Handle slides homeomorphism

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?

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Let $P$ be the real projective plane and $T$ the two-torus. Let $\#$ denote connect sum. You are asking how one proves that $P \# T$ is homeomorphic $P \# P \# P$, correct? ---- If that is the case you can find the answer in Jeff Weeks book "Shape of space". However it might be more useful to you to start drawing pictures of what handle slides do to the arrangement. – Sam Nead Apr 20 '11 at 8:55
Yes, the handle slide operation is enough to show this. – Jim Conant Apr 20 '11 at 18:35
Any reference I can look at? There are too few materials online about this topic. – Hu Yi Chen Apr 20 '11 at 19:24

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!