Timeline for do spectra have diagonal maps?

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8 events

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Jan 2, 2012 at 1:12	comment	added	Tom Goodwillie		Jeff, there certainly seem to be coalgebras that are not bounded below: start with a commutative differential graded coalgebra over $\mathbb Q$.
Dec 29, 2011 at 1:09	comment	added	Jeff Smith		So, suspension spectra are commutative coalgebras in spectra. Is there a converse? Is the homotopy theory of $E_\infty$ coalgebras in spectra the homotopy theory of $HZ$ local spaces.
Dec 28, 2011 at 20:54	comment	added	Peter May		Thanks, Tom, of course that is another typo. It would be nice if comments could also be edited.
Dec 28, 2011 at 19:51	comment	added	Tom Goodwillie		By $X\wedge X\to X\times X$ Peter presumably meant $X\vee X\to X\times X$.
Dec 28, 2011 at 18:43	comment	added	Fernando Muro		@Peter You're right about the existence of an unstable diagonal map to the smash product of based spaces, but I don't know what you mean with 'the canonical map $X\wedge X\rightarrow X\times X$'
Dec 28, 2011 at 18:34	comment	added	Peter May		This is an edit of my comment above (since I don't know how to edit it properly): $\sma$ should be $\wedge$, $\rtarr$ should be $\to$.
Dec 28, 2011 at 18:26	comment	added	Peter May		In based spaces, $X\sma X$ is a quotient of $X\times X$, so we do have a diagonal, namely the composite of the diagonal $X\to X\times X$ and the quotient map $X\times X\to X\sma X$. Of course, it is not a categorical diagonal, but it is often useful. Since the suspension spectrum functor commutes with smash products, this does give suspension spectra a diagonal. It is used all the time in duality theory. A relevant conceptual point is that, in spectra, the canonical map $X\wedge X\rtarr X\times X$ is an equivalence, just as in Abelian categories.
Dec 28, 2011 at 18:17	history	answered	Fernando Muro	CC BY-SA 3.0