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May 28, 2016 at 19:57 history edited Dylan Thurston CC BY-SA 3.0
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Mar 31, 2011 at 22:56 answer added Charles Rezk timeline score: 8
Mar 31, 2011 at 10:33 vote accept Dylan Thurston
Mar 31, 2011 at 2:09 answer added Tom Goodwillie timeline score: 9
Mar 29, 2011 at 16:50 answer added Sam Isaacson timeline score: 15
Mar 29, 2011 at 16:35 comment added Dylan Thurston @Sam @Tyler: I would like an actual space, not a spectrum.
Mar 29, 2011 at 16:34 history edited Dylan Thurston CC BY-SA 2.5
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Mar 29, 2011 at 13:42 comment added Tyler Lawson @Sam: If you take a mod-2 Moore space and suspend it, you should be able to do that unstably.
Mar 29, 2011 at 5:53 comment added Sam Isaacson Sorry; that should be $X = M \vee \Sigma M$
Mar 29, 2011 at 5:52 comment added Sam Isaacson @Mariano, thanks for pointing out the subtlety I missed. I think you can produce a counterexample stably by looking at $X = M \wedge \Sigma M$ where $M$ is the mod $2$ Moore spectrum. Let $f:X \to X$ be the map which on $\Sigma M$ is the inclusion of the $\Sigma M$ summand and on $M$ is the sum of the inclusion of the $M$ summand and the essential composition $g:M\to S^1 \to \Sigma M$. The map $f$ is the identity on $H\mathbb{Z}$, but since $g$ is nonzero on $H\mathbb{Z}/2$, the map $f$ is not the identity on $H\mathbb{Z}/2$.
Mar 29, 2011 at 5:09 comment added Mariano Suárez-Álvarez @Sam: the question is: is that isomorphism also the identity? Notice that an endomorphism of short exact sequences which has identites on the left and on the right does not necessarily have an identity in the middle.
Mar 29, 2011 at 4:46 comment added Sam Isaacson BTW, the simplest example of the failure of the UCT to split naturally that I know of is the map from a mod $n$ Moore space to a sphere that collapses the bottom cell.
Mar 29, 2011 at 4:43 comment added Sam Isaacson Doesn't the UCT imply that no such counterexample exists? The splitting isn't natural, but the short exact sequence is. More simply, the multiplication by $2$ map on $\mathbb{Z}$ induces a LES in the homology of $X$ and if $f$ is a homology iso, the 5 lemma shows it is an iso with mod $2$ coefficients. Or am I missing something?
Mar 29, 2011 at 4:11 history asked Dylan Thurston CC BY-SA 2.5