Alphabeta Math
False statementConstruction: AI-adaptedVerification: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-01
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The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

  • Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
  • AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
  • AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

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FALSE: the left and right internal homs agree in every monoidal category

Statement

The left and right internal homs agree in every monoidal category.

Facts & Assumptions

Given: Let R be the ring of upper-triangular 2×2 matrices over a field k, let N=ke12⊂R, and work in the monoidal category of R-bimodules with tensor product ⊗R. Set X:=R/N and Y:=R.

[L1]

Left and right internal homs are defined separately as right adjoints to X⊗− and −⊗X (Left-closed, right-closed, and biclosed monoidal categories).

[L2]

An (R,R)-bimodule has commuting left and right actions, and balanced pairings induce unique maps from tensor products ((S,R)-bimodules and commuting left and right scalar actions, Universal property of the tensor product for balanced maps into abelian groups).

Refutation

technique · direct
1.1L1L2givenconstructalgebra

For bimodules A,X,Y, a bimodule map h:A⊗RX→Y corresponds to the bimodule map a⟼(x⟼h(a⊗x)) from A to Hom⁡−R(X,Y), where (rφs)(x)=rφ(sx). The inverse is evaluation, (a,x)↦φa(x); balancing and the two bimodule actions make both constructions well defined by [L2]. Thus Hom⁡−R(X,−) is right adjoint to −⊗RX. Similarly, maps X⊗RA→Y correspond to maps A→Hom⁡R−(X,Y), with (rψs)(x)=ψ(xr)s. Hence [L1] identifies these as the right and left internal homs respectively.

2.1step 1.1givenalgebra

A right R-linear map f:R/N→R is determined by a:=f(1+N), and it is well defined exactly when aN=0. Thus [X,Y]≅ann⁡ℓ(N)={0b0c:b,c∈k}.

2.2step 1.1givenalgebra

A left R-linear map g:R/N→R is determined by b:=g(1+N), and it is well defined exactly when Nb=0. Thus ⟨X,Y⟩≅ann⁡r(N)={ab00:a,b∈k}.

3.1step 2.1step 2.2L1algebra

These two bimodules are not isomorphic: every element of ann⁡r(N) is annihilated on the left by e22, while e22⋅e22=e22≠0 in ann⁡ℓ(N). By [L1], the left and right internal homs therefore need not agree.

4.1step 3.1∎

Hence the statement is false.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

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Sources