I am working on ALgebraic Topology in my past Quals.
"Prove that no closed orientable 3-manifold is homotopy
equivalent to $S_g\vee S^3$, where $S_g$ is the orientable
surface of genus $g\geq 0$"
My plan is prove that some homotopy equivalent invariant of these two spaces are not equal.
With wedge sum, I believe that $\pi_1(S_g\vee S^3)=\pi_1(S_g) * \pi_1(S^3)$ and we know both. I don't have a rigorous proof for this, what I think is that I take $U=S_g \cup $some nbd of $S^3$ that deformation retracts to the intersection point. And similarly I can construct $V$ s.t. $S_g\vee S^3=U\cup V$ and $U\cap V$ retracts to a point. Is this right?
I also think of using homology, but I'm not sure about the formula of $H_n(S_g\vee S^3)$.
Either way, I am stuck at the closed orientable manifold. I'm reading Hatcher, and some results like Propositiion 3.25, compute the homology of a closed orientable CONNECTED manifold. So there is no result here that can help with this problem.
Am I approaching it right? What is the right way to solve this?
Best Answer
Though the other answer mentions the key point (Poincare duality), there's a slick, quick, and instructive argument to be made here that deserves to be spelled out.
A consequence of Poincare duality is that the Euler characteristic (alternating sum of ranks of homologies) of any odd-dimensional closed, orientable manifold is $0$. However, using how homology and wedge sum interact, we can compute $$H_3(S^3 ∨ S_g) \cong \mathbb{Z}, \quad H_2(S^3 ∨ S_g) \cong \mathbb{Z}, \quad H_1(S^3 ∨S_g) \cong \mathbb{Z}^{2g}, \quad H_0(S^3∨S_g) \cong \mathbb{Z}, $$ so that $\chi(S^3∨S_g) = 2g -1$, which is odd hence nonzero. This obstructs a homotopy equivalence between $S^3 ∨ S_g$ and any closed orientable $3$-manifold.