Abstract:Leighton's graph covering theorem says that two finite graphs with a common cover have a common finite cover. We present a new proof of this using groupoids, and use this as a model to prove two generalisations of the theorem. The first generalisation, which we refer to as the ball-restricted version, restricts how balls of a given size in the universal cover can map down to the two finite graphs when factoring through the common finite cover -this answers a question of Neumann from [11]. Secondly, we consider… Show more
“…Alternative proofs and various generalisations of this result can be found, e.g., in [Neu10], [BaK90], [SGW19], [Woo21], [BrS21], and references therein. Does a similar result hold for any CW-complexes, i.e.…”
How many two-cells must two finite CW-complexes have to admit a common, but not finite common, covering? Leighton's theorem says that both complexes must have two-cells. We construct an almost (?) minimal example with two two-cells in each complex.
“…Alternative proofs and various generalisations of this result can be found, e.g., in [Neu10], [BaK90], [SGW19], [Woo21], [BrS21], and references therein. Does a similar result hold for any CW-complexes, i.e.…”
How many two-cells must two finite CW-complexes have to admit a common, but not finite common, covering? Leighton's theorem says that both complexes must have two-cells. We construct an almost (?) minimal example with two two-cells in each complex.
“…Alternative proofs and generalisations of this result can be found, e.g., in [Neu10], [BaK90], [SGW19], [Woo21], [BrS21], and references therein. For two-dimensional CW-complexes (and cell coverings) the similar assertion is false:…”
How many 2-cells must two finite CW-complexes have to admit a common, but not finite common, covering? Leighton’s theorem says that both complexes must have 2-cells. We construct an almost (?) minimal example with two 2-cells in each complex.
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