homotopy equivalences $$\mathcal{X}({\mathscr{C}})\simeq\mathcal{X}({\overline}{{\mathscr{C}}})\simeq\mathcal{X}({\mathscr{C}}_S).$$

Framings of 1–dimensional moduli spaces {#subsec:framings}
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In order to compute cohomology operations such as $\operatorname{Sq}^2$ we need to understand how the framings of the 1–dimensional moduli spaces behave after a handle slide. If we keep in mind the cancellation from the intermediate flow category ${\overline}{{\mathscr{C}}}$ to ${\mathscr{C}}_S$ we get $$\begin{aligned}
{\mathcal{M}}(a',x') &= {\mathcal{M}}(f,e) \times {\mathcal{M}}(a,x), \\
{\mathcal{M}}(a',y') &= {\mathcal{M}}(a,y) \sqcup \left({\mathcal{M}}(f,y) \times {\mathcal{M}}(a,x)\right).\end{aligned}$$ Furthermore, $${\mathcal{M}}(x',b') = {\mathcal{M}}(x,b) \sqcup {\mathcal{M}}(y,b),$$ but with the framing of ${\mathcal{M}}(y,b)$ depending on whether we do a $(+)$ or a $(-)$-handle slide.

Since ${\mathcal{M}}(f,e)$ is a negatively framed point, we get that the framing on ${\mathcal{M}}(a',x')$ corresponds to the old framing on ${\mathcal{M}}(a,x)$ by the framings in the cancelled category, compare [@JLSmorse]. Similarly, if we use the $(-)$–handle slide the framings on ${\mathcal{M}}(a',y')$ agree with the original framings on ${\mathcal{M}}(a,y)$ and ${\mathcal{M}}(a,x)$. In the case of the $(+)$–handle slide however, the copy of ${\mathcal{M}}(a,x)$ in ${\mathcal{M}}(a',y')$ has an extra reflection in one coordinate.

Recall that interval components $J$ in ${\mathcal{M}}(a,b)$ with $|a|=|b|+2$ can be framed in two ways (up to fixed boundary framing). The standard framings are described in [@JLSmorse §3.2] and [@MR3252965 §3], and we write $fr(J)\in {\mathbb{Z}}/2{\mathbb{Z}}$ with $fr(J)=0$ if 