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Apr 6, 2020 at 16:26 comment added Gustavo Granja The mapping telescope of a sequence of maps is defined in section 3F in Hatcher and a relevant special case is used in the proof of Lemma 2.34 in Hatcher. It's homology groups or homotopy groups are computed by applying the homology or homotopy group functor to the sequence of maps and taking the colimit. Thus when $M$ has the homotopy type of a cell complex, the canonical map from the mapping telescope to $M$ is a homotopy equivalence.
Apr 6, 2020 at 16:17 answer added Gustavo Granja timeline score: 6
Apr 6, 2020 at 16:08 history edited erz CC BY-SA 4.0
added a special case to the question
Apr 6, 2020 at 16:05 comment added erz @GustavoGranja could you please elaborate? Especially for somebody who is used to general topology as opposed to the constructions common in algebraic topology.
Apr 6, 2020 at 16:02 comment added erz @MarkGrant Fubini-Study metric is probably also highly non-compact being а quotient of the Hilbert sphere with respect to $S^1$.
Apr 6, 2020 at 15:19 comment added Gustavo Granja You can replace $\mathbb CP^\infty$ by the telescope of the inclusions $\mathbb CP^n \hookrightarrow \mathbb CP^{n+1}$ which is locally compact and $\sigma$-compact.
Apr 6, 2020 at 9:46 comment added Mark Grant There is however a metric on $\mathbb{C}P^\infty$ for which each inclusion $\mathbb{C}P^n\subseteq\mathbb{C}P^\infty$ is an isometric embedding (Fubini-Study). One could ask if the construction of the phantom maps still works with this coarser topology.
Apr 6, 2020 at 9:41 comment added Mark Grant @erz Ah, OK, it is more subtle than I thought. It appears that $\mathbb{C}P^\infty$ with the weak (CW) topology may not be metrizable, which would imply that it's also not locally compact. I haven't checked, but I assume that the constructions of these phantom maps use the weak topology.
Apr 6, 2020 at 3:29 comment added Wlod AA A trivial remark: homomorphisms of homotopy groups which are induced by $\ \phi\ $ are trivial.
Apr 5, 2020 at 15:54 comment added erz @MarkGrant Unfortunately, I am not competent to see if this answers my question. Is $\mathbb{C}P^\infty$ locally compact and $\sigma$-compact? Is every compact subset of it included in a finite skeleton?
Apr 5, 2020 at 11:34 comment added Wlod AA ANRs are locally contractible.
Apr 5, 2020 at 6:36 comment added Mark Grant There exist phantom maps $\mathbb{C}P^\infty\to S^3$. These are non-null homotopic maps which become null-homotopic when restricted to each finite skeleton. Doesn't this answer your question in the negative?
Apr 5, 2020 at 6:20 history asked erz CC BY-SA 4.0