What is the homotopy type of $S^3 \times S^3$ - 4 points?

Question

What is the homotopy type of $S^3 \times S^3$ - 4 points? Is there a compact CW-complex of $S^3 \times S^3$ - 4 points, which is a deformation retract of $S^3 \times S^3$ - 4 points and has dimension < 6?

Answer 1

The homology of $X=S^3 \times S^3- \{1,2,3,4\}$ is nontrivial only for $H_3(X)=\mathbb{Z}\oplus\mathbb{Z}$ coming from the wedge of 3-spheres and $H_5(X)=\mathbb{Z}^3$ coming from the spheres enclosing each of the first three points (this can also be seen with Poincare duality for manifolds with boundary). The sphere enclosing the fourth point is homologous to a combination of the first three.

Let's assume $\{1,2,3\}$ are chosen near the wedge of 3-spheres. Then we may take it so that each sphere enclosing one of the $\{1,2,3\}$ touches the wedge exactly once. Let $A$ be the union of the wedge of the two 3-spheres and the three 5-spheres.

The inclusion $A \rightarrow X$ is clearly a homology equivalence, so it is a homotopy equivalence since $A$ and $X$ are simply connected. Since $X$ can be given a CW structure with $A$ a subcomplex, we can apply the result that if the inclusion of a subcomplex is a homotopy equivalence, the subcomplex is a deformation retract.

Hence, $Y$ is what you're looking for. It is homotopy equivalent to a bouquet of two 3-spheres and three 5-spheres. This is the lowest dimensions CW complex possible, since the homology of $X$ is nontrivial in dimension 5.

I recommend reproducing this argument explicitly for $S^1 \times S^1$ and removing two points.