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The notion of a fibration has a nice geometric intuition of one topological space (a fiber) being parametrized by another topological space (the base) -- this is taken from the Wikipedia entry on Fibration.

Now, I would like to know if there is an analogous geometric picture for the situation of cofibrations. Like that of a parametrized inclusion of a space into another space (apologies if this is nonsense).

camilo
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1 Answers1

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Yes, there are precisely the dual diagrams defining a cofibration. The way to think about them -- the way I think about them, at least -- is just that they're sufficiently nice inclusions, such that the subspace has a bit of "wiggle room". This is made precise in the definition of a "neighborhood deformation retract (NDR) pair", which you can read about in May's Concise Course in Algebraic Topology. Maybe the best example to keep in mind is when you turn an arbitrary map into a cofibration with the "mapping cylinder" construction; then it's obvious that the subspace has plenty of wiggle room.

Aaron Mazel-Gee
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  • From this perspective, the important aspect of a fibration is not that it has a well-defined fiber (as a homotopy type), but rather that it behaves well with respect to *lifting* maps *through* the morphism. Dually, the important aspect of a cofibration is that it behaves well with respect to *extending* maps *along* the morphism. – Aaron Mazel-Gee Nov 13 '12 at 22:47
  • Thanks for the answer. So for cofibration there is not a "dual" geometric intuition to that of the fibration? I mean, I understand the algebraic duality between fibration and cofibration, but cannot grasp the geometric idea (if any) so easily. – camilo Nov 14 '12 at 15:02
  • (cont.) An instance of the nice intuition for fibration is for fiber bundles. Here one may have non-trivial bundles, as is always exemplified by the Moebius band. Is there a "dual" geometric picture to this situation? that is, where you define a cofibration "locally" which is nevertheless globally nontrivial? – camilo Nov 14 '12 at 15:06
  • @camilo: I'm not sure; there might be some connection that I just don't see though. However, here is another duality-type statement. A fibration sits in a fiber sequence, which loops backwards indefinitely (look up the "Puppe sequence", or maybe Dold-Puppe) and induces a lexseq for homotopy classes of maps *in* (e.g. from spheres -- this is the lexseq in homotopy). On the other hand, a cofibration sits in a cofiber sequence, which suspends forward indefinitely, and this induces a lexseq for homotopy classes of maps *out* (e.g. to Eilenberg-MacLanes -- this is the lexseq in cohomology). – Aaron Mazel-Gee Nov 15 '12 at 01:25
  • In this setup, if a fiber sequence $X \rightarrow Y \rightarrow Z$ is trivial (so $Y \simeq X \times Z$) then the connecting maps in homotopy will be trivial; dually, if a cofiber sequence $A \rightarrow B \rightarrow C$ is trivial (so $C \simeq B \vee \Sigma A$) then the connecting maps in cohomology will be trivial. I think these facts get at what you're asking. – Aaron Mazel-Gee Nov 15 '12 at 01:29
  • That's nice, thanks. If you ever come across the geometric situation for nontrivial cofibration, as the example of a Moebius band is for nontrivial fibration, and pay a visit to this question, I will be very grateful! – camilo Nov 15 '12 at 15:51
  • I'll let you know if I come across anything, but I think I've read enough at this point that I would've seen something if it existed (beyond what I've already said). The point of both of these notions -- from a model category point of view, where one converts (up to weak equivalence) any morphism to a fibration or a cofibration willy-nilly -- is that they give us different measurements the "difference" of a morphism, i.e. its failure to be a homotopy equivalence. [cont] – Aaron Mazel-Gee Nov 15 '12 at 22:56
  • ... Perhaps the you should read about model categories to see these two notions on a more balanced footing. Although I will readily admit that this is going in a *less* geometric direction. Also along the lines of this "difference" perspective is that when you pass to the *stable* homotopy category, these notions of difference become equivalent; indeed, that's kind of the entire point. – Aaron Mazel-Gee Nov 15 '12 at 22:58
  • @camilo (I just realized I didn't @ you on the previous comments): Anyways, many maps that you come across will be cofibrations -- for instance, any inclusion of CW-complexes -- and so anytime this induces a nontrivial long exact sequence in cohomology (i.e. it doesn't split into a bunch of short exact sequences, i.e. the connecting homomorphisms aren't all zero), then it must have been a nontrivial cofibration. The most basic example is perhaps the standard inclusion $S^1 \hookrightarrow D^2$, or even $S^0 \hookrightarrow D^1$ for that matter... – Aaron Mazel-Gee Nov 15 '12 at 23:01
  • These two examples are trivial cofibrations right? If yes, what is the simplest example of a nontrivial cofibration? – camilo Nov 19 '12 at 17:16
  • No, they are not trivial! A trivial cofibration is a cofibration which is a weak homotopy equivalence. Equivalently, the homotopy groups of the cofiber (i.e. the mapping cone) all vanish. – Aaron Mazel-Gee Nov 19 '12 at 19:02
  • I just saw this randomly, and noticed that I made a mistake in my previous comment. You can't check weak homotopy equivalences by taking mapping cones; for instance, there are noncontractible ("acyclic") spaces whose suspensions are contractible. – Aaron Mazel-Gee Dec 22 '20 at 18:32