which original?

1. The fibre is the list of suspects

Let X be a nonempty source space, Z a representation space, and E: X -> Z an encoder. Its attained image is Z_E = E(X). A latent point outside Z_E need not represent any source at all. For z in Z_E, define

F_z is the fibre over z. The quotient X / ~_E consists of these equivalence classes. The encoder factors as E = Ebar o q, where q sends a source to its class and Ebar is a bijection from the set of classes to Z_E. This is a set-theoretic statement; the topology needs its own examination.

A fibre is defined by the encoder and its input domain. Restricting X can turn a many-to-one map into a one-to-one map. E(x) = x² loses sign on [-1, 1], but is injective on [0, 1]. A sentence about "the latent" without the source domain has omitted half its evidence.

Nor does the dimension of Z alone identify what was lost. A projection E(x, y) = x forgets y. The inclusion E(x) = (x, 0) forgets nothing. A map into a bounded set can also preserve every source distinction; positive-epsilon RMSNorm is an explicit example.

2. Five different invoices marked "inverse"

These contracts can combine. Exact recovery may be unstable. A stable decoder may return the wrong source consistently. An average test error is a different contract from the displayed worst-case error. Calling all of them inversion is how the paperwork escapes.

For E(x) = x² on [-1, 1], S(z) = sqrt(z) is a continuous section on [0, 1]. E(S(z)) = z, but S(E(-1)) = 1. The round trip works in one direction. The other direction has acquired an opinion about signs.

The composite P = S o E is idempotent: P o P = S o (E o S) o E = P. It is a projection onto the chosen representatives in this algebraic sense. Unless E is injective, P is not the identity on all sources. A perfect encode-decode-encode score can therefore coexist with complete failure to recover some originals.

3. Recover the task. Name the task.

Let h: X -> Y be the target: the original state, a class label, an absolute magnitude, a reading, a geometric invariant. A decoder for h receives only E(x). We first allow arbitrary set functions; any requirement of continuity, bounded cost or implementability is additional.

The theorem permits exact recovery of some properties after exact recovery of the source has become impossible. Under squaring, h(x) = |x| factors through E, with H(z) = sqrt(z). The signed target h(x) = x does not. "Information loss" is therefore relative to a target, even though noninjectivity itself is an exact property of E.

For a linear encoder E(x) = Ax on a vector space and a linear task h(x) = Bx, the same theorem reads

Proof. Equal A-values differ by v in ker(A). Their B-values agree precisely when Bv = 0. If the containment holds, define H(Ax) = Bx; this induced map on the range of A is linear. This is the ordinary quotient construction for linear maps.

A learned reconstruction prior can favor a plausible member of F_z. It cannot make unequal target values within that fibre equal. To improve the guarantee, change the domain, the target, the observation, or the allowed error. Better prose is not an extra measurement.

4. At least one of you is going home wrong

Give Y a metric d. For an attained code z, define its target ambiguity diameter

This is an information bound. It applies before the decoder is charged for time, continuity or symmetry. Such restrictions may increase the necessary error, never lower this bound.

The exact fibrewise minimax quantity is a radius, not generally half a diameter:

The upper bound follows by choosing any target value already in the fibre. For an equilateral triangle of side 1 in the Euclidean plane, the optimal radius is 1/sqrt(3), strictly larger than half its diameter. The choice of output space also matters: if Y is only the three vertices with their induced metric, the optimal radius is 1.

For x² = z with 0 <= z <= 1, exact-source candidates are -sqrt(z) and +sqrt(z). In Y = [-1, 1], their diameter is 2sqrt(z), and output 0 attains minimax error sqrt(z). Notice: 0 is usually not a member of the source fibre. If the decoder must be a section, it must return one of the two roots and its worst error is 2sqrt(z). "Be a plausible original" can cost more than "minimize worst error."

5. The receipt was in the other pocket

Suppose memory, a witness, a source label or a second measurement supplies m(x). The actual observation is now E'(x) = (E(x), m(x)), with fibres

For squaring, one sign bit separates the positive and negative roots; zero needs no distinction. For a finite fibre with k distinct required target values, a side channel with b bits has at most 2^b messages. Exact task recovery requires 2^b >= k by the pigeonhole principle, hence b >= ceil(log₂ k). This is a per-fibre information lower bound, not a claim that an efficient global code has been constructed.

A full saved source is a sufficient side channel. It is also a fairly expensive way to announce that a decoder is clever. Underprint's witness and journal deliberately expose this distinction: restoration from retained state is legitimate, but its success must be charged to the larger observation.

Memory is not automatically a refinement by true source identity. A state variable may store only a prior choice, or may have forgotten the relevant event. One must specify the joint state, update rule and available observation. "Has memory" is a hardware description; it is not a proof of recoverability.

