cardinality of final coalgebras in Top - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T04:37:09Z http://mathoverflow.net/feeds/question/1266 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/1266/cardinality-of-final-coalgebras-in-top cardinality of final coalgebras in Top Reid Barton 2009-10-19T19:25:44Z 2009-10-20T16:28:01Z <p>Let P be a polynomial functor from Top to Top, by which I mean a functor of the form P(X) = ∐<sub>i &ge; 0</sub> S<sub>i</sub> &times; X<sup>i</sup> where the S<sub>i</sub> are finite sets, all but finitely many of which are empty. Corresponding to P there is an ordinary polynomial p(x) = &Sigma;<sub>i &ge; 0</sub> |S<sub>i</sub>| x<sup>i</sup>. I believe that there is a final P-coalgebra X<sub>P</sub> in Top; the structure map X<sub>P</sub> -> P(X<sub>P</sub>) is a homeomorphism; the underlying set of X<sub>P</sub> is the final P-coalgebra in Set; and a basis for the open sets of X<sub>P</sub> is given by the preimages of the points of the targets of the maps X<sub>P</sub> -> P<sup>(n)</sup>(X<sub>P</sub>) -> P<sup>(n)</sup>(&bull;) as n ranges over all nonnegative integers.</p> <p>Examples are the Cantor space, corresponding to p(x) = 2x, and the space of binary trees, corresponding to p(x) = x^2 + 1.</p> <p>My question is: if P and Q are two polynomial functors, such that neither p(x) nor q(x) is of the form x + k, and the spaces X<sub>P</sub> and X<sub>Q</sub> are homeomorphic, does it follow that the polynomials p(x) - x and q(x) - x have a common root?</p> <p>More generally, consider all the spaces that can be formed starting with the collection of spaces X<sub>P</sub> (p(x) not of the form x + k) by taking disjoint unions and products. Can I assign an algebraic number to each of these spaces in a way which is a homeomorphism invariant and commutes with sums and products?</p> http://mathoverflow.net/questions/1266/cardinality-of-final-coalgebras-in-top/1449#1449 Answer by Tom Leinster for cardinality of final coalgebras in Top Tom Leinster 2009-10-20T16:28:01Z 2009-10-20T16:28:01Z <p>I don't know, but one could begin to investigate your first question by considering polynomials of the form <i>ax + k</i>, where <i>a</i> is a natural number greater than 1. Maybe you already know that the answer to the question is "yes" in that case? (If so, I'd be interested to see your answer.) </p> <p>A point related to this, and to the general idea about algebraic numbers, is discussed here:</p> <p><a href="http://golem.ph.utexas.edu/category/2007/04/reportback_on_bmc.html#c009149" rel="nofollow">http://golem.ph.utexas.edu/category/2007/04/reportback_on_bmc.html#c009149</a></p> <p>There are also close relations with the theory described in <a href="http://arxiv.org/abs/math.DS/0411343" rel="nofollow">arXiv:math.DS/0411343</a> (an improved version of which appears in <i>Real and Complex Singularities</i>, ed. Paunescu et al, World Scientific 2007). But there are differences too, e.g. you allow products but I didn't.</p>