Clarification about the Computation of the Homology of the Connected Sum in degree $n-1$.

Question

There are plenty of questions about the homology of the connected sum of two $n$-manifolds, but I didn't find an explicit explanation of the computation done in degree $n-1$. Let's show some examples of such questions:

1) In this question it is outlined the way to solve it, which is perfectly clear, but I don't know how to prove the result claimed in degree $n-1$.

In the comments it's written that the boundary map $$H_n(M) \to H_{n-1}(B\setminus B') = H_{n-1}(S^{n-1})$$ is an iso, but being a purely algebraic map, I cannot visualise it, and hence I don't know how to prove that it is in fact an isomorphism.

My Idea ($M,N$ orientable): if the map were induced by inclusion we would be OK, because we are mapping the fundamental class of the connected sum in a local orientation, and this is an iso. But I don't know how to formalise it properly. Moreover I don't know how to deal with the boundary map, which is purely algebraic, so not good to handle in general. In the case that one of them is not orientable, I don't know how to proceed.

2) The same problem arises in this question where there is written:

[...] In the case that both are orientable, the above sequence turns into $$0\to \mathbb{Z} \to \mathbb{Z}\oplus\mathbb{Z} \to \mathbb{Z} \to \widetilde{H}_{n-1}(M\# N)\to \widetilde{H}_{n-1}(M\vee N) \to 0$$ as their connected sum is also orientable. From this, we see that $$\widetilde{H}_{n-1}(M\# N)\to \widetilde{H}_{n-1}(M\vee N)$$ must be an isomorphism.

I'm not able to find a reason on why it is true. A similar (mutatis mutandis) result it's stated in the case that only one of them is orientable.

Here you can find the same question, with the additional "hint" that:

the so called bounding sphere is homologous to zero, (which should means that the map $H_n-1(S^n-1)$ into $H_n-1(M-p) \oplus H_n-1(N-p)$ is the zero map) .

But even here without any further references on why this result is true.

I'm starting thinking that this should be a kind of triviality, but I'm facing several problems in finding a proof of it. So can someone explain the isomorphism between $H_{n-1}(M \sharp N)$ and $H_{n-1}(M)\oplus H_{n-1}(N)$ in the cases that both manifold are orientable and only one of them is orientable?

Answer 1

You need to know a couple of facts. Namely, that the homology of a compact manifold is finitely generated (see Homology of a compact manifold is finitely generated) and Corollary 3.28 from Hatcher. Basically this allows us to conclude that $(*)$ for an $n$-dimensional closed manifold $M$ which is either orientable or non-orientable, $H_{n-1}(M) = \mathbb{Z}^k$ or $\mathbb{Z}/2\oplus \mathbb{Z}^k$, respectively.

Case where $M,N$ are both orientable.

We have the sequence $$0 \to \mathbb{Z} \to \mathbb{Z}^2 \overset{\partial}\longrightarrow \mathbb{Z} \to H_{n-1}(M\# N) \to H_{n-1}(M\vee N) \to 0,$$ and we know that $$0 \to \mathbb{Z}/\text{im}\,\partial \to H_{n-1}(M\# N) \to H_{n-1}(M\vee N)\to 0$$ is short exact. But by $(*)$ the $(n-1)$st homology of an $n$-dimensional orientable manifold is free, so that $\mathbb{Z}/\text{im}\,\partial$ is free as well (subgroups of free are free). To see that $\mathbb{Z}/\text{im}\,\partial$ does not have rank 1, we just need to see that $\partial$ is not the zero map. This is trivial because we know $$0\to\mathbb{Z} \to \mathbb{Z}^2 \to \text{im}\,\partial\to0$$ is short exact so that $\text{im}\,\partial \not= 0$.

Case where $M$ is orientable, and $N$ is not.

This time the sequence is $$0 \to \mathbb{Z} \overset{\partial}\longrightarrow \mathbb{Z} \to H_{n-1}(M\# N) \to H_{n-1}(M\vee N) \to 0$$ and we just need to check that $\partial$ is an isomorphism.

Again by $(*)$ we know $$0\to \mathbb{Z}/\text{im}\, \partial \to \mathbb{Z}/2\oplus\mathbb{Z}^m \to \mathbb{Z}/2\oplus\mathbb{Z}^k \to 0$$ is short exact. By injectivity of $\partial$, we know $\mathbb{Z}/\text{im}\,\partial$ is finite. However, the only non-trivial finite subgroup of $\mathbb{Z}/2\oplus\mathbb{Z}^m$ is the $\mathbb{Z}/2$ summand. This can't be the kernel of the right map because $\mathbb{Z}/2\oplus\mathbb{Z}^k$ is not free. So $\mathbb{Z}/\text{im}\,\partial$ must be trivial.

Answer 2

There is a basic theorem for the case of triangulable orientable compact n-manifolds without boundary that the top dimensional Z-homology is Z and is generated by an oriented sum of the n-simplices of the triangulation.

The picture is that each n-simplex shares each of its (n-1)-faces with exactly one other n-simplex and with proper orientation they cancel under the boundary operator. This oriented sum of n-simplices is called the fundamental cycle.

If you remove one of these n-simplices(this corresponds to your n-ball) then its (n-1)- faces do not cancel and so the boundary of the remaining n-simplices is the boundary of the removed n-simplex. This corresponds to your sphere or spherical shell. So the spherical shell is homologous to zero.

If the manifold is not orientatble there is no fundamental cycle and the bounding sphere is not null homologous as the example of the Mobius band shows.

If you choose the triangulation so that the small ball that has been removed lies in the interior of one of the n-simplices, then it is easy to show that the manifold minus the n-simplex is a strong deformation retract of the manifold minus the ball.

• I am not sure how the proof goes if the manifold can not be triangulated.