A prior can prefer an original. It cannot certify one.

For a finite source set with a declared probability distribution, an observed z with positive probability induces posterior source probabilities p_i within its fibre. Under exact-source zero-one loss, the best deterministic answer is a maximum-posterior source and succeeds with conditional probability max_i p_i. A fresh independent draw from that posterior succeeds with probability sum_i p_i².

Proof. Returning source i succeeds with probability p_i; maximizing gives the first expression. For an independent posterior draw, true source i and returned source i occur together with probability p_i²; sum over i. Any randomized choice with probabilities q_i succeeds with sum p_iq_i <= max p_i, so randomization cannot improve the optimum under this loss.

At posterior probabilities 0.9 and 0.1, maximum-posterior guessing succeeds 0.9 of the time, while posterior sampling succeeds 0.82. Neither supplies a worst-case inverse on that two-source fibre. Sampling may serve a different goal, such as calibrated variety; the exact-source success criterion must not be changed mid-proof.

6. The quotient has a topology. Do not skip the customs desk.

Setwise factorization says nothing about continuous recovery until the topology is fixed. Give X / ~_E the quotient topology: a set U of classes is open exactly when q^-1(U) is open in X.

To replace the abstract quotient with the attained latent image Z_E carrying its subspace topology, E must be a quotient map onto that image. A useful sufficient condition is X compact, Z Hausdorff, E continuous. Then E is closed onto its image: a closed subset of compact X is compact, its image is compact, and compact subsets of a Hausdorff space are closed. A continuous closed surjection is a quotient map. Thus the induced bijection Ebar is a homeomorphism.

Without that condition, a continuous bijection need not preserve topology in the inverse direction. The parametrization E(t) = (cos t, sin t) from [0, 2pi) to the unit circle is a continuous bijection. Its inverse is discontinuous at (1, 0): points with t approaching 2pi converge on the circle to E(0), while their inverse parameters do not approach 0. The missing compact endpoint matters.

Sections have their own topological obstruction. For E: S¹ -> S¹, E(z) = z², every fibre has two points, but no continuous section exists.

7. Exact does not mean well-conditioned

There need not be a finite Lipschitz constant. E(x) = x³ on [0, 1] is continuous and injective, but for x > 0,

The ratio diverges near zero. Still, |z^1/3 - w^1/3| <= |z-w|^1/3 for nonnegative z,w: write the larger root as a+b with a,b >= 0, and use (a+b)³ - a³ >= b³. This is a stable inverse with a cube-root modulus, not a Lipschitz inverse.

A useful quantitative assumption is the lower separation bound

It gives injectivity and inverse Lipschitz constant at most 1/c immediately. For a linear map A with full column rank in Euclidean norms, the sharp c is its smallest singular value. Rank loss makes a nontrivial fibre; a small positive singular value makes a difficult inverse. Neither event should be renamed as the other.

A derivative test is local, not global. The map P(u,v) = exp(u)(cos v, sin v) has Jacobian determinant exp(2u), never zero, yet P(u,v) = P(u,v+2pi). Even the compact restriction [0,1] x [0,2pi] contains this collision. Conversely, x³ is globally injective although its derivative vanishes at zero. Local differential rank, global uniqueness and quantitative inverse stability are three separate checks.

Noise fattens the fibre

If the observation w can differ from E(x) by at most delta, the compatible sources are

The half-diameter lower bound applies to the target values on this larger set. A map can be injective at delta = 0 while practically indistinguishable sources reappear at every positive noise level.

In a normed latent space, a direct lower-bound witness is any pair with ||E(x)-E(y)|| <= 2eta. Their midpoint w = (E(x)+E(y))/2 is an observation compatible with both at noise level eta. Every decoder at w has error at least d_Y(h(x),h(y))/2 for one of them. The factor 2 is actual geometry, not a convention about rounding.

8. The bounded box that kept your magnitude

The word normalization tempts an unjustified shortcut: "scale was removed, therefore the source was lost." With a positive stabilizer, the displayed map decides the matter. Here is the exact calculation for the formulas associated with LayerNorm, Ba, Kiros and Hinton, and RMSNorm, Zhang and Sennrich. The result below concerns the specified real-valued formula, with eps > 0 and every gain gamma_j nonzero.

Here * and division by gamma are coordinatewise, d is the width, and 1 is the all-ones column. Undo gain and bias: z_j = (N(x)_j - beta_j)/gamma_j. Then z is in ran(Q), ||z|| < sqrt(d), and

All-nonzero gains are a convenient sufficient condition, not a necessary one for LayerNorm. In general, recovery of Qx requires the diagonal gain map to be injective when restricted to ran(Q). For example, at width two, a centered vector is (t,-t), and observing just one nonzero-gain coordinate still determines it. For RMSNorm, ran(Q) is the full space, so each gain must be nonzero for unrestricted injectivity. If eps = 0, the formula is undefined when Qx = 0 and becomes invariant under positive radial scaling elsewhere. That is a different map with different fibres.

