Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

$X$ is said to be the homotopy direct limit of the sequence of subsets $X_1\subset X_2\subset ...$ if the projection $\cup_i X_i\times [i,i+1] \rightarrow X$ is a homotopy equivalence.

The following is true:

Suppose $X, Y$ are homotopy direct limits of the sequences $X_1\subset X_2 \subset ...$ and $Y_1 \subset Y_2 \subset ...$ respectively. Then if $f: X\rightarrow Y$ and each $f|_{X_i}: X_i \rightarrow Y_i$ is a homotopy equivalence then $f$ itself is a homotopy equivalence.

My Question: Does anyone know off the top of his or her head whether the map $f$ can be replaced by a sequence of homotopy equivalences $f_i: X_i \rightarrow Y_i$ where the resulting diagram is homotopy commutative (i.e. $(Y_i\subset Y_{i+1})\circ f_i$ is always homotopic to $f_{i+1}\circ (X_i\subset X_{i+1})$)? I.e. are the spaces still homotopy equivalent? Thanks!

share|improve this question
add comment

1 Answer

up vote 3 down vote accepted

Yes, the spaces are still homotopy equivalent.

Your situation with homotopy direct limits is a special case of a homotopy colimit and the defining property of homotopy colimits is exactly what you're asking for. See, for example, Theorem 1.4 of Rainier M. Vogt's "Homotopy limits and colimits". Be warned, there is a lot to read and to prove in order to get to this result.

However, since you are interested only in directed systems we should be able to find a more direct (no pun intended) proof. At our disposal we have the homotopy equivalences between $X_i$ and $Y_i$ and the homotopies assuring associativity of our diagram. With these tools can you see a way of constructing a natural-looking continuous map between $\cup_i X_i\times [i,i+1]$ and $\cup_i Y_i\times [i,i+1]$? Can you construct a homotopy inverse?

I did the former part, but I failed to find the homotopy inverse. I hope you'll have more luck than me, though!

EDIT: See the comments for a discussion on how such a homotopy inverse might be constructed.

share|improve this answer
Thanks Roar! I'm quite relieved. Yeah the obvious thing to do is $(x, t) \mapsto (f_i(x), 2t), 0\leq t\leq 1/2, (x,t) \mapsto (h_i(x,t), i+1)$ where $h$ is the homotopy between $\iota \circ f_i$ and $ f_{i+1} \circ \iota$. I'll have to think about an inverse. Thanks very much. –  Tim kinsella May 19 '12 at 20:30
some typos in my definition but I hope we have in mind the same map "move with double speed for half the interval and use the second half to homotope" –  Tim kinsella May 19 '12 at 20:41
Yes, that's exactly the map I had in mind, Tim! This answer to another question is related to our situation, but there the existence of a homotopy inverse is guaranteed by an algebraic argument. –  Roar Stovner May 19 '12 at 22:16
OK I think that the same thing will work for an inverse, $G$ (I am basically just adapting the proof from the case where the $f_i$s are restrictions): If $g_i$ is an h-inverse to $f_i$ then in fact the "inverse diagram" is also h-commutative: $\iota \circ g_i =_h g_{i+1}\circ f_{i+1} \circ \iota \circ g_i =_h g_{i+1}\circ \iota \circ f_i \circ g_i =_h g_{i+1}\circ \iota$. Now if $X_{\Sigma} := \cup_i X_i \times [i, i+1]$ then I think $X_{\Sigma}$ is a h-direct limit of the spaces $U_i := \cup_{k\leq i} X_k \times [k,k+1]$. And each $U_i$ def retracts onto $X_i \times {i}$ on which... –  Tim kinsella May 19 '12 at 22:32
$G\circ F$ has the form $f_i \circ g_i$ which is homotopic to the $id|_{X_i}$. And I think the same argument will show that $F\circ G|_{U_i}$ is also homotopic to the identity. So if we invoke the proposition for the case when the $f_i$s are restrictions of some big $f$ we should get our generalization? Does this make any sense? –  Tim kinsella May 19 '12 at 22:40
show 2 more comments

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.