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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?

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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.

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  • $\begingroup$ do you think there exists any compact manifold with such properties(what OP initially asked)? $\endgroup$ – Anubhav Mukherjee May 26 '16 at 15:22
  • $\begingroup$ @Anubhav.K With what properties? $\endgroup$ – user98602 May 26 '16 at 15:23
  • $\begingroup$ I still do not understand your question. Can you spell it out for me? $\endgroup$ – user98602 May 26 '16 at 15:28
  • $\begingroup$ actually leave it...what initially I was thinking turns out to be wrong... $\endgroup$ – Anubhav Mukherjee May 26 '16 at 15:30
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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.

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  • $\begingroup$ I think you mean to have $H_2(P) \oplus H_2(N)$ on the right side of that isomorphism. $\endgroup$ – user98602 May 25 '16 at 3:52
  • $\begingroup$ In general, yes, but in this case $H_2(N)=0$ since I am taking $N$ to be the example OP mentioned. $\endgroup$ – Eric Wofsey May 25 '16 at 3:54
  • $\begingroup$ I misread. Thanks for clarifying. $\endgroup$ – user98602 May 25 '16 at 3:55

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