Can we cut/sew path integrals along surfaces which aren't everywhere spacelike? Usually when we work with the path integral, we specify boundary data on spacelike hypersurfaces. But, at least formally, we can also write down path integrals where the boundary has timelike parts. Indeed, there are respectable-looking papers [1,2] proposing that amplitudes can be assigned even in these cases. However, considering such path integrals leads me to the following paradox:
Fix a globally hyperbolic spacetime $(\mathcal{M},g)$. Take it to be 1+1D, to make the diagrams easier. Let $\Sigma_1,\Sigma_2$ be Cauchy surfaces, with $\Sigma_2$ lying to the future of $\Sigma_1$. Suppose QFT amplitudes are given by the path integral with some external source $J$:
$$\langle \phi_2|U_{1\to 2}|\phi_1\rangle = \int_{\phi_{\Sigma_i}=\phi_i} D\phi e^{i\left(S[\phi]+\int dV J(x)\phi(x)\right)}.$$
Suppose further that supp$(J)$ is compact and lies between $\Sigma_1$ and $\Sigma_2$. This setup is depicted in Fig. 1:

Let's fix an initial state on $\Sigma_1$ and evolve it to $\Sigma_2$. Choose an arbitrary point $p\in\Sigma_2$, and let $\Sigma_2^L$ be the part of $\Sigma_2$ lying to the left of $p$. Now construct a surface $\Sigma$ lying between $\Sigma_1$ and $\Sigma_2$, such that 1) The causal domain of $\Sigma$ is the full spacetime, 2) supp$(J)$ lies entirely in the region between $\Sigma$ and $\Sigma_2$ , and 3) $\Sigma_2^L\subset\Sigma$. Note: we don't require $\Sigma$ to be spacelike everywhere. This is depicted in Fig. 2.

Note: the middle surface has a typo: it should be labelled $\Sigma$, not $\Sigma_2$
Now, by the sewing/gluing property of the path integral, the overall evolution $\Sigma_1\to\Sigma_2$ equals the composition of the evolutions $\Sigma_1\to\Sigma$ and $\Sigma\to\Sigma_2$.$^\dagger$. Evolving $\Sigma_1\to\Sigma$ doesn't depend on $J$, since $J=0$ in this region. Then, evolving $\Sigma\to\Sigma_2$ doesn't change the state on $\Sigma_2^L$. Hence we  conclude that the state on $\Sigma_2^L$ doesn't depend on $J$ in any way.
But this can't be right! Since we're allowing $\Sigma$ to have timelike parts, we can make $\Sigma_2^L$ as big as we like - in particular, part of it can lie in the causal future of supp$(J)$, where the state should surely depend on $J$.
What was my mistake?

$\dagger$ The reader might worry about calling e.g. $\Sigma_1\to\Sigma$ an "evolution" if $\Sigma$ isn't spacelike, and hence can't be interpreted as a constant-time slice. But regardless of nomenclature, the path integral does assign complex numbers to field data on $\Sigma_1\cup\Sigma$, and to field data on $\Sigma\cup\Sigma_2$, which sew/glue in the usual way.
[1] https://arxiv.org/abs/hep-th/0306025
[2] https://arxiv.org/abs/hep-th/0509123
 A: A simple and direct answer to the question would be to say that the timelike boundary can't be the issue, because the same paradox can be constructed using a combination of past and future spacelike boundaries:

The full answer is that a timelike boundary is perfectly reasonable, and neither of these situations are paradoxes, as long as you're careful.
The path integral is a function of the boundary conditions. When you're sewing / composing the two regions, the boundary $\Sigma$ between those two regions stops being a boundary, and becomes just more points internal to the region. So to compose them you need to integrate over all the possible field states on $\Sigma$, with the integrand being the product of the amplitudes for each of the two regions:
$$
\langle\phi_2|U_{1\to2}|\phi_1\rangle = \int \mathcal D \phi(\Sigma)\langle\phi_2|U_{\Sigma\to2}|\phi_\Sigma\rangle\langle\phi_\Sigma|U_{1\to\Sigma}|\phi_1\rangle
$$
That's how the influence of $J$ on $\Sigma$ interacts with the influence of $\Sigma$ on $\Sigma_2^L$. This does involve the influence going "backwards" from the second region to the first region, but that's fine. Something like that has to happen given the spacetime diagram, and as you said, it works fine formally if you just think of it in terms of "fields in, amplitudes out".
To take a concrete situation, say that $|\phi_1\rangle$ is a vacuum state, and $J$ produces some photons which will pass through $\Sigma_2^L$. So $U_{\Sigma\to2}$ tells us that there is a large amplitude for $\phi_\Sigma$ to have photons passing through it (informally speaking). If instead $J$ were 0, there would be a large amplitude for $\phi_\Sigma$ to be undisturbed. And if $U_{1\to\Sigma}$ has a boundary condition where photons are coming through $\Sigma$, it will give large amplitudes for photons passing through $\Sigma_2^L$. Otherwise, if $\Sigma$ is undisturbed, there will be large amplitudes for vacuum. So when you perform the integral over all the possible boundary conditions, whether there's a large value for vacuum or for photons at $\Sigma_2^L$ depends on whether there is a large amplitude vacuum or for photons at $\Sigma$. Which in turn depends on $J$.
The timelike part of the boundary is acting like a weird mixture of initial-value and final-value. But the integral of the product of amplitudes is symmetrical, so it doesn't really care which one comes before the other, and doesn't care whether they have a causally messy relationship, like in this case.
One thing to note is that $U_{1\to\Sigma}$ and $U_{\Sigma\to2}$ aren't necessarily unitary, despite the naming. One other thing to note is that you can specify fields on $\Sigma$ that are actually inconsistent with each other, which should cause the path integral to spit out infinitesimal amplitudes in those cases.
A: I've realised that my argument fails for a very simple but slightly subtle reason, so I'm answering my own question here for posterity.
Let the $\Sigma_1\to\Sigma$ map be $B$, which is independent of $J$. 
Let the $\Sigma\to\Sigma_2$ map be $\mathbb{I}\otimes A[J]$, where $\mathbb{I}$ is the identity on $\Sigma_2^L$, and $A$ depends on $J$ in some way.
Then if the state on $\Sigma_1$ is $|\psi\rangle$, the state on $\Sigma_2$ can be written as $(\mathbb{I}\otimes A[J])\circ B|\psi\rangle$.
My mistaken argument in the original question was:

*

*$B|\psi\rangle$ is independent of $J$

*Applying $\mathbb{I}\otimes A[J]$ doesn't change the reduced density matrix on the first tensor factor (since we just applied the identity to the first factor, right?!)

*Hence the final state $(\mathbb{I}\otimes A[J])\circ B|\psi\rangle$, when reduced to the first factor (i.e. reduced to $\Sigma_2^L$) is independent of $J$.

The mistake is in step 2. It's true applying a unitary of the form $\mathbb{I}\otimes U$ doesn't change the reduced state on the first factor. Indeed, this is the standard situation in QFT, quantum info, etc. But in this strange case with timelike boundaries, $A[J]$ is not unitary. Hence, applying $(\mathbb{I}\otimes A[J])$ can change the reduced state on the first factor.
I think that's all there is to it!
