# The $H_{n-1}$ of the connected sum of two closed, non-orientable manifolds

I am attempting to show that the connected sum $$M_1 \# M_2$$ of two closed, non-orientable, connected $$n$$-manifolds has $$H_{n-1} \cong H_{n-1}(M_1)\oplus H_{n-1}(M_2)$$, IF one swaps a $$\mathbb{Z}_2$$ summand for a $$\mathbb{Z}$$. This is the last part of exercise 3.3.6 of Hatcher:

His hint is to use Euler characteristics. But, I do not see how they enter into this problem. I know that $$M_1 \# M_2$$ is non-orientable, thus its $$H_{n-1}$$ has torsion part $$\mathbb{Z}_2$$. I also know that the Euler characteristic of an odd-dimensional, non-orientable manifold is zero. But how can I use the Euler characteristic in this case to say something about the homology?

I am going to assume you already know $$H_k(M_1\sharp M_2;R) \cong H_k(M_1;R)\oplus H_k(M_2;R)$$ for any ring $$R$$ and any $$0 < k < n-1$$. I'm also going to assume all the cohomology groups are finitely generated.
Since Euler characteristic is independent of the field coefficients, let's take $$R = \mathbb{Z}/2\mathbb{Z}$$ and see what happens. I'm going to let $$c_i$$ denote the $$i$$-th Betti number with $$\mathbb{Z}/2\mathbb{Z}$$ coefficients, reserving $$b_i$$ for the usual rational Betti numbers. Note that $$M_1\sharp M_2$$ has Euler characteristic $$\chi(M_1)+\chi(M_2) - (1+(-1)^n)$$. Thus, $$\sum_{k=0}^n (-1)^k c_k(M_1\sharp M_2) = \left(\sum_{k=0}^n (-1)^k c_k(M_1) + (-1)^kc_k(M_2)\right) - (1+(-1)^n).$$
However, $$c_0(M_1\sharp M_2) = c_0(M_1) + c_0(M_2) - 1$$, while $$c_k(M_1\sharp M_2) = c_k(M_1) + c_k(M_2)$$ for $$0 < k < n-1$$. In addition, as you've already noted, $$c_n(M_1\sharp M_2) = c_n(M_1)+c_n(M_2)-1$$. Substituting this into the above, we deduce that $$c_{n-1}(M_1\sharp M_2) = c_{n-1}(M_1) + c_{n-1}(M_2)$$.
Now, think about universal coefficients theorem. Writing $$H_{n-1}(M_1\sharp M_2)$$ as a free part summed with the torsion $$\mathbb{Z}/2\mathbb{Z}$$, we find that $$c_{n-1}(M_1\sharp M_2) = b_{n-1}(M_1\sharp M_2) + 1$$, while $$c_{n-1}(M_i) = b_{n-1}(M_i) + 1$$.
So, we have $$b_{n-1}(M_1\sharp M_2) + 1 = b_{n-1}(M_1) + 1 + b_{n-1}(M_2) + 1$$, so $$b_{n-1}(M_1\sharp M_2) = b_{n-1}(M_1)+b_{n-1}(M_2) + 1.$$ In other words, $$H_{n-1}(M_1\sharp M_2)$$ has an extra $$\mathbb{Z}$$-summand, and, by the result you already mentioned, one less $$\mathbb{Z}/2\mathbb{Z}$$ summand.
• @DescartesBeforetheHorse: Good question! That is definitely a mistake on my part: the formula $c_{n-1} = b_{n-1} + 1$ is only valid if we know that $H_{n-2}$ has no $2$ -torsion. However, the contribution from $Tor(H_{n-2}(M_1\sharp M_2), \mathbb{Z}/2\mathbb{Z})$ cancels with the contribution from $Tor(H_{n-2}(M_1),\mathbb{Z}/2\mathbb{Z})\oplus Tor(H_{n-2}(M_2), \mathbb{Z}/2\mathbb{Z})$ because $Tor$ distributes over direct sums in the first slot and we already know $H_{n-2}(M_1\sharp M_2)\cong H_{n-2}(M_1)\oplus H_{n-2}(M_2)$. Oct 5 at 1:40