Homotopic maps out of cofibration sequences - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T16:19:54Z http://mathoverflow.net/feeds/question/80538 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/80538/homotopic-maps-out-of-cofibration-sequences Homotopic maps out of cofibration sequences Thomas Belulovich 2011-11-10T00:37:32Z 2011-11-10T12:21:19Z <p>Consider a model category $\mathcal{C}$ and a sequence of cofibrations $0 \to X_0 \to X_1 \to X_2 \to \dots$ lying in $\mathcal{C}$. Let $X$ be the colimit of this sequence. Suppose furthermore that we have two maps $f,g : X \to Y$, where $Y$ is fibrant, such that the restrictions $f_n : X_n \to Y$ and $g_n : X_n \to Y$ are homotopic for every $n$. </p> <p>Does it follow that $f$ is homotopic to $g$? </p> <p>A simpler version: if $h : X \to Y$ is a morphism such that the restrictions $h_n$ are equivalences, then $h$ is an equivalence, since $X$ is the homotopy colimit of the $X_i$. </p> <p>Context: I saw this result when studying rational homotopy theory, in the context of Sullivan algebras in the category of CDGAs. The result was stated in terms of a filtration of <em>minimal</em> Sullivan algebras, however. Perhaps some additional condition is required above, then. </p> http://mathoverflow.net/questions/80538/homotopic-maps-out-of-cofibration-sequences/80540#80540 Answer by Jeff Strom for Homotopic maps out of cofibration sequences Jeff Strom 2011-11-10T01:27:58Z 2011-11-10T12:04:52Z <p>No. </p> <p>For example, if $X$ is a CW complex with skeleta $X_n$ and $f_n\simeq*$, then $f$ is a phantom map. Their homotopy classes are in bijective correspondence with $\lim^1 [\Sigma X_n, Y]$, and are frequently nonzero. For example, $\mathbb{C}P^\infty$ is the domain of nontrivial phantom maps.</p> <p>It is true that there are no nontrivial phantom maps between (simply-connected at least) rational spaces.</p> <p>MORE:</p> <p>It goes back to Milnor in the early 1960s, but in full generality I think it came a little later (perhaps Bousfield and Kan) that if $X$ is the colimit of your telescope diagram, then there is a natural short exact sequence of pointed sets $$* \to {\lim}^1 [\Sigma X_n, Y] \to [X, Y]\to \lim [ X_n, Y]\to * .$$ For phantom maps, we take the telescope to be the skeleta of a CW decomposition of $X$, and we get the identification $\mathrm{Ph}(X, Y) \cong \lim^1 [\Sigma X_n, Y]$. This comes from just looking at the long cofiber sequence of the big fold/inclusion map $\bigvee X_n \to X$, whose cofiber $\Theta_X : X\to \bigvee \Sigma X_n$ is known as the universal phantom map (it is phantom and every other phantom factors (nonuniquely) through it; see the paper "Universal phantom maps" by Gray and McGibbon).</p> <p>It is not too hard to show that if $\Sigma X$ is not a retract of a wedge of finite-dimensional spaces, then $\Theta_X\not\simeq *$, so there are nontrivial phantoms out of $X$. This is the case, for example, when the Steenrod algebra action takes elements to arbitrarily high dimension (as for $\mathbb{C}P^\infty$ or $\mathbb{R}P^\infty$ or $B\mathbb{Z}/p$, etc.)</p> http://mathoverflow.net/questions/80538/homotopic-maps-out-of-cofibration-sequences/80549#80549 Answer by John Klein for Homotopic maps out of cofibration sequences John Klein 2011-11-10T03:00:04Z 2011-11-10T03:00:04Z <p>I once read in a paper of McGibbon about the following example: </p> <p>Let $X = \Bbb RP^\infty$ and let $X_n$ be the $n$-skeleton. Then there is a canonical map $\vee_n X_n \to X$. The mapping cone of this map is identified with $\vee_n \Sigma X_n$, and the map $$X \to \bigvee_n \Sigma X_n$$ is a non-trivial phantom map in the sense that it is essential and the restriction to each $n$-skeleton is null-homotopic. </p> <p>This gives an explicit counterexample.</p> http://mathoverflow.net/questions/80538/homotopic-maps-out-of-cofibration-sequences/80586#80586 Answer by Tom Goodwillie for Homotopic maps out of cofibration sequences Tom Goodwillie 2011-11-10T12:21:19Z 2011-11-10T12:21:19Z <p>Tom, </p> <p>In partial atonement for steering you totally wrong when you asked me this in person yesterday, let me amplify on the answers. </p> <p>The obvious thing to try in seeking (in vain) a positive answer is: given two maps $X_{n+1}\to Y$ which are homotopic, and given a homotopy between the two restrictions $X_n\to Y$, try to extend the homotopy to $X_{n+1}$. Do this over and over and you have what you want. But this step cannot always succeed. For example, what if the maps $X_{n+1}\to Y$ both restrict to $0$ on $X_n$ and the existing homotopy is $0$? Then you are trying to show that two maps $X_{n+1}/X_n\to Y$ must be homotopic if the two composed maps $X_{n+1}\to Y$ are homotopic, which is clearly false. (What if $X_{n+1}$ is contractible, for example?)</p> <p>So maybe your next attempt is to show that even if you can't fix things up at every $n$ as you go along, maybe you can leave it till later: plan to somehow match up the homotopies through $X_n$ before you get to $\infty$. But that's just the sort of thing that $lim^1$ interferes with.</p> <p>This goes wrong even rationally: Take $Y$ to be a rational Eilenberg-MacLane space, so that maps into $Y$ are cohomology classes. There is an exact sequence $$0\to lim^1 H^{k-1}(X_n;\mathbb Q)\to H^k(X;\mathbb Q)\to lim H^k(X_n;\mathbb Q)\to 0,$$ which comes from the long exact sequence associated to a short exact sequence of cochain complexes $$0\to lim C^\star(X;\mathbb Q)\to \Pi_nC^\star(X_n;\mathbb Q)\to\Pi_nC^\star(X_n;\mathbb Q)\to 0.$$</p>