Let $L=L_1 \cup L_2 \cup \cdots \cup L_n $ be a framed link in $ S^3 $. I want to perform the surgery along $L$ to get a new manifold $M$.
By definition, to perform this surgery, I must perform the surgery along each component $L_i$. Why isn't this dependent on the order in which I perform the surgeries?
I also have another question: Let $M$ be the manifold obtained by surgery along $L$. It is known that $M$ bounds a $4$-manifold $X$. Why is $X$ of the form $D^4 \cup 2$-handles?
Here is what I know about this second question. Let $K$ is a knot in $S^3$ and $M$ the manifold obtained by surgery. Let $W$ be the manifold obtained by gluing appropriately a $2$-handle on $ S^3 \times \lbrace 1 \rbrace \subset S^3\times [0,1]$. $W$ is a cobordism from $S^3$ to $M$. Hence $X=W \cup_{S^3} D^4$ is a $4$-manifold with boundary $M$. Is it clear that $X$ is of the form $D^4 \cup$ a $2$-handle? If the answer is yes, does this adapt to the link case?
Edit: I will try to answer my second question. Start from the $4$-ball and glue $2$-handles $D^2 \times D^2$ as follows. To glue the $2$-handle I must specify a homeo $ \partial D^2 \times D^2 $ to some solid torus in $\partial D^4=S^3$. For that, use the tubular neighbourhoods of the link components so that the meridian of $ \partial D^2 \times D^2 $ goes to the curve specified by the framing of the components. In other words, the effect on $S^3=\partial D^4$ of gluing these $2$-handles is precisely to change do the surgery of $S^3$ along $K$. Hence the boundary of this $ D^4 \cup 2$-handles is precisely $M$. This seems like working, so why do textbooks usually bother with this $S^3 \times [0,1]$ construction?