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I'm reading R.E. Gompf and A.I. Stipsicz, 4-Manifolds and Kirby Calculus. There is something I don't understand on page 116 (Google Books link to page 116; alternatively, here are images of page 115 and page 116)

Now, we consider compact 4-manifolds, and they have handle decomposition. 0-handle and $m$ 1-handles form $X_{1}$, which is diffeomorphic to $\natural m S^1 \times D^3$. Let $X_{2}$ be $X_{1}\cup $ 2-handles. I understand those facts, but I can't understand $$\partial X_{2}=\partial(\natural m S^1 \times D^3).$$ If you can help me, then please teach me the reason for this.

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Sorry, I edited the naturals to sharps, but then I realized Gompf and Stipsicz used naturals! So I reverted to Zev Chonoles's edit. –  Grumpy Parsnip Jun 23 '12 at 18:34
    
Apparently I earned the "cleanup" badge for cleaning up my own mistake. :) –  Grumpy Parsnip Jun 23 '12 at 19:38

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The idea is to use duality to notice that the union of $3$ and $4$ handles must be isomorphic to some handlebody $\natural nS^1\times D^3$. (Here $m$ doesn't have to equal $n$. If you look at the book carefully, it uses both $n$ and $m$.) Recall that a $k$-handle can be thought of as an $(n-k)$-handle if you read the decomposition backward. (This is the dual decomposition.) Thus the $3$ and $4$ handles represent $0$ and $1$-handles in the dual decomposition. The union of the rest of the handles must attach to the boundary of this, so their boundary must equal $\partial(\natural nS^1\times D^3)$.

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Okay,I see that,but I can't understand that 3-,4-handles and 0-,1-,2-handles must have the same boundary.Why? –  Takahiro Oba Jun 24 '12 at 1:12
    
Because you are gluing them together to get a closed manifold. –  Grumpy Parsnip Jun 24 '12 at 3:27
    
Oh,I see.Thanks:-) –  Takahiro Oba Jun 24 '12 at 11:11

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