The exercise on Hatcher's Algebraic Topology goes:
6.$~$Given two disjoint connected $n$-manifolds $M_1$ and $M_2$, a connected $n$-manifold $M_1\# M_2$, their connected sum, can be constructed by deleting the interiors of closed $n$-balls $B_1\subset M_1$ and $B_2\subset M_2$ and identifying the resulting boundary spheres $\partial B_1$ and $\partial B_2$ via some homeomorphism between them. (Assume that each $B_i$ embeds nicely in a larger ball in $M_i$.)
(a) Show that if $M_1$ and $M_2$ are closed then there are isomorphisms $H_i(M_1\# M_2;\mathbb{Z})\approx H_i(M_1;\mathbb{Z})\oplus H_i(M_2;\mathbb{Z})$ for $0<i<n$, with one exception: If both $M_1$ and $M_2$ are nonorientable, then $H_{n-1}(M_1\# M_2;\mathbb{Z})$ is obtained from $H_{n-1}(M_1;\mathbb{Z})\oplus H_{n-1}(M_2;\mathbb{Z})$ by replacing one of the two $\mathbb{Z}_2$ summands by a $\mathbb{Z}$ summand. [Euler characteristics may help in the exceptional case.]
(b) Show that $\chi(M_1\# M_2)=\chi(M_1)+\chi(M_2)-\chi(S^n)$ if $M_1$ and $M_2$ are closed.
I'm having some trouble about (a) on proving the case where $M_1$ and $M_2$ are both nonorientable. My attempt is to consider the long exact sequence of the pair $(M_1\# M_2, S^{n-1})$, and find that we have exact sequence $$0\to H_{n-1}(S^{n-1})\to H_{n-1}(M_1\# M_2)\to H_{n-1}(M_1\# M_2,S^{n-1})\to 0$$ We have $H_{n-1}(S^{n-1})=\mathbb{Z}$ and $H_{n-1}(M_1\# M_2,S^{n-1})=H_{n-1}(M_1)\oplus H_{n-1}(M_2)$. The exercise suggests that the map $H_{n-1}(S^{n-1})\to H_{n-1}(M_1\# M_2)$ is multiplication by $2$ into a summand $\mathbb{Z}$ of $H_{n-1}(M_1\# M_2)$ that is mapped by the quotient $\mathbb{Z}\to \mathbb{Z}_2$ onto one of the two $\mathbb{Z}_2$ summands of $H_{n-1}(M_1)\oplus H_{n-1}(M_2)$, but where does this conclusion comes from?
Also, the description about "with one exception" confused me. Does it mean that if one of $M_1$ and $M_2$ is orientable, then we still have $H_{n-1}(M_1\#M_2;\mathbb{Z})\cong H_{n-1}(M_1;\mathbb{Z})\oplus H_{n-1}(M_2;\mathbb{Z})$？ Using the long exact sequence of pair $(M_1\# M_2, S^{n-1})$ , we need to show that the map $H_n(M_1\# M_2,S^{n-1})\to H_{n-1}(S^{n-1})$ is surjective, but this map is a puzzle to me.
Showing the surjectivity of $H_n(M_1\# M_2,S^{n-1})\to H_{n-1}(S^{n-1})$ seems to be necessary: With (a) proved, the homology groups of dimension less than $n$ are clear to us. But in order to prove (b), to calculate the Euler characteristic we need to look at the homology groups of dimension $n$. If both $M_1$ and $M_2$ are non-orientable, then $H_n(M_1\# M_2)=H_n(M_1)\oplus H_n(M_2)=0$ by the LES, hence $$\chi(M_1\# M_2)=\chi(M_1)+\chi (M_2)-1+(-1)^{n-1}=\chi(M_1)+\chi (M_2)-(1+(-1)^n)=\chi(M_1)+\chi (M_2)-\chi(S^n)$$ But if one of $M_1$ and $M_2$ is orientable, the homology group $H_n(M_1\# M_2)$ is again a puzzle, unless we can show that the map $H_n(M_1\# M_2,S^{n-1})\to H_{n-1}(S^{n-1})$ is surjective, after which the equation follows.
To summarize, there remains two questions:
- Show that the maps in the exact sequence $$0\to H_{n-1}(S^{n-1})\to H_{n-1}(M_1\# M_2)\to H_{n-1}(M_1\# M_2,S^{n-1})\to 0$$ behave as described.
- Show that the map $H_n(M_1\# M_2,S^{n-1})\to H_{n-1}(S^{n-1})$ in the LES is surjective.
I'm sorry if I'm having too many questions. Thanks in advance.
EDIT: I think I've came up with a solution to the first question: Since $M_1\# M_2$ is also an $n$-manifold, we can discuss the cases if it is orientable or not. If $M_1\# M_2$ is orientable, then the torsion of $H_{n-1}(M_1\# M_2)$ is trivial, hence $H_{n-1}(M_1\# M_2)$ is a direct sum of $\mathbb{Z}$'s, i.e., it is free. Modulo the image of the map $H_{n-1}(S^{n-1})=\mathbb{Z}\to H_{n-1}(M_1\# M_2)$ is only able to produce at most one summand of $\mathbb{Z}_2$, contradicting to that the result of the quotient is isomorphic to $H_{n-1}(M_1\# M_2,S^{n-1})=H_{n-1}(M_1)\oplus H_{n-1}(M_2)$, which has two summands of $\mathbb{Z}_2$. Therefore $M_1\# M_2$ is non-orientable, hence $H_{n-1}(M_1\# M_2)$ has torsion $\mathbb{Z}_2$, and a similar argument gives our desired result.
Still I don't have a clue to the second question.
