Singular chains as an HZ-module spectrum - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T05:12:54Z http://mathoverflow.net/feeds/question/67165 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/67165/singular-chains-as-an-hz-module-spectrum Singular chains as an HZ-module spectrum Urs Schreiber 2011-06-07T17:18:33Z 2011-06-07T18:10:53Z <p>For $R$ any ring and $H R$ its Eilenberg-MacLane spectrum -- a ring spectrum -- there is an equivalence between the $\infty$-categories of $H R$-module spectra and that of unbounded chain complexes of $R$-modules - a stable version of the Dold-Kan correspondence. </p> <p>At least in good cases such as $R = \mathbb{Z}$ this refines also to an equivalence between $H \mathbb{Z}$-algebra spectra and unbounded dg-rings -- a stable version of the monoidal Dold-Kan correspondence. (A presentation of this by a Quillen equivalence has been given by Shipley, see <a href="http://ncatlab.org/nlab/show/algebra+spectrum" rel="nofollow">http://ncatlab.org/nlab/show/algebra+spectrum</a>).</p> <p>Now for $X$ a sufficiently nice topological space, one would hope that under this equivalence the "spectrum of integral homology chains" $(\Sigma^\infty_+ \Omega X) \wedge H \mathbb{Z}$ on the loop space of $X$ is identified, up to equivalence, with the ordinary chain complex of singular homology chains $C_\bullet(\Omega X, \mathbb{Z})$, both as $\infty$-modules and as $\infty$-algebra objects.</p> <p>This must be easy to see, but I am a little stuck. Can anyone help?</p>