Understand a one-sided inverse with rectangular maps

A one-sided inverse can restore every input in one direction while failing to restore inputs in the reverse direction. This original exact-integer example uses a map that drops a third coordinate and another that appends zero. Starting with any pair, appending zero and then dropping it returns the pair. Starting with a triple whose third coordinate is seven, dropping that coordinate and appending zero changes the triple.

Keep two domains visible throughout the exercise: ordered integer pairs and ordered integer triples. A pair has two coordinates in a fixed order; a triple has three. The integers are stipulated mathematical inputs, with no measurements, device output or recorded learner response. Equality of two pairs requires both corresponding coordinates to agree. Equality of two triples requires all three, including the last one.

Define the maps and their order

Call the first map A. For any integer triple (r, s, t), define A(r, s, t) = (r, s). Its output contains the first two coordinates in their original order. The definition supplies an output for every integer triple, regardless of the value of t. Many triples can consequently have the same output because changing only the third coordinate does not change the returned pair.

Call the second map B. For any integer pair (u, v), define B(u, v) = (u, v, 0). Its output is a triple with a supplied zero in the third position. Every output of B therefore has third coordinate zero. The first two coordinates retain the pair that was supplied. Both maps are fully defined before either composition is evaluated.

Use AB to mean A after B: apply B first, then apply A to its output. Use BA to mean B after A. The rightmost map acts first under this convention. This order gives AB a pair as its starting input and BA a triple. Write the convention beside the formulas so the same letters do not conceal a change of starting domain.

Verify restoration for every pair

Start with the specific pair (4, -2). B appends zero, giving (4, -2, 0). A then keeps the first two coordinates, returning (4, -2). The complete path begins and ends with the same pair. Its intermediate triple is defined by B; we have not guessed a missing entry or looked up a saved earlier triple to fill it.

For arbitrary integers u and v, the same calculation reads A(B(u, v)) = A(u, v, 0) = (u, v). This argument covers every pair in the declared domain. Call the identity on integer pairs I2: it returns each pair unchanged. We have proved AB = I2 by evaluating the composition on an arbitrary input rather than relying only on the four-minus-two example.

The word identity names this whole-domain behavior. A single restored pair would establish a result for that input, but would leave other pairs unchecked. The arbitrary-coordinate calculation supplies the stronger statement here. It also records the type of input being restored. I2 concerns pairs, so its formula cannot simply be copied as a claim about every triple without a separate check.

Compute the reverse path on a triple

Now start with (4, -2, 7) and apply BA. A first returns (4, -2), discarding the third coordinate seven. B then appends its specified zero, giving (4, -2, 0). The first two coordinates agree with the starting triple, but the third does not. This is a defined calculation with a different output, sufficient to reject BA as the identity on all triples.

Nothing in B asks for the third coordinate that A removed. Its definition always appends zero. Replacing that zero with seven because seven appeared in the saved starting record would apply a different rule. The current question concerns the two maps already defined, including what information each receives. B receives only the pair that A supplies in the reverse composition.

The triple (4, -2, 0) does return unchanged under BA. A gives (4, -2), and B returns (4, -2, 0). Keep this successful control alongside the failed seven example. It identifies a set of inputs on which the reverse path happens to restore its starting point, without turning that restricted success into a claim about every triple.

For an arbitrary triple, B(A(r, s, t)) = B(r, s) = (r, s, 0). This equals (r, s, t) exactly when t = 0. The first two coordinates already agree, leaving only that last equality to check. Thus the reverse path restores precisely the triples with third coordinate zero, while the forward path restores every pair in its declared domain.

Represent the same maps with rectangular matrices

Write A as a matrix with two rows and three columns. Its first row is 1, 0, 0 and its second row is 0, 1, 0. Multiplying this matrix by a column with entries r, s, t returns a column with entries r and s. The supplied zeros make the third coordinate contribute zero to each output entry, matching the defined map.

Write B with three rows and two columns. Its rows are 1, 0; then 0, 1; then 0, 0. Multiplying B by a column with entries u and v returns u, v and zero. These matrices represent the exact maps already used. Preserve the row breaks and the order of the coordinates; an unordered list of their entries would not supply the same calculation.

The product AB has shape two by two because a three-entry row of A pairs with a three-entry column of B. Its first output row is 1, 0. The first entry comes from 1 times 1 plus 0 times 0 plus 0 times 0. The second comes from 1 times 0 plus 0 times 1 plus 0 times 0.

Using the second row of A gives the second output row 0, 1. Therefore AB has rows 1, 0 and 0, 1, the two-by-two identity matrix. On any integer pair, the first output coordinate is u and the second is v. This entry calculation agrees with the earlier coordinate proof and keeps the identity size attached to the result.

The product BA is also defined. A two-entry row of B now pairs with a two-entry column of A, producing a three-by-three matrix. Its first row is 1, 0, 0 and its second row is 0, 1, 0. Those rows copy the first two coordinates of a triple. Both are complete calculated rows, including their zero entries in the third column.

