The compositor for π₯,π¦,π§ βπ‘0 has components which are themselves natural isomorphisms, namely
πΎπ,π:πβπββ(ππ)β:π€π§βπ€π¦for π₯πβπ¦πβπ§ in π‘.
These are constructed using the fact that the composite of cartesian morphisms is cartesian
and the uniqueness of cartesian lifts:
In the diagram

we see that both πβπ§β²(ππ) and πβπβπ§β²π are cartesian lifts of ππ,
therefore they must be related by an isomorphism (πΎπ,π)π§β².
Uniqueness arguments show that the entire diagram commutes, and we see that πΎπ,π is natural.
Since π‘opβββ is locally discrete, naturality of πΎ is free.
The unitor for π₯ βπ‘0 is a natural isomorphism
π:1π€π₯β(1π₯)β:π€π₯βπ€π₯.This can be obtained quite directly from the diagram

which constructs ππ₯β² using the cartesian lift of 1π₯ as the universal factoring of the displayed identity.
The inverse is given by the cartesian lift itself, and we also witness naturality.
Let ππβππβπββπ in π‘.
Unwrapping the hexagon, the left hand side reads
πβπβββπΎπ,πββββββββββββββ(ππ)βββπΎππ,ββββββββββ(β(ππ))βπΌββΉ((βπ)π)β:π€πβπ€πwhere πΌ is the unique witness that β(ππ) =(βπ)π in π‘; and the right hand side reads
πβπβββidβΉπβπβββπββπΎπ,πβββββββββββπβ(βπ)βπΎπ,βπβββββββββ((βπ)π)β:π€πβπ€π.Let πβ² be an object over π.
Then at πβ² the left hand side is constructed by the diagram

and so it is the unique factorization of (πββ)(πβπ)(πβπ) via the cartesian morphism πβ((βπ)π).
Similarly, the right hand side is constructed by the diagram

so it is must be the same unique factorization.
Therefore the hexagon holds.
Now let π₯πβπ¦ in π‘ and π¦β² be an object over π¦.
The nontrivial path of the left of the unitality coherences reads
πβπβπββββββββββ(1π₯)βπβπΎπ₯,πββββββββ(π1π₯)βπββΉπβ:π€π¦βπ€π₯where π is the unique witness that π1π₯ =π in π‘.
Then at π¦β² this path is constructed from

so it uniquely factors πβπ via the cartesian morphism πβπ,
so it is equal to the trivial path.
Similarly, the nontrivial path on the right of the unitality coherences reads
πβπββπβββββββββπβ(1π¦)βπΎπ,π¦βββββββ(1π¦π)βπββΉπβwhere π is the unique witness that 1π¦ π =π.
Then at π¦β² this path is constructed from

so it uniquely factors 1π¦β²πβπ via the cartesian morphism πβπ,
so it is equal to the trivial path.