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i forgot that immersions between equidimensional disks are diffeos. So yes, you are correct that there do not exist expanding diffeos from a disk to smallet disks.
@Leo Moos. Yes i will elaborate my above answer. But briefly, if the cut locus had a point ("vertex") which had infinite "degree", then the maximal disk centred at that point would intersect the boundary at infinitely many points, and then we'd conclude the boundary would consist of circular arcs (quarter circles, etc.). But this leads to contradiction because the cut locus of circular arcs are points, not edges. Maybe thats still not clear....
I was mistaken. It is incorrect to write $\psi_*$, since we are speaking of the restriction of $\psi$ to fibres. What is needed is a self map $f$ of the base $\phi(U)$ for which $f_*\circ j_0 = j_1 \circ f_*$. The bundle morphism from Prop. 2 is identity map $f=id$ on the base. So Prop. 2 does not yield a holomorphic map $f$.
From Proposition 2, you conclude the tangent bundle of $\phi(U)$ is isomorphic with tangent bundle of $\phi(U)$, i.e. homeomorphism on the base and fibrewise linearly isomorphic. But the isomorphism needs not transport $j_0$ to $j_1$, i.e. inequality generally holds $\psi_* \circ j_0 \neq j_1 \circ \psi_*$ along the fibres.
Moreover you might compare Ricci-Levi Civita's memoir eudml.org/doc/157997, especially pp.162, where the nonexistence of first order differential invariants on Riemannian manifolds is established. This nonexistence is old, long forgotten result, and which plays interesting role in proving the nonexistence of any tensorial expression for Einstein's gravitational energy density $T_{00}$. .
For the Hermitian manifold $(M, I, g)$, the $2$-form $g(\cdot, I \cdot)$ defines the symplectic $2$-form $\omega$ on $M$. Some argue that Darboux Theorem proves $\omega$ contains zero local geometric information. Moreover, if a riemannian manifold $(M,g)$ had a canonical $1$-form $\alpha$, then it would equivalently have a canonical vector field $X$ defined by $g(\cdot, X)=\alpha(\cdot)$. Except riemannian manifolds do not have canonical vector fields, e.g. euclidean space has no canonical vector field except the constant zero vector field.
If the immersion is non injective, then I think it's quite plausible the volume of the image can decrease, even if the immersion is locally expanding. And to be specific, by "volume of the image" i mean the unit-normalized Lebesgue volume of the set-theoretic image $f(U)$.
Of course the round disk is diffeomorphic to smaller disks in $\mathbb{R}^n$. And immersions are identical with local diffeos and submersions because we are studying maps between open subsets in finite dimensions. Even Alexander trick suffices to rescale the entire space, and immerse/diffeo/submerse the image $f(U)$ into arbitrarily small ball. What is impossible is to find a locally expanding injective immersion (i.e. locally expanding diffeomorphism) of the round disk to a smaller disk. Is that what you meant?
The above immersion is strictly locally length increasing, or am I mistaken? (where distance on the image is induced path length as a subset of $D$). My definition of expanding-embedding is incomplete: it needs hypothesis that $f$ is injective. And is it clear that your map is continuous? Namely, when you "arrange these pieces" and when you "connect the corresponding cuts..." how to ensure a continuous map is being defined? It's interesting that the volume of the image $f(U)\subset D$ can perhaps(!) be made arbitrarily small by your construction, an "expanding collapse".