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Let $(A_i,f_i)$ be a directed system of CW-complexes with colimit $A$. Further, let $g_i:A_i\to A_{i+1}$ be maps such that $g_{i+1}f_i=f_{i+1}g_i$ and $f_i\simeq g_i$.

This might or might not be the smartest way to write this data into a diagram

$$\begin{array} AA_i & \stackrel{f_i}{\longrightarrow} & A_{i+1}& \stackrel{f_{i+1}}{\longrightarrow} & A_{i+2} \\ \downarrow{=} & \stackrel{g_i}{\searrow} & \downarrow{=}& \stackrel{g_{i+1}}{\searrow} & \downarrow{=} \\ A_i & \stackrel{f_i}{\longrightarrow} & A_{i+1} & \stackrel{f_{i+1}}{\longrightarrow} & A_{i+2} \end{array} $$

Here, the rhombi are strictly commutative and the triangles up to homotopy.

Now the $g_i$ clearly induce a morphism $g$ from the colimit $A$ to itself. In fact the maps $A_i\xrightarrow{g_i}A_{i+1}\to A$ and $A_i\to A$ are even homotopic. Does this imply that $g$ is a homotopy equivalence?

If it matters, the homotopies $f_i\simeq g_i$ are not compatible.

Edit: I've found a result that says that $colim(A_i,f_i)\simeq colim(A_i,g_i)$. It feels like this should be helpful.

Edit: Maybe I am mistaken but it might actually be as easy as this: filtered colimits preserve homotopy groups. On homotopy the incuded maps ${g_i}_*$ and ${f_i}_*$ are actually equal and thus $g_*$ is the identity on homotopy groups. Since we are working with CW-complexes we are done. But maybe my head is just spinning after chasing through diagrams for two days. Is this idea correct?

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  • $\begingroup$ There's one problem with your diagram: the triangles including the two $A_i$s, the identity arrow, and $g_i$ and $f_i$ imply, if they commute, that $g_i=f_i$. Your $=$ arrows should instead be the homotopies $f_i\simeq g_i$. $\endgroup$ Commented Jun 19, 2013 at 19:31
  • $\begingroup$ @KevinCarlson, I know that the diagram doesn't automatically contain all information. It is only homotopy commutative, but including the homotopies would make it too complicated, in particular since the homotopies don't need to be compatible. $\endgroup$ Commented Jun 19, 2013 at 19:51
  • $\begingroup$ You may be able to replace the homotopy commutative diagrams with strictly commutative ones. As you seem to know this cannot always be done. This straightening problem seems quite tractable since one might be able to do this inductively. $\endgroup$ Commented Jun 20, 2013 at 13:28
  • $\begingroup$ @BabyDragon, I am not particularly familiar with all the model category issues, but I think what you say is possible, provided the $A_i$ are fibrant (true since CW-complexes) and the $f_i,g_i$ are cofibrations (here inclusions as subcomplexes). I think this is essentially the proof for the result I quoted in the edit. $\endgroup$ Commented Jun 20, 2013 at 13:57
  • $\begingroup$ @SimonMarkett was not explicitly using anything having to do with model categories (although one may be able to generalize to this context). Instead, what I was suggesting is that given any of the $A_i$, we may find a homotopy equivalence, $h_i:A\to A_i'$ so that $h_i f_{i-1}=h_i g_{i-1}$, so that you do have the triangles commuting. $\endgroup$ Commented Jun 20, 2013 at 14:56

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I think I found an answer myself and am going to add it here for completeness, and so that somebody can stop me if I am talking nonsense.

Filtered colimits preserve homotopy groups, in other words

$$\pi_n(A)=colim\ \pi_n A_i$$

Further the homomorphisms ${g_i}_*: \pi_nA_i\to \pi_nA_{i+1}$ induce the map $g_*: \pi_nA\to \pi_nA$. Since $g_i\simeq f_i$ they induce the same map on homotopy and thus $g_*=id$. Since $g_*$ is a homotopy isomorphism of CW-complexes it is a homotopy equivalence.

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  • $\begingroup$ Colimits do not preserve homotopy groups. Think about the pushout diagram of $D^{n}\leftarrow S^{n-1}\rightarrow D^n$. The pushout of this is $S^n$ but if we hit this with $\pi_n$ we get a diagram of groups, $0 \leftarrow\pi_nS^{n-1}\rightarrow 0$. $\endgroup$ Commented Jun 26, 2013 at 17:14
  • $\begingroup$ @BabyDragon Uhm, I forgot the word filtered. Then the colimit is actually a homotopy colimit. $\endgroup$ Commented Jun 26, 2013 at 17:31

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