Cancelling Handle Attachments Let $(W, \partial W)$ be an $n$-dimensional manifold with boundary. Suppose that $(W', \partial W')$ is obtained from $(W, \partial W)$ by attaching a $k$-handle via an embedding $\phi: S^{k-1}\times D^{n-k} \hookrightarrow \partial W$ and suppose that $(W'', \partial W'')$ is obtained from $(W', \partial W')$ by attaching a $(k+1)$-handle via an embedding $\psi: S^{k}\times D^{n-k-1} \hookrightarrow \partial W'$. If the belt sphere of the $k$-handle intersects the attaching sphere of the $(k+1)$-handle transversely in a single point then the handles cancel.
Does the result still hold if the second handle attachment is of index $(k-1)$ instead of index $(k+1)$? 
 A: If I understood you correctly, you can always choose to attach handles in increasing order, i.e., start by attaching 0-handles, then move on to 1-handles, etc. until you attach the top k-handles (if any). Then the condition you stated, i.e., that $\partial h^k=(+/-1)h^{k-1}$, i.e., the algebraic (instead of geometric) intersection  is $+/-1$ (and $h^k$ has algebraic intersection number $0$ with the belt spheres of all the other (k-1)-handles that you have attached) is necessary for canceling. This means $h^k$ goes algebraically * . STILL, you must be able to actually find a way of finding a way to(having an embedding so that you)have the attaching sphere intersect exactly once  the k-handle $h^k$ .
This is taken from my understanding and from a few different sources.


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*This ultimately means that there is an embedding of the handle which has these properties, and not that ( the case of geometric intersection) that the intersection can be avoided through a different choice of embedding.  

