Gluing 1-handles to 0-handle

Question

In Gompf-Stipsicz p.115, they claim, for a 4-manifold handlebody, that

I wanted to ask how do we show/see that attaching 1-handles to a 0-handle is diffeomorphic to $\natural n S^1\times D^3$?

As far as boundaries go, I understand that the boundary of the end manifold is a connect sum of $S^1\times S^2$'s, since we're doing 0-surgeries on the boundary $S^3$ repeatedly. That is: drill two balls $S^0\times D^3$ out of $S^3$, giving $I\times S^2$, then attach $D^1\times S^2$ to it along $S^2\times\{0,1\}$, giving $S^1\times S^2$. But I'm not sure about what's happening on the trace of this surgery.

Update: If we are only gluing one 1-handle I can see it this way: just like the 1-handle is $D^1\times D^3$, so we can see the 0-handle as a $D^1\times D^3$. Since we are gluing both handles along a common $S^0\times D^3$ in their boundaries, we use the $S^0\times D^3\subset D^4$ to revisualize it as a $D^1\times D^3$ (it doesn't matter much how the rest of the boundary is broken down), and thus $D^4\cup_{h} D^1\times D^3\cong D^1\times D^3\cup_{h:S^0\times D^3\to S^0\times D^3} D^1\times D^3 = S^1\times D^3$.

Answer 1

I think I see an easy way for this. If we're gluing just one 1-handle, we knot we get $S^1\times D^3$. If we are gluing $n$ 1-handles, we can partition $D^4$ into $n$ balls first, so that $D^4 = \natural^n D^4$ (see picture below). In this way, we can either see a $D^4$ with $n$ 1-handles attached, or n copies of $S^1\times D^3$, which are progressively glued to each other along a common $D^3$.

Answer 2

You can prove it in two steps.

Step 1 (easy) : find a Morse function $f:S^1\times D^3\to \mathbb{R}$ with only two critical points with indices $0$ and $1$. Hint:

if $f_i\in Morse(X_i) $ then $f_1 + f_2\in Morse(X_1\times X_2)$

Consequently $S^1\times D^3$ is THE manifold (up to diffeomorphism) obtained by attaching a single 1-handle to a $0$-handle .

Step 2 (tedious): show that if $X_1$ and $X_2$ are two smooth $n$-manifolds with non-empty connected boundary, then the boundary connected sum $X_1\natural X_2$ is diffeomorphic to $(X_1\sqcup X_2)\cup h^1$, i.e. the manifold obtained by attacching a 1-handle to the disjoint union $X_1\sqcup X_2$ in such a way that one foot of the 1-handle lies in $X_1$ and the other foot lies in $X_2$.

At this point we conclude: $$(S^1\times D^3)\natural (S^1\times D^3) \cong \left((S^1\times D^3)\bigsqcup (S^1\times D^3)\right)\bigcup h^1\cong (h^0\cup h^1) \cup (h^0\cup h^1)\cup h^1 $$ $$\cong h^0\cup h^1\cup h^1$$

where in the last equality we have cancelled the last $h^1$ with one of the two 0-handles (since are complementary handles). The same proof clearly works for any number of $1$-handles.