The final row of BA is 0, 0, 0 because the final row of B contains only zeros. The resulting matrix has diagonal entries one, one and zero. The three-by-three identity I3 would instead have a one in its final diagonal position. Apply BA to (0, 0, 1): it returns (0, 0, 0), whereas I3 returns (0, 0, 1).

Both products passed their shape requirements. AB is two by two and BA is three by three; neither is an undefined reverse product. Their different identity behavior follows from their calculated entries and domains. A correct shape permits calculation, while the actual output entries determine whether the resulting map returns every input unchanged. Keep these two questions separate in the written explanation.

Name the side without losing the domain

For A, a right inverse is a map B with AB = I2 under our rightmost-first convention. B appears on the right in the written product and acts first. The identity belongs to A’s output domain, the integer pairs. Our B satisfies that condition for every pair, so it is a right inverse of A in the declared sense.

The same equality AB = I2 also says that A is a left inverse of B. Here A appears on the left of B, and the identity belongs to B’s input domain. This is another description of the equality we proved, with the role of the map being discussed changed. It supplies no new calculation of BA and makes no identity claim about triples.

A two-sided claim for this pair would also require BA = I3. That is the reverse identity already rejected by the triple with third coordinate seven and by the last row of BA. Writing only I in both places could hide which inputs the identity must preserve. Retain I2 and I3, or spell out identity on pairs and identity on triples.

Show why another return rule cannot restore A completely

The loss in A prevents any function from undoing it on every triple. Compare (1, 2, 3) and (1, 2, 4). These are different triples, but A sends both to the same pair (1, 2). Suppose a function C could restore every triple after A. Applied to that one pair, C would have to return both of those different triples.

A function assigns one output to a given input. It cannot send the same pair to two different triples as required by those two restoration demands. Therefore A has no left inverse on the full integer-triple domain. Consulting a separate saved third coordinate would enlarge the input available to the return rule and create a different problem; it is not a function of A’s pair output alone.

B has a corresponding limitation in the other direction. Whatever pair is supplied to B, the resulting triple has last coordinate zero. No function D acting before B can make BD return (0, 0, 1), because B still appends zero to D’s pair output. Thus B has no right inverse that makes BD the identity on all integer triples.

These arguments concern the explicitly defined A and B and their full declared domains. They do not claim that every pair of rectangular matrices has the same entries or that every reverse composition loses a third coordinate. The particular definitions establish the restoration and loss here. No general matrix-rank theorem is needed for either the collision argument or the unreachable-output argument.

Keep working controls within their scope

Starting with zero inputs gives successful results in both directions: the zero pair returns as a pair, and the zero triple returns as a triple. That control agrees with the formulas. It does not remove the failed nonzero-third-coordinate case. The statement BA = I3 would have to include that case as well, so testing only zeros would miss the decisive distinction.

The first two coordinates are not the source of the reverse failure. For example, starting with (4, -3, 7) returns (4, -3, 0), retaining the changed second coordinate exactly. Negative values require no rounding or sign adjustment in these maps. The reverse output differs only at the third position, whose supplied replacement is fixed by B rather than calculated from the discarded value.

A finite control sheet can use the seven integers from minus three through three. They give forty-nine ordered pairs and 343 ordered triples. Exact enumeration restores all forty-nine pairs under AB. Under BA, precisely the forty-nine triples with third coordinate zero return unchanged; the other 294 do not. These finite controls agree with the arbitrary-coordinate proof, which also covers integers outside that small sheet.

Keep the two matrix products beside these controls. AB has four entries to check and BA has nine, with their different shapes preserved. The pair restoration formula and the triple output formula explain what those entries do. A collection of favorable individual inputs would not replace the full-domain identity argument, while the one failed triple already disproves the proposed reverse identity.

Bound the actual dispenser accounts

The Ounona 22069454 dispenser account has eight images showing a clear-front wall unit, a lid-opening key and dedicated mounting fasteners. Bathroom fixtures and towels are styling props. Printed capacity, dimensions and waste-control language are unverified claims; no installation or dispensing result is established here.

The FurnOrama FU2600736 account names a silver, polished-chrome dispenser. Its single image shows an ornate pump, threaded shank and reservoir bottle with a warmer gold-toned appearance. Those observations remain separate; no finish or compatibility claim is verified. Neither account supplies the maps or their proof.

Distinguish the neighboring questions

The matrix-shape compatibility guide has a defined AB and an undefined BA because its reverse inner sizes differ. Here both products are defined. The question is which domain each calculated composition restores, with a failed identity demonstrated by an actual triple.

The inverse position-map guide constructs a return list for a complete permutation and separately rejects a repeated-index input. Here a valid one-sided inverse already restores every pair. The additional decision is whether its reverse composition also restores every triple, without borrowing a permutation return list.

Save AB = I2 beside the pair path (u, v) to (u, v, 0) to (u, v). Save BA(r, s, t) = (r, s, 0) beside the condition t = 0 for restoration. Together these records show both the successful one-sided inverse and the precise failure of the reverse identity on the original full triple domain.

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