Proof. We first show that vanishes on the image of , that is,
Indeed, from the composition law (5) we see that the left hand side is associated to the surface cobordism and the right hand side to . However, inside the 3-handle , the sphere gets filled with a core , and therefore and are isotopic rel boundary. It follows that the two cobordism maps are the same.
Therefore, factors through a map
We need to prove that is bijective. For this, we construct its inverse . Given a lasagna filling of with boundary , observe that the cocore of the 3-handle is one-dimensional, and therefore we can isotope to be disjoint from this cocore; after this, we can push it into , to obtain a lasagna filling there, called , with boundary . We set
To see that is well-defined, we need to check that if two lasagna fillings and are equivalent in , then the corresponding fillings and differ (up to equivalences in ) by an element of . We use Lemma 2.1, in which we fix balls away from the 3-handle, and consider the equivalences listed in the lemma (with the ball replacements happening in ). Then, the equivalences in give rise to equivalences in , with one exception: an isotopy of the surfaces may intersect the one-dimensional cocore of (which is an interval). Generically, this happens in a finite set of points, each point at a different time during the isotopy. Every time the isotopy meets the cocore, the corresponding surfaces in differ by replacing a hemisphere of (with boundary some closed curve ) with its complement in . Up to an isotopy supported near , we can assume that is the equator with its chosen orientation. (For example, if is with the opposite orientation, we can rotate it by about a transverse axis to get with the original orientation.) Then, the hemispheres being interchanged are and and hence the classes of and differ by an element in the image of .
This shows that is well-defined, and its definition makes it clear that it is an inverse to . It follows that is bijective, and the conclusions follow. ∎