The homomorphism problem for trace monoids is undecidable in gen- eral, or even if the second monoid is nontrivial free or free commutative. We can also ...
It is proved to be decidable, for any given finite subset F of X ∗ , dependency alphabet (Y,τ) and mapping ϕ : F → Y ∗ , whether or not ϕ can be extended to ...
Oct 22, 2024 · We prove that two Lefschetz fibrations with multisections are isomorphic if and only if their monodromy factorizations in the relevant mapping ...
Mar 14, 2011 · If M is a monoid, M2=M×M is a monoid with coordinate-wise multiplication: that is, if (a,b), (a′,b′)∈M2, then the product is such that (a ...
Dec 28, 2016 · The problem with monoids is that unlike in groups, you can't take inverses, so the kernel doesn't give you enough information.
Missing: trace | Show results with:trace
Abstract (EN): It is proved to be decidable, for any given finite subset F of X* dependency alphabet (Y, tau) and mapping rho : F --> Y*, whether or not rho ...
Oct 4, 2006 · Problem 1: Let S be a monoid. Find a subgroup G of S with the property that any monoid homomorphism f: H-->S (H any group) has its image in G.
... Given monoids M; N , the homomorphism problem for M and N consists in deciding, given a mapping ' : F → N with F ⊆ M ÿnite, whether or not ' can be ...
$S$ is a homomorphism from the monoid $M$ to the monoid $S^S$ ; a right operation is a homomorphism into the opposite monoid of $S^S$ . We may also say that ...
Nov 1, 2023 · Monoids have richness when you ask why they aren't like groups. For example, there are Green's relations, which have an interesting dynamic, but are trivial ...
Missing: trace | Show results with:trace