Drawable 3-manifolds

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

Drawable 3-manifolds

One thought on “Drawable 3-manifolds

  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.

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