The push-up rule: p << q and q <= r imply p << r, together with the timelike-piece criterion

(Finite local subdivision.) If q=r then p<<r is immediate. Otherwise fix a future causal curve gamma:[0,1]->M from q to r. Cover its compact image by finitely many open sets C_j, each a strictly convex normal neighborhood contained in an adapted chart. A Lebesgue-number subdivision gives 0=t_0<...<t_m=1 such that gamma([t_i,t_{i+1}]) lies in one such C_i. The adapted chart time is strictly increasing on each subarc, so x_i:=gamma(t_i) and x_{i+1}:=gamma(t_{i+1}) are distinct.

(Normalization.) At any base point x use the specified smooth future unit timelike vector n_x=N_x/sqrt(-g_x(N_x,N_x)); no unit-length hypothesis on the original time-orientation field N is needed.

(Pushing up one causal segment.) Let x,y be distinct points in one C_i with a future causal curve in C_i from x to y, and suppose p<<x. By [L4] the position vector v=exp_x^{-1}(y) is future causal and nonzero. If v is timelike, the radial geodesic gives x<<y and hence p<<y. Suppose v is null. For small epsilon>=0 let x_epsilon=exp_x(-epsilon n_x); this is defined and lies in C_i for epsilon small, and x_epsilon tends to x. Define H(epsilon)=D^2(x_epsilon,y). By [L5], H is differentiable at zero, H(0)=g_x(v,v)=0, and H'(0)=2g_x(P(y,x),-n_x)<0, because the reversed null geodesic has past-directed null terminal tangent P(y,x), while -n_x is past timelike ([L6],[L7]). Thus H(epsilon)<0 for all sufficiently small positive epsilon. The inverse vector exp_{x_epsilon}^{-1}(y) varies continuously with epsilon and at epsilon=0 is future null, so for small positive epsilon it is future timelike: its timelike character follows from H(epsilon)<0 and its future orientation follows by continuity of its pairing with the smooth future unit field n. By [L4] the radial segment in C_i gives x_epsilon<<y. Since I^+(p) is open and x belongs to it, choose epsilon small enough that x_epsilon also belongs to I^+(p). Then p<<x_epsilon<<y, so p<<y. This proves the local push-up step.

(Finite induction along gamma.) Put x_i=gamma(t_i). Initially p<<x_0=q. If p<<x_i, apply step 1.3 to the causal subarc from x_i to x_{i+1} in C_i to obtain p<<x_{i+1}. Finite induction gives p<<x_m=r, proving (a). No uniform perturbation radius along the entire arc is used.

(Conclusion of (a).) Step 2.1 proves that p<<q and q<=r imply p<<r, including the reflexive case handled in step 1.1.

(Clause (b).) Consider the time reverse: $(M,g,-N)$ is again a time-oriented smooth Lorentzian four-manifold, its chronological and causal relations are the reverse relations ${\ll }^{\mathrm{r}\mathrm{e}\mathrm{v}}=\{(x,y):y\ll x\}$ and ${\le }^{\mathrm{r}\mathrm{e}\mathrm{v}}=\{(x,y):y\le x\}$, and every object used in the proof of (a) is defined by the same formulas with $N$ replaced by $-N$. Suppose $p\le q$ and $q\ll r$; in the reversed orientation $r{\ll }^{\mathrm{r}\mathrm{e}\mathrm{v}}q$ and $q{\le }^{\mathrm{r}\mathrm{e}\mathrm{v}}p$, so (a) applied in the reversed orientation gives $r{\ll }^{\mathrm{r}\mathrm{e}\mathrm{v}}p$, that is, $p\ll r$.

(Clause (c).) Let $\gamma$ and $s<t$ be as in (c). The subcurve $\gamma {\mathrm{\mid }}_{[s,t]}$ is a future timelike curve, so $\gamma (s)\ll \gamma (t)$ by definition; the segment $\gamma {\mathrm{\mid }}_{[a,s]}$ gives $\gamma (a)\le \gamma (s)$, and clause (b) yields $\gamma (a)\ll \gamma (t)$; the segment $\gamma {\mathrm{\mid }}_{[t,b]}$ gives $\gamma (t)\le \gamma (b)$, and clause (a) yields $\gamma (a)\ll \gamma (b)$.

(Clause (d).) Let $\mathrm{\Omega }\subseteq M$ and $z\in I^{+}(J^{+}(\mathrm{\Omega }))$: there is $y\in J^{+}(\mathrm{\Omega })$ and $x\in \mathrm{\Omega }$ with $x\le y$ and $y\ll z$. Then $x\ll z$ by (b), so $z\in I^{+}(\mathrm{\Omega })$; the reverse inclusion is trivial because $x\le y$ for $y=x$. This proves the first identity, the second is its time reverse, and $J^{+}(I^{+}(\mathrm{\Omega }))=I^{+}(\mathrm{\Omega })$ follows from the same argument with $x\ll y$, $y\le z$.