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Are ##S^1 ## -knots (meaning homeomorphisms of ##S^1 ## into ## R^3## or ## S^3## , classified under isotopy) that are just C

^{1}- knots orientable? If we were working " in the smooth category " , then we could just say that we can pushfoward the orientation form ##dx## by a diffeomorphism, and the diffeomorphism would give us a nowhere-zero form, which would be positive or negative depending on whether the diffeo. is orientation-preserving or not. But if we don't know if the embedding of the knot is a smooth embedding , can we still guarantee orientability? I think we can say that if ##S^1## is embedded, then it is a submanifold, and submanifolds admit tubular neighborhoods, meaning the normal bundle of the knot is trivial , which implies orientability. What if we only know that the embedding is a homeomorphism, but we do not address issues of degree of smoothness, i.e., we do not know whether the embedding is even ##C^1## , can we determine orientability?

Also: outside of the "smooth category", where we cannot work with differential forms, do we define orientability in terms of the fundamental class (i.e., the generator of the top homology class )? Then I guess we would have to decide whether the map induced in top homology preserves this class ?

Thanks.