Teruaki Kitano, Masaaki Suzuki, Masaaki Wada
Published 2005-10-11Version 1
If phi: G-->G' is a surjective homomorphism, we prove that the twisted Alexander polynomial of G is divisible by the twisted Alexander polynomial of G'. As an application, we show non-existence of surjective homomorphism between certain knot groups.
Comments: Published by Algebraic and Geometric Topology at http://www.maths.warwick.ac.uk/agt/AGTVol5/agt-5-51.abs.html
Journal: Algebr. Geom. Topol. 5 (2005) 1315-1324
Twisted Alexander polynomial of links in the projective space
Twisted Alexander polynomials and a partial order on the set of prime knots
Twisted Alexander polynomials of 2-bridge knots associated to metacyclic representations
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