Boundary orientation of the compactified one-dimensional Morse moduli space

Choose entry and exit levels $f(r)\pm \epsilon$ in the chart of [F5]. The incoming sheet of $W^u(p)$ in the entry level is a $k$-disk transverse to the stable sphere: quotienting the Morse--Smale transversality by the common flow direction gives an isomorphism from its tangent space to the $u$-space. By [F6] it is $D=\{(u,h(u))\}$, with $\mathrm{\mid }h(u){\mathrm{\mid }}^2=\epsilon +\mathrm{\mid }u{\mathrm{\mid }}^2$. Integrating [F5] gives $(u,v)\mapsto (e^{2t}u,e^{-2t}v)$; writing $u=s\theta$ and $g=h\mathrm{/}\mathrm{\mid }h\mathrm{\mid }$, its passage image is $H(s,\theta )=(\sqrt{\epsilon +s^2}\theta ,sg(s\theta ))$. This extends smoothly to $s=0$. There ${\mathrm{\partial }}_{s}H=(0,g(0))$ is nonzero and independent of the angular derivatives, which span the unstable sphere. For $s\ge 0$ its parameters are recovered by $s=\mathrm{\mid }v\mathrm{\mid }$ and $\theta =u\mathrm{/}\mathrm{\mid }u\mathrm{\mid }$, so its image $Q$ is an embedded collar of that sphere.

At the outgoing crossing, the stable slice of $q$ has codimension $\lambda (q)=k-1$ in the exit level. Transversality of $W^u(r)$ and $W^s(q)$, after removing their common flow line, says that its defining equations restricted to the unstable sphere have invertible angular derivative. Extend $H$ to negative $s$ locally and apply [F6] to obtain a unique smooth solution $\theta =\theta (s)$; for $k=1$ there are no angular equations. Thus $Q\cap W^s(q)$ is a smooth half-interval $H(s,\theta (s))$ transverse in $Q$. For $s>0$ it represents ordinary trajectories, and its entry and exit points converge to those of the given broken pair. It is the collar end in [F2], by that collar's uniqueness and entry-radius parameter.

Orient the $u$-space by $o{r}_r$. At the incoming crossing, [F3] gives $o{r}_p={ϵ}_1(X_{\mathrm{i}\mathrm{n}},o{r}_r)$, since the normal quotient to $W^s(r)$ is the $u$-space. Removing the positive flow direction orients $D$ as ${ϵ}_{1}o{r}_r$. Write $\sigma$ for the sphere orientation with $(\text{positive radial direction},\sigma )=o{r}_r$; then polar coordinates orient $D$ as ${ϵ}_1({\mathrm{\partial }}_s,\sigma )$ for $s>0$. Flow passage preserves this slice orientation: its derivative is $D{\mathrm{\Phi }}_t$ plus a multiple of $X$, and the added term disappears in the quotient by $X$. Hence $Q$ has orientation ${ϵ}_1({\mathrm{\partial }}_s,\sigma )$, extending nonvanishingly to $s=0$ in the embedding of step 1.1.

At the outgoing crossing the positive flow direction on $W^u(r)$ is positive radial. If $N_q$ is the oriented normal quotient to $W^s(q)$, [F3] gives $o{r}_r={ϵ}_2(X_{\mathrm{o}\mathrm{u}\mathrm{t}},N_q)$. Thus $\sigma$ maps to ${ϵ}_2{N}_q$ in that quotient. Step 2.1 yields $o{r}_Q={ϵ}_1{ϵ}_2({\mathrm{\partial }}_s,N_q)$ at the boundary. The kernel-first exact sequence for the transverse intersection in step 2.1 therefore orients its tangent as ${ϵ}_1{ϵ}_2{\mathrm{\partial }}_s$; a lift tangent to the solution curve differs from ${\mathrm{\partial }}_s$ by angular directions, which do not change this determinant. For $s>0$ this agrees with the flow-first moduli orientation, because the ordered splittings of $T{W}^u(p)$ are $(X,Q)$ and $(X,T\mathcal{M}(p,q),N_q)$.

The orientation in step 3.1 extends nonvanishingly to every collar endpoint of [F1]. Comparing its positive tangent ray ${ϵ}_1{ϵ}_2{\mathrm{\partial }}_s$ with the outward ray $-{\mathrm{\partial }}_s$ in [F4] gives boundary sign $-{ϵ}_1{ϵ}_2$. On an oriented interval the two outward boundary signs are opposite, so the products are opposite as well. Only unstable orientations and their normal quotients were used; no orientation of $M$ was used.