= Solution
Put $A=\Sigma_h^\partial$. This <surface with boundary> deformation retracts onto a wedge of $2h$ circles, so $H^1(A;\mathbb R)$ has dimension $2h$ and $H^2(A;\mathbb R)=0$. If $r$ were a <retraction>, $\iota^*r^*=\mathrm{id}$ would make $r^*$ injective. Its image $W\subset H^1(\Sigma_g;\mathbb R)$ would have dimension $2h$.
Every pair of classes $u,v$ in $H^1(A;\mathbb R)$ has $u\smile v=0$, since $H^2(A;\mathbb R)=0$. Naturality gives
$$
r^*u\smile r^*v=r^*(u\smile v)=0.
$$
Thus $W$ is an <isotropic subspace of a symplectic vector space> for the nondegenerate skew <Poincare duality pairing> on the $2g$-dimensional space $H^1(\Sigma_g;\mathbb R)$. The stated linear-algebra bound gives $2h\leq g$. Hence
$$
\boxed{h>g/2\quad\Longrightarrow\quad\text{no retraction exists}.}
$$
\b[The bound is sharp.] Double $A$ along its boundary: $D(A)=A\cup_{\partial A}A$ is the closed oriented surface of genus $2h$. Identify each copy with $A$ and fold them onto one copy. The two maps agree on the joining circle, so they give a continuous <retraction> fixing the first copy pointwise. This includes $h=0$, where the double of a disc is a sphere.
More generally, for $g\geq2h$, add $g-2h$ handles in the interior of the second copy. Pinch those extra handles onto their connecting point, keeping the boundary fixed, and then fold onto $A$. This is still a <retraction>. In fact the construction works for any embedding in the question: the connected complement has one boundary component and genus $g-h$ by Euler-characteristic additivity, and the <classification theorem for surfaces> identifies it, relative to that boundary, with a copy of $A$ having $g-2h$ additional handles. Thus the exact existence criterion is $2h\leq g$, as expressed by <retraction onto a punctured oriented surface>.
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