I am currently working through Scorpan's Wild World of 4-Manifolds specifically the section on Casson Handles. On page 78, he says if $D$ is the core of the handle after $n$ stages we may assume $\pi_1(M\setminus D)$ is a perfect group. Why? Here $M$ is a smooth closed simply-connected 4-manifold.
Clearly this says $H_1(M\setminus D) = 0$ by the Hurewicz theorem, and thus every loop in the complement bounds a surface in the complement. If a loop is contractible in $M$ then the graph of its contraction homotopy function will be a disk in $M$ which we can perturb to be immersed. If that disk does not intersect the partial Casson Handle, $D$, then it will remain in the complement, $M\setminus D$, and we can contract the loop to a point along it. Thus every non-trivial element of $\pi_1(M\setminus D)$ must bound an immersed disk in $M$ which has non-null intersection with $D$. I don't understand how removing $D$ and therefore removing a portion of that immersed disk does not create any 1-homology. If we can perturb the immersed disk so it does not intersect $D$ then we can contract along it and we get a trivial element of $\pi_1(M\setminus D)$, but if we can't perturb it off $D$ then we can shrink the loop down to the boundary of the intersection where it no longer bounds a surface in $M\setminus D$ and must therefore contribute to $H_1(M\setminus D)$. This would indicate either $\pi_1(M\setminus D) = 0$ or $H_1(M\setminus D) \ne 0$.
What am I missing here?