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I don't know where you took the picture from, but that link is not symmetric (and it likely does not describe the correct manifold). Indeed, the 0-framed component of the link you drew is a trefoil: i …

Yes, there is a simple way. Below is a sequence of pictures illustrating the procedure (created using Kirby calculator). $5_2$: Blowup at the clasp: Isotopy: Blowdown the purple unknot:

There are many examples of the sort, in effect. As far as I know, Akbulut and Kirby (Mazur manifolds, Michigan Math. J. 26 (1979)) proved that $\Sigma(2,5,7)$, $\Sigma(3,4,5)$, and $\Sigma(2,3,13)$ bo …

When the decomposition of $X$ has no 3-handles, this is often feasible. The trick is to turn the handle decomposition of $X$ upside down (see Example 5.5.5 in Gompf and Stipsicz's 4-manifolds and Kirb …

3-handles do matter. The reason why they're not usually there is because people often care about closed 4-manifolds; in this case, there is an essentially unique way of attaching all 3-handles to the …

I guess that this is as explicit and low-tech as it gets: if $X$ is a K3 surface (i.e. a non-singular quartic hypersurface in $\mathbb{CP}^3$, with the complex orientation), then $X \# \overline{\math …

Yes, you can always do this. In fact, I think you can do it for any framed link, regardless of the framings, the number of components, and the slice assumption. (For simplicity, I will write the answe …