Let $L$ be a framed link of $m$ components and $(L_i,f_i)$ be the component-framing pair. The linking matrix $A=[a_{i,j}]$ where $a_{i,i}=f_i$ and $a_{i,j}=lk(L_i,L_j)$ when $i\ne j$. I have read that this matrix is a presentation matrix for the first homology group $H_1(M; \Bbb Z)$ and that the nullity of $A$ is equal to the first betti number of $M$, where $M$ is the 3-manifold obtained from $S^3$ by surgery on $L$. Can someone explain this fact to me or point me to a proof or a reference?
EDIT: I am reading Nicolaescu's Reidemeister Torsion of 3-manifolds p.77 and there I made some progress but I have some questions as well. Suppose the $c_i$ are the meridians of the solid tori we glue back on the link complement $E$ to obtain $M$. This means that we send $c_i \to f_iμ_i + λ_i$, where $μ_i, λ_i$ are the meridians and longitudes in the regular neighbourhood of $L$. It is clear to me that $μ_i$ form a basis for $H_1(E)$ and taking $K$ to be the free abelian generated by the $c_i$'s and sending them to $f_iμ_i + \sum_{k \ne i} a_{i,j}μ_k$ we have a morphism $K \to H_1(E)$ with matrix the linking matrix mentioned above. There Nicolaescu says that by taking the natural map $a:H_1(E) \to H_1(M)$ we complete the presentation as this map is onto. Here is when it's unclear to me. By $a$ I guess he means the map that sends $μ_i $ in $E$ to $μ_i $ in $M$. If this is the case then I don't understand why it's onto and while it's clear to me that the $f_iμ_i + \sum_{k \ne i} a_{i,j}μ_k$ will be in the kernel because now they bound a disk in $M$ I don't see why they should generate it.
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