie algebra homomorphism> is $\boxed{\Phi(f)=X_f,\quad[X_f,X_g]=X_{\{f,g\}}}$]. It is linear and onto the space of <Hamiltonian vector fields> by definition. For completeness, the <Jacobi identity> for the bracket on functions follows from closure of $\omega$: evaluating $d\omega=0$ on $X_f,X_g,X_h$ and using the displayed commutator relation gives the cyclic Jacobi sum zero. Thus this really is a map of <Lie algebras>, not just a bracket-preserving notation.
Its kernel is determined by nondegeneracy:
$$
\boxed{\ker\Phi=\{f:df=0\}=\{\text{locally constant smooth functions}\}.}
$$
On a connected manifold these are precisely the real constants; on a disconnected manifold the constant may differ on each component. Therefore the quotient by these functions is isomorphic to the <Lie algebra> of <Hamiltonian vector fields>. The printed “homeomorphis” is interpreted as homomorphism: the map before taking this quotient is not an isomorphism because of its nontrivial kernel. The alternative convention $\iota_{X_f}\omega=df$ requires a corresponding bracket sign change to retain this homomorphism.
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