= Solution
Suppose $F:\Sigma_g\to\Sigma_h$ has degree one. For degree-one classes $u,v$ on the target, naturality of the <cup product> and the definition of the <degree of a map between oriented manifolds> give
$$
\left\langle F^*u\smile F^*v,[\Sigma_g]\right\rangle=\left\langle u\smile v,F_*[\Sigma_g]\right\rangle=\left\langle u\smile v,[\Sigma_h]\right\rangle.
$$
If a nonzero $u$ had $F^*u=0$, nondegeneracy of the <Poincare duality pairing> would supply $v$ with nonzero right side, a contradiction. Thus $F^*$ injects the $2h$-dimensional real degree-one <cohomology> into the $2g$-dimensional source. Hence $g\geq h$. This is the <cohomological injectivity of a degree-one map> in this setting.
Conversely, for $g\geq h$, express $\Sigma_g$ as $\Sigma_h\#\Sigma_{g-h}$. Collapse the second punctured summand and the joining circle to a point. The quotient of the retained punctured summand by its boundary is homeomorphic to $\Sigma_h$, giving a continuous map to that surface. Its restriction to a small oriented disc away from the collapsing region is an orientation-preserving homeomorphism, and a point in this disc has exactly one preimage. The induced map on local top homology, and hence on the <fundamental class>, has coefficient $+1$. The map therefore has degree one. For $h=0$, the same construction is the familiar collapse of the complement of a disc to obtain $S^2$.
Consequently
$$
\boxed{\text{A degree-one map }\Sigma_g\longrightarrow\Sigma_h\text{ exists exactly when }g\geq h.}
$$
This proves both directions of the <degree-one maps between closed oriented surfaces> criterion. The case $g=h$ also admits the identity map.
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