the framing corresponds to the standard choice, and $fr(J)=1$ if not.

Similarly, if $|a|=|b|+1$ and $A\in {\mathcal{M}}(a,b)$ is a point, we write $\varepsilon_A\in {\mathbb{Z}}/2{\mathbb{Z}}$ for the framing sign of this point.

\[prop:framechanges\] Let ${\mathscr{C}}$ be a framed flow category containing two objects $x,y$ with the same grading, and let ${\mathscr{C}}_S$ be the framed flow category obtained from ${\mathscr{C}}$ by sliding $x$ over $y$. Let $a,b$ be objects in ${\mathscr{C}}$ with $|a|=|b|+2$.

1.  If $|a|=|x|+2$, then ${\mathcal{M}}(a',b')={\mathcal{M}}(a,b)$ for all $b\not=y$ with the same framing. Furthermore $${\mathcal{M}}(a',y') = {\mathcal{M}}(a,y) \sqcup {\mathcal{M}}(a,x)$$ with components from ${\mathcal{M}}(a,y)$ identically framed. If $J\subset {\mathcal{M}}(a,x)$ is an interval component, denote $J'$ for the same component when viewed as a subset of ${\mathcal{M}}(a',y')$. Then

    1.  If ${\mathscr{C}}_S$ is obtained by a $(-)$–handle slide, then $$fr(J')= fr(J).$$

    2.  If ${\mathscr{C}}_S$ is obtained by a $(+)$–handle slide, then $$fr(J')= 1 + fr(J).$$

2.  If $|a|=|x|+1$, then $${\mathcal{M}}(a',b') = {\mathcal{M}}(a,b) \sqcup {\mathcal{M}}(y,b)\times [0,1] \times {\mathcal{M}}(a,x)$$ and every component of ${\mathcal{M}}(a,b)$ is framed the same way. For each pair of points $(B,A)\in {\mathcal{M}}(y,b)\times {\mathcal{M}}(a,x)$, denote $I_{B,A}$ the corresponding interval component in ${\mathcal{M}}(a',b')$.

    1.  If ${\mathscr{C}}_S$ is obtained by a $(-)$–handle slide, then $$fr(I_{B,A})=\varepsilon_B.$$

    2.  If ${\mathscr{C}}_S$ is obtained by a $(+)$–handle slide, 