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. 



Homotopic maps out of cofibration sequences
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
$$
0\to lim C^\star(X;\mathbb Q)\to \Pi\_nC^\star(X\_n;\mathbb Q)\to\Pi\_nC^\star(X\_n;\mathbb Q)\to 0.
$$


