# Drawable 3-manifolds

For an embedded compact 3-manifold $M \subseteq \mathbb{R}^3$, $H_1(M)=0$ implies $\pi_1(M)=0$.

1. Notice that the boundary of $M$ is a disjoint union of compact, orientable surfaces. Call them $T_i$ and thus $\partial M = \coprod T_i$. It suffices to show that: $H_1(M) = 0$ iff $\partial M$ contains only 2-spheres iff $\pi_1(M) = 0$.
Let us prove that if the boundary does not contain only 2-spheres then the homology, and thus also the fundamental group, are non-trivial. For this look at $M$ embedded in $\mathbb{S}^3$. Call $A$ to the complement of $M$ in $\mathbb{S}^3$ and enlarge $M$ to an open neighborhood that retracts to $M$, call this neighborhood $M'$. Thus $M' \cap A$ is homotopically equivalent to $\partial M$. Now using the Alexander duality with coefficients in $\mathbb{Q}$ and the universal coefficients theorem we deduce $H_1(A) \simeq H_1(M) = H_1(M')$.
Doing Mayer-Vietoris with $M'$ and $A$ we get an exact sequence: $\cdots \to 0 \to H_1(\partial M) \to H_1(A) \oplus H_1(M') \to \cdots$ (The zero comes form the fact that the union of $M'$ and $A$ is $\mathbb{S}^3$, which has trivial second homology). And thus if $\partial M$ contains some surface of positive genus $H_1(M)$ cannot be trivial (because $H_1(A)$ is isomorphic to $H_1(M')$).
For the other implication it suffices to show that if all the components of the boundary are 2-spheres then both $H_1(M)$ and $\pi_1(M)$ are trivial. For this we use a strong result that says that a *smoothly* embedded 2-sphere in the 3-sphere divides the 3-sphere in two 3-balls (an instance of the generalized Schoenflies theorem). Now by induction on the number of connected components of the boundary of $M$ we can “fill” each 2-sphere to a 3-ball, which is just adjoining a 3-cell, and thus the $H_1$ and the $\pi_1$ remain invariant untill the end, where we get the 3-sphere. This means that both the first homology and the fundamental group were trivial.