Let $L$ be a framed link in an integral homology 3-sphere $M$. I read in this paper that if $L$ is algebraically split (pairwise linking number is zero) and unit-framed (framing $\pm 1$), then by doing Dehn surgery along $L$ again yields an integral homology sphere $M_L$, the author says it's easy to confirm the result. Does this follow from a direct computation of $H_1$ of the resulting manifold and use Poincaire duality to say $H_2=H_1=0$, or is there a more general formula on homology group of the manifold obtained from the Dehn surgery, so that we can easily deduce that the first homology is unchanged under $\pm 1$ surgery in this case?
Conversely, If the resulting manifolds $M_{L'}$ by doing surgery along sublink $L'\subset L$ are homology spheres, then $L$ must be algebraically split and unit-framed. I donnot understand why $M_K$ for a componet knot $K\subset L$ is homology sphere implies $K$ is unit framed? And why $M_{K_i\cup K_j}$ is homology sphere implies the linking number of $K_i,K_j$ is $0$?
(I know little about surgery theory, before this I only read Adams' knot book and I didn't find any theorem indicating this.)
Any reference is appreciated!