Given two disjoint connected \(n\)-manifolds \(M_1\) and \(M_2\), a connected \(n\)-manifold \(M_1 \sharp M_2\), 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 \(M_i\).)
Show that if \(M_1\) and \(M_2\) are closed then there are isomorphisms \(H_i(M_1 \sharp M_2; \mathbb{Z}) \approx H_i(M_1; \mathbb{Z}) \oplus H_i(M_2; \mathbb{Z})\) for \(0 < i < n\), with one exception: If both \(M_1\) and \(M_2\) are nonorientable, then \(H_{n-1}(M_1 \sharp M_2; \mathbb{Z})\) 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.
Show that \(\chi(M_1 \sharp M_2) = \chi(M_1) + \chi(M_2) - \chi(S^n)\) if \(M_1\) and \(M_2\) are closed.
Solution
Let \[ N_i=M_i-\operatorname{int}B_i, \qquad M=M_1\#M_2=N_1\cup_{S^{n-1}}N_2. \] For \(0<i<n-1\), removing an open \(n\)-ball does not change homology in degree \(i\), and \(H_i(S^{n-1})=0\). Mayer–Vietoris therefore gives \[ H_i(M)\cong H_i(M_1)\oplus H_i(M_2) \qquad(0<i<n-1). \]
It remains to analyze degree \(n-1\). Excision gives \[ H_j(M_i,N_i)\cong H_j(D^n,S^{n-1}), \] so the long exact sequence of \((M_i,N_i)\) contains \[ 0\longrightarrow H_n(M_i)\longrightarrow \mathbb Z \xrightarrow{\partial_i}H_{n-1}(N_i) \longrightarrow H_{n-1}(M_i)\longrightarrow0. \] If \(M_i\) is orientable, \(H_n(M_i)\cong\mathbb Z\) and the first map is an isomorphism, so \[ H_{n-1}(N_i)\cong H_{n-1}(M_i) \] and the boundary sphere represents zero in \(H_{n-1}(N_i)\).
If \(M_i\) is nonorientable, \(H_n(M_i)=0\). Corollary 3.28 gives a distinguished \(\mathbb Z_2\) torsion summand in \(H_{n-1}(M_i)\). On the other hand Lefschetz duality for the compact manifold with boundary \(N_i\) gives \[ H_{n-1}(N_i;\mathbb Z)\cong H^1(N_i,\partial N_i;\mathbb Z), \] and the universal coefficient theorem shows that the group on the right is torsion-free. Hence the extension above cannot split on the \(\mathbb Z_2\) summand; there it is necessarily \[ 0\to\mathbb Z\xrightarrow{\times2}\mathbb Z\to\mathbb Z_2\to0. \] The free part is unchanged. Thus \(H_{n-1}(N_i)\) is obtained from \(H_{n-1}(M_i)\) by replacing its \(\mathbb Z_2\) summand by \(\mathbb Z\), and the boundary sphere class is twice the new generator.
Now use Mayer–Vietoris for \(M=N_1\cup N_2\). Since the intersection is \(S^{n-1}\), the relevant quotient is \[ H_{n-1}(M) \cong \frac{H_{n-1}(N_1)\oplus H_{n-1}(N_2)} {\langle([S],-[S])\rangle}, \] with the usual harmless choice of sign.
If both \(M_i\) are orientable, both boundary classes vanish, and the usual top-dimensional Mayer–Vietoris term supplies the expected direct sum in degree \(n-1\). If exactly one, say \(M_2\), is nonorientable, the relation is \(2z_2=0\) on its new \(\mathbb Z\) summand, restoring precisely the original \(\mathbb Z_2\) summand. Thus again \[ H_{n-1}(M)\cong H_{n-1}(M_1)\oplus H_{n-1}(M_2). \]
If both are nonorientable, the two \(\mathbb Z_2\) summands have first been replaced by \(\mathbb Z z_1\oplus\mathbb Z z_2\), and Mayer–Vietoris imposes the single relation \[ 2z_1-2z_2=0. \] Hence \[ (\mathbb Z z_1\oplus\mathbb Z z_2)/\langle2z_1-2z_2\rangle \cong \mathbb Z\oplus\mathbb Z_2. \] Thus, compared with \[ H_{n-1}(M_1)\oplus H_{n-1}(M_2), \] one of the two \(\mathbb Z_2\) summands is replaced by a \(\mathbb Z\) summand, exactly as claimed.
For Euler characteristic, write again \(M=N_1\cup N_2\) with intersection \(S^{n-1}\). Removing the interior of one \(n\)-cell changes Euler characteristic by \(-(-1)^n\), so \[ \chi(N_i)=\chi(M_i)-(-1)^n. \] Therefore \[ \begin{aligned} \chi(M) &=\chi(N_1)+\chi(N_2)-\chi(S^{n-1})\\ &=\chi(M_1)+\chi(M_2)-2(-1)^n-(1+(-1)^{n-1})\\ &=\chi(M_1)+\chi(M_2)-(1+(-1)^n)\\ &=\boxed{\chi(M_1)+\chi(M_2)-\chi(S^n)}. \end{aligned} \]