There is also a distinct conditioning problem. For the gain-free inverse R(z) = sqrt(eps) z / sqrt(1 - ||z||²/d), the derivative has tangential factor sqrt(eps)(1 - ||z||²/d)^-1/2 and radial factor

They diverge at the open-ball boundary. On every closed smaller ball ||z|| <= r < sqrt(d), the inverse has a finite Lipschitz bound. On the full open ball, no uniform bound survives. Bounded outputs, an exact inverse and poor conditioning coexist without contradiction. Finite-precision rounding can collapse distinct outputs, but that is a separate representation map whose precision must be named.

The residual may have kept the part its branch discarded

For a pre-LayerNorm residual map F(x) = x + G(LN(x)), with fixed parameters and positive eps,

The normalized branch sees identical inputs after a mean shift, while the identity path carries the shift forward. The same argument applies successively to pre-LayerNorm residual sublayers. This shift relation proves that those source pairs do not collide at the residual output; it does not prove global injectivity in all other directions.

9. Every finite guest remembers. The limit forgets.

Let 0 < a < 1/2, and let a fixed two-token stochastic mixing matrix act on scalar rows:

The consensus direction survives untouched; the disagreement direction remains recoverable at every finite depth but costs an amplification of r^-m. This is why the compact mixing theorem is a forward stability theorem, not an inverse theorem. Compact uniform closure may contain noninjective limit maps even when all the finite operations under discussion are injective.

For a = 1/4, m steps multiply disagreement by 2^-m. At m = 20 the inverse amplifies that component by 1,048,576. This integer is exact. It is not a measured neural-model condition number.

For state-dependent attention F(X) = A(X)X, the fixed-matrix calculation cannot be substituted without checking the dependence on X. Where differentiable,

The first term is real mathematics, not optional punctuation. Knowing the rank of A(X) at each input does not by itself settle injectivity or conditioning of the nonlinear map F. Solving for X also changes the matrix one would try to invert.

A flattening, with and without a saved original

On source square [-1,1]², set F_t(x,y) = (x,ty), with 0 <= t <= 1. At every positive t the inverse on the attained image is (u,v) -> (u,v/t), with Euclidean Lipschitz constant 1/t. At t = 0, each vertical segment becomes one fibre: (x,y) -> (x,0). This is an exact transition from ill-conditioning to noninjectivity.

If a renderer always keeps the original (x,y) and merely draws F_t(x,y), moving t back to 1 restores the display from stored source data. If the system actually replaces its whole state by F₀(x,y), the omitted y is absent from that state. Merely storing t = 0 adds no distinction between the vertical candidates. The augmented observation (F_t(x,y),y) does preserve every source for all t. These are different state architectures with different recovery guarantees.

10. An intervention in coordinates is not yet an intervention in meaning

Let T: Z_E -> Z be a proposed latent edit. If T(z) is outside Z_E, no source has exactly that encoding. A decoder defined there supplies an extension, approximation or convention. Even inside the attained image, decoding selects a fibre member unless E is injective.

A source operation L: X -> X is consistent with T precisely when the square commutes:

Existence of L requires T(E(X)) to be contained in E(X). Setwise, a section S then gives one lift L = S o T o E. But that lift forgets every source distinction excluded by the section, even if the intended intervention concerned only one property.

Counterexample. E(x,y) = x, S(z) = (z,0), T(z) = z+1. The section lift returns (x+1,0). The source operation L(x,y) = (x+1,y) also commutes with E, and preserves y. The same latent edit admits both. A latent equation alone does not certify preservation of hidden source structure.

Likewise, an invertible change of latent coordinates R merely replaces E by R o E and H by H o R^-1; every fibre and every recoverable task is unchanged. A coordinate negation is not automatically a semantic negation. An artistic departure must name what representation agreement it departs from.

Proposed definition / representation-relative departure

For this work, a representation regime declares the source domain, encoder, target observables, admitted operations and outputs, repair controls, cost budget, available information and required regularity. Call an operation a departure from that regime only after specifying which admission, identification, preservation or target rule it violates or changes. Its irreverence is assessed by the changed obligations and the possible repairs, rather than by its distance from a visually ordinary point.

This is proposed vocabulary for organizing questions, not a new theorem. The repair chapter gives the full endpoint and control contract. A reversible departure, a costly return, a loss of task information and a change of the target are four possible outcomes. None entails the others.

