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Let $w(M)$ be the total Stiefel-Whitney classes of a manifold $M$. If we know $w(M)$ and $w(N)$, can we obtain $w(M\# N)$? The following simple relation: $$ w(M \# N) = w(M) + w(N) . $$ seems not right, at lest for the leading term 1. But for $w_1$, it seems right.

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According to the answer below, we have a simple result: $$ w(M \# N) = w(M) + w(N) -1. $$

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This is true except in bottom degree.

1) Let $M^\circ = M \setminus \{pt\}$. Then the inclusion induces an isomorphism $H^*(M^\circ) \cong H^*(M)$ in degrees $* < n$. Because $TM^\circ = TM\big|_{M^\circ}$, we get that $w_k(M^\circ) = w_k(M)$ for $k<n$. By Mayer-Vietoris one gets an isomorphism $H^k(M \# N) \cong H^k(M^\circ) \oplus H^k(N^\circ)$ in degrees less than $n$. Because again $TM^\circ = T(M \# N)\big|_{M^\circ}$ and similarly with $N$, we see that $w_k(M^\circ)$ is the projection of $w_k(M \# N)$ in the above isomorphism. This is precisely what it means to say that $w_k(M \# N) = w_k(M) + w_k(N)$.

2) In top degree, $w_n(M) = \chi(M) \mod 2$, so $$\begin{align}w_n(M \# N) &= \chi(M \# N) \mod 2\\&= \chi(M^\circ) + \chi(N^\circ) - \chi(S^{n-1}) \mod 2\\ &= \chi(M) - 1 + \chi(N) - 1 - (1+ (-1)^{n-2})\mod 2\\&= \chi(M)+\chi(N) \mod 2\\&=w_n(M)+w_n(N).\end{align}$$ This could be obtained cohomologically as in (1) by being more careful about the Mayer-Vietoris sequence in top degree.

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  • $\begingroup$ Are you assuming that $M$ and $N$ are closed? If not, how does one make sense of the argument for $w_n$? $\endgroup$ – Michael Albanese Dec 31 '16 at 8:59
  • $\begingroup$ Yes, I am. Of course, if they're not closed, that class is automatically zero (since the cohomology group it lives in is zero). $\endgroup$ – user98602 Dec 31 '16 at 9:00
  • $\begingroup$ Yeah, I just realized that, thanks. Great answer by the way. $\endgroup$ – Michael Albanese Dec 31 '16 at 9:08
  • $\begingroup$ Weird. It doesn't hold if one of them is closed and the other isn't, e.g. $\mathbb{RP}^2\#\mathbb{R}^2$. $\endgroup$ – Michael Albanese Dec 31 '16 at 9:18
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    $\begingroup$ I know it makes sense, we just don't have $w_2(\mathbb{RP}^2) + w_2(\mathbb{R}^2) = w_2(\mathbb{RP}^2\#\mathbb{R}^2)$ - one side is zero, the other isn't. Of course, that's because equality is not literal here, it just being used to mean 'corresponds to' under a certain map. $\endgroup$ – Michael Albanese Jan 1 '17 at 12:35

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