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Apr 4, 2013 at 4:01 comment added Andy Putman Properness is certainly needed; it is not hard to construct counterexamples without it. I do not know whether assuming that $f$ is cellular is good enough.
Apr 4, 2013 at 3:47 comment added Cusp @Andy: Thanks. I am a little curious about the property that the map $f$ should proper. How important this condition really is? What I mean is can I replace this condition by a more simple condition like taking $f$ to be cellular or something of that sort.
Apr 4, 2013 at 3:42 vote accept Cusp
Apr 4, 2013 at 3:42 vote accept Cusp
Apr 4, 2013 at 3:42
Apr 4, 2013 at 3:31 comment added Andy Putman @Vel Nias : It depends on your conventions. All right-thinking people consider the empty set non-connected (and certainly not contractible), but just in case the OP has other ideas I thought I'd avoid this issue by assuming surjectivity.
Apr 4, 2013 at 3:02 comment added Vidit Nanda Andy: this is how I have seen (and stated) this theorem also, but does one really need to say "surjective"? The empty set does not have the homotopy type of a point, does it?
Apr 3, 2013 at 22:13 comment added Ricardo Andrade I just want to add a small, tangential refinement. In Andy's conclusion for finite dimensional complexes, we can actually do a bit better. If the fibres of a map between finite CW-complexes are all contractible and locally contractible, then the map is actually a "cell-like map", and thus a simple homotopy equivalence.
Apr 3, 2013 at 21:10 history answered Andy Putman CC BY-SA 3.0