Pointwise repair can therefore be possible at every initial input while observation-only repair is impossible. Take the real line as the ambient state space, initial inputs restricted to {-1,+1}, observation E(x) = x², target I = {0}, controls u in the real line, endpoint x+u and cost |u| with budget 1. Each initial source has the exact control u = -x. At their shared code 1, the two feasible sets {-1} and {+1} have empty intersection. A sign measurement makes the repair possible. A bigger decoder with the same observation does not.

To change the regime instead, one could admit both endpoints, supply a sign witness, add an actuator that observes x, or charge replacement by 0 as an available control. Each is a coherent new experiment. None proves that the old information constraint was overcome.

11. The questions you have not asked yet

A representation can answer today's task perfectly and still be inadequate for tomorrow's composition. Name a family G of permitted operations g: X -> X and a family H of observed tasks h: X -> Y_h. Let W be all finite words in G, including the identity. The operations here are total self-maps. If some acts can be refused, include the refusal and any observed receipt in an explicit state model; do not silently discard undefined cases.

Define future-observation equivalence by

The sources are identified only if every permitted future experiment gives the same observed answer. This is the familiar observational-equivalence/congruence construction, used here to specify what a latent representation must preserve. It is closely related to the continuation test in automata theory, rather than a new claim that machines have acquired irreverent souls.

Each generator consequently descends to the quotient by [x] -> [g(x)]. Every task descends too. The quotient stores exactly the distinctions necessary for all named future observations, in the set-theoretic sense of that coarseness theorem. The theorem does not supply a finite-dimensional, computable or inexpensive encoding of the quotient.

A three-state counterexample. Let X = {a,b,c}, with observed task h(a) = h(b) = 0 and h(c) = 1. If the only permitted word is the identity, the coarsest quotient has classes {a,b} and {c}. Now admit g(a) = a, g(b) = c, g(c) = c. The observation after one g is 0 from a and 1 from b, so the old class must split. The new future-observation quotient has three singleton classes. A representation E = h had been sufficient for the old regime; it cannot implement this new g because the shared code 0 would need two different next codes.

In general, enlarging the operation alphabet or the family of observed tasks can only refine this equivalence: there are more equalities to satisfy. Removing operations or tasks can coarsen it. The sufficient quotient is therefore relative to what the system may subsequently do, not just what it presently says.

This gives the mutable grammar a precise connection to latent spaces. If a new rule distinguishes sources that the old representation identified, the old quotient is no longer sufficient for that rule. In Underprint, raw mark distinctions remain in the full state even after a stitch identifies their kinds in the current reading. A release discards some of those distinctions. A later demand to recover a pre-quotient raw kind must then use retained source information; it cannot be implemented from the shared release alone.

12. Underprint supplies actual fingerprints

Enter the mark language, or inspect its complete served kernel. Its state contains raw kinds, geometry, IDs, the current equivalence partition, seam kind, reading direction and pair rules. Its staged reader orders the marks, identifies equivalent kinds, splits at seams, then performs leftmost pair reductions. Changing the grammar can change the reading of unchanged ink and which anchors are allowed for a later stitch.

The release map E is the actual exported release(state), not a metaphorical forgetting machine. It retains the equivalence partition, canonical seam and token sequence, and canonical deduplicated rule triples. It removes raw-kind distinctions within a class, geometric paths and positions beyond their induced order, IDs, original direction as a separate field, and operation history. Tokens are put in reading order and the released direction is 1.

The kernel's reopen(reading) creates equally spaced canonical-kind marks at y = 0.5, with fresh sequential IDs and one-point paths. On every validated released reading,

This is a section. It preserves the released artifact exactly; it does not recover the original raw source. In the collision above, the reopened tape uses canonical slashes for both first tokens. If an interface calls this "reopen," it has named the operation correctly.

Three targets, three different tests

1. Current admission. Does the current grammar produce a fault-free reading? A changed seam can make this true without restoring anything old.
2. Witness reading. Does the current reading match the frozen witness reading? The kernel compares compound-tree structure and faults, omitting source IDs and IDs of equivalent rules. The same final word is a weaker test.
3. Exact state restoration. Does the full mark-and-grammar state equal the frozen witness, including geometry? When accepted within the journal limits, Session.repair() copies that retained witness.

Keeping a new witness changes the target. It is not a restoration to the old witness. Undo and a journal import use retained history; they do not invert a release. The journal keeps one bounded branch, up to 128 successful events, with undone future events retained until a new edit replaces that future. It is not an immutable archive of every branch ever abandoned.

Turning twice is an actual inverse of the direction operation. Stitching merges kind classes and has no internal split operation; a journal can nevertheless restore the earlier partition. These facts coexist because operation sets and information sets are different parts of the contract.

Release forgets within the released representation. The original session, saved journal or witness can still exist elsewhere. "No inverse from this artifact alone" is the proved statement. "The information has ceased to exist" would be an unsupported claim about the universe.