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The usual embedding of the closed orientable surface $M_g$ of genus $g$ in $\mathbb{R}^3$ bounds a compact orientable $3$-manifold which is homotopically equivalent to a bouquet of $g$ circles.
If a compact orientable $3$-manifold $N$ has $M_g$ as its boundary, is it true that $N$ is homotopically equivalent to a bouquet of $g$ circles?
Definitely not. For instance, you could take your example $N\subset\mathbb{R}^3$ and form a connected sum $N\mathbin{\#} P$ where $P$ is any closed oriented 3-manifold. By some simple long exact sequence computations you can find that $H_2(N\mathbin{\#}P)\cong H_2(P)$, so $N\mathbin{\#}P$ cannot be homotopy equivalent to a wedge of circles if $H_2(P)$ is nontrivial.
No. As an obnoxious counterexample, take $g=0$ and $N$ to be the complement of some small open ball in a 3-manifold $M$; then $\pi_1(N) = \pi_1(M)$, so if $M$ was not simply connected (whence $S^3$, thanks to Perelman), then $N$ is not homotopy equivalent to the point. More generally, delete (a small neighborhood of) a wedge of $g$ circles from some 3-manifold $M$; the result has boundary $\Sigma_g$, but if $H_1(M)$ is large enough, you couldn't possibly have $H_1(N) = \Bbb Z^g$.
Actually, if $N$ is homotopy equivalent to a wedge of $g$ circles, then it's homeomorphic to the handlebody with $g$ handles, $H_g$; pre-Perelman, one could prove that it's homeomorphic to $H_g \# Y$, $Y$ some simply connected closed 3-manifold.
