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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 $n$.

Does it follow that $f$ is homotopic to $g$?

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 $X_i$.

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 minimal Sullivan algebras, however. Perhaps some additional condition is required above, then.

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A phrase to look up in this context is Milnor's $lim^1$-sequence. –  Tyler Lawson Nov 10 '11 at 3:20
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up vote 6 down vote accepted

Tom,

In partial atonement for steering you totally wrong when you asked me this in person yesterday, let me amplify on the answers.

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?)

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.

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 $$ 0\to lim C^\star(X;\mathbb Q)\to \Pi_nC^\star(X_n;\mathbb Q)\to\Pi_nC^\star(X_n;\mathbb Q)\to 0. $$

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To make it go wrong even rationally, you have to start with $X_n$'s that have infinite-dimensional rational cohomology. $lim^1$ of an inverse sequence of finite dimensional vector spaces $V_i$ over a field is always zero, because the inverse sequence satisfies the Mittag-Leffler condition. That is, for every $n$, the images of $V_{n+i}$'s in $V_n$ eventually stabilize as $i\to\infty$. –  Sergey Melikhov Nov 10 '11 at 15:46
    
Yes. Also, to make it go wrong when the $X_n$ are the skeleta of $X$ you have to let $Y$ be something other than an Eilenberg-MacLane space. –  Tom Goodwillie Nov 11 '11 at 4:54
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No.

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.

It is true that there are no nontrivial phantom maps between (simply-connected at least) rational spaces.

MORE:

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).

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.)

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Is there a salvage (in this case, some property separating the categories Top and CDGA)? Or do these computations need to be done on a case-by-case basis? –  Thomas Belulovich Nov 10 '11 at 4:58
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Tom, I think you should accept this answer rather than mine. –  Tom Goodwillie Nov 11 '11 at 4:55
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I once read in a paper of McGibbon about the following example:

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.

This gives an explicit counterexample.

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