Collapsing the boundary of a handlebody While trying to compute the homology of a handlebody, I started wondering what space is obtained if one collapses the boundary of a handlebody of genus g (therefore a surface of genus g) to a point. My suspicion is that the resulting space is a 3-ball but I do not know how to prove this.
 A: Suppose $H_g$ is a region of $\mathbb{R}^3$ bounded by an embedding of $\Sigma_g$, the surface of genus $g$, and let $X_g = H_g/\Sigma_g$. Like I said in my comment, this space is homeomorphic to $H_g \cup_{\Sigma_g} C\Sigma_g$ (where $C$ denotes the cone), and there is a non-trivial result which states that if $M$ is a manifold then $CM$ is a manifold iff $M$ is a sphere. Therefore $X_g$ is a manifold iff $g=0$ and in fact $X_0 \cong S^3$, but if $g>0$ then for local point-set reasons it cannot be homeomorphic to any manifold.
But we can also see $X_g$ is not even homotopy equivalent to the $3$-ball by computing homology groups. There is a result (in Hatcher for example) which states that if $(X, A)$ is a pair of spaces such that $A$ is the deformation retract of some open neighbourhood $U\subset X$ then $H_*(X, A) \cong \tilde{H}_*(X/A)$, so we can use this to compute the homology of $X_g$. Consider the long exact sequence for the pair $(H_g, \Sigma_g$):
$$ \dots 0 \to H_3(H_g, \Sigma_g) \to H_2(\Sigma_g) \to H_2(H_g) \to  $$
$$H_2(H_g, \Sigma_g) \to H_1(\Sigma_g) \to H_1(H_g) \to $$
$$ H_1(H_g, \Sigma_g) \to H_0(\Sigma_g) \to H_0(H_g) \to 0.$$
The groups $H_*(\Sigma_g)$ are well-known (and I'm assuming you've seen the computation already); note that $H_g\simeq \vee^g S^1$, a wedge-sum of $g$ circles, so $H_*(H_g)\cong H_*(\vee^g S^1)$; and finally note that the map $H_0(\Sigma_g) \to H_0(H_g)$ is an isomorphism. Plugging in the information we know, we get
$$0 \to H_3(X_g) \to \mathbb{Z} \to 0 \to H_2(X_g) \to \mathbb{Z}^{2g} \to \mathbb{Z}^g \to H_1(X_g) \to 0. $$
Immediately we see $H_3(X_g)\cong \mathbb{Z}$, so in particular $X_g$ cannot be homotopy-equivalent to the $3$-ball. If $g>0$ it also doesn't have the same homology as the $3$-sphere since $H_2(X_g) \cong \mathbb{Z}^g$ (to study the map $\mathbb{Z}^{2g} \to \mathbb{Z}^g$, look at what happens to the circles generating $H_1(\Sigma_g)$ when you include them into $H_g$). Then finally $H_1(X_g)=0$ since $\mathbb{Z}^{2g} \to \mathbb{Z}^g$ is surjective.

With a little bit more work you can show that in fact that when $g>0$ the homology and cohomology of $X_g$ fail to satisfy Poincare Duality, and therefore it cannot be homotopy-equivalent to any closed manifold.
