I am reading Ranicki's Algebraic and Geometric Surgery, and I believe I'm starting to understand the surgery obstructions, but I'd like a little clarification.
It's my understanding that when we have a (k-1)-connected normal map $M \rightarrow X$, where $M$ is a 2k-dimensional manifold and $X$ is a 2k-dimensional Poincare complex, then the homology kernel (the homology of map on universal covers) has a quadratic form given by counting intersections "equivariantly".
I believe it is true that taking the connect sum of this map with $S^k \times S^k$ has the effect of direct summing a hyperbolic form on the homology kernel, which again, to my understanding, is that a hyperbolic form is a 2x2 matrix with 0's on the diagonal and a $1$ in the top right and a $\pm1$ in the bottom left, depending on the parity of $k$.
Now the even dimensional surgery obstruction is the equivalence class of this quadratic form after I stabilize with respect to adding on a hyperbolic form. Now surgery theory tells us that we are zero in this group precisely when we are normal bordant to a homotopy equivalence.
$\bf Question:$ Is it correct to deduce that in this even dimensional case that normal bordant to a homotopy equivalence is the same as stably a homotopy equivalence (by this I mean a homotopy equivalence after connect summing with enough $S^k \times S^k$)? And is there an analogous geometric interpretation of the surgery obstruction in the odd dimensional case?