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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:

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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?

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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.

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  • $\begingroup$ One question. How are we using the UCT where you mention? $\endgroup$ Oct 5 at 1:30
  • $\begingroup$ @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 $H_{n-2}(M_1\sharp M_2)\cong H_{n-2}(M_1)\oplus H_{n-2}(M_2)$. $\endgroup$ Oct 5 at 1:40

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