For handle decomposition of surface, suppose I have a twisted 1-handle(twisted only once) adjacent to an isolated pair of linked handles, is the handle slide operation enough to move the previous one to become 3 twisted 1-handle(each is the same as the previous twisted 1-handle), which proves that these two are homeomorphic to each other?



Handle slides homeomorphism
I'm too lazy to draw this right now, but I think I can describe it anyway. Consider a twisted band attached right next to a dual pair of untwisted bands. I'll write this $A\bar{A}BCBC$. Here the overbar represents a twist and the bands attach to the same letters on each end. Now slide the B band into the middle of the A band to get $AB\bar{A}C\bar{B}C$. Slide the left foot of the $C$ band over the B band into the middle of the A band to get $ABC\bar{A}\bar{C}\bar{B}$. Now slide the right end of the $A$ band along the connected boundary until it is adjacent to the left end of the $A$ band. It slides once over the $B$ and $C$ bands, so picks up two half-twists. So we get $A\bar{A}BC\bar{C}\bar{B}$. Now take the entire $C$ band and slide it along the connected boundary until it lies next to the $B$ band on the right, yielding the desired $A\bar{A} B\bar{B} C\bar{C}$. Ta da!


