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CounterexampleConstruction: AI-generatedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-07-31
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On a two-element chain, an incidence function with a zero diagonal value is not convolution-invertible

Statement refuted

Every incidence function on a finite poset is convolution-invertible.

Facts & Assumptions

Given: A two-element chain 0<10<1, a nonzero commutative ring RR, and the incidence function ff with f(0,0)=0Rf(0,0)=0_R, f(0,1)=0Rf(0,1)=0_R, and f(1,1)=1Rf(1,1)=1_R.

[L1]

An incidence function is convolution-invertible exactly when every diagonal value is a unit (An incidence function is convolution-invertible if and only if every diagonal value is a unit).

[F1]

The convolution identity has δ(0,0)=1R\delta(0,0)=1_R (The delta and zeta incidence functions).

Counterexample

technique · direct
1.1

If gg were a convolution inverse, evaluation at (0,0)(0,0) would give (f∗g)(0,0)=f(0,0)g(0,0)=0R(f*g)(0,0)=f(0,0)g(0,0)=0_R.

given
2.1

But an inverse equation requires (f∗g)(0,0)=δ(0,0)=1R(f*g)(0,0)=\delta(0,0)=1_R, and 0R≠1R0_R\ne1_R because the ring is nonzero.

step 1.1F1
3.1

Therefore ff is not invertible, in agreement with [L1] because its diagonal value 0R0_R is not a unit.

step 2.1L1∎

Depends on

Used by

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Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 23 results over 15 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

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