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Prove or disprove the following statement:

Suppose $X,Y,$ and $Z$ are simply connected $CW$ complexes and that $X \rightarrow Y \rightarrow Z$ is simultaneously a cofiber sequence and a fiber sequence. Show that either $X$ or $Z$ is homotopy equivalent to a point.

Could anyone help me in answering this question?

EDIT:

I think this question is about sections 11.2 and 11.3 of the book named "Modern Classical Homotopy Theory" but still I do not know how to answer it.

Emptymind
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  • Can you provide some more context? Was this assigned as an exercise/from a particular book, or is it a problem you came up with yourself? – William Mar 06 '20 at 15:55
  • it is not an excercise from a particular book but we are working from "AT". And "Modern Classical Homotopy Theory" by Jeffery Strom. @William – Emptymind Mar 06 '20 at 16:50
  • Which sections of these books have you covered so far? It would be helpful to know what tools you have available. – William Mar 06 '20 at 17:10
  • @William the exercise may also be on the second cube theorem .... I do not know. – Emptymind Mar 07 '20 at 00:47
  • we took a statement that said that "fibre sequence behave like a cofibre sequence from n to 2n " and we calculated the connectivity of $F \wedge \Omega B$ to be $n + m$ and that of $\sum (F \wedge \Omega B)$ to be $n+m+1$ ...I do not know if this is related @William – Emptymind Mar 07 '20 at 01:07
  • May be we wanted to relate homology to cohomology in that context .... I do not know @William – Emptymind Mar 07 '20 at 01:19
  • I'm not sure I belive that the statement is true. Can you give a reference to exactly where it appears in the book? – Tyrone Mar 07 '20 at 08:24
  • No, I can not, I do not think this is a question from a book.@Tyrone do you have a counterexample that disprove it? – Emptymind Mar 07 '20 at 09:40

1 Answers1

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I claim that there do exist sequences $X\rightarrow Y\rightarrow Z$ of simply connected spaces (even CW complexes) which are both fibration and cofibration sequences. Here is my example.

For an abelian group $A$ and an integer $n\geq2$ we denote by $M(A,n)$ the degree $n$ Moore space, characterised by the property that it is a simply connected CW complex satisfying $$\widetilde H_*M(A,n)\cong\begin{cases}A&\ast=n\\0&\text{otherwise}.\end{cases}$$

Now choose distinct primes $p,q$ and integers $n,m\geq 2$. Let $M(\mathbb{Z}_p,m)$ and $M(\mathbb{Z}_q,n)$ be the Moore spaces in the indicated degrees. These are simply connected and we can assume they are pointed CW complexes. Then $$M(\mathbb{Z}_p,m)\xrightarrow{i} M(\mathbb{Z}_p,m)\vee M(\mathbb{Z}_q,n)\xrightarrow{\xi} M(\mathbb{Z}_q,n)$$ is a cofibration sequence, where the first map is the inclusion and $\xi$ is the pinch map. We also have a fibration sequence $$M(\mathbb{Z}_p,m)\xrightarrow{j} M(\mathbb{Z}_p,m)\times M(\mathbb{Z}_q,n)\xrightarrow{\pi} M(\mathbb{Z}_q,n)$$ where the first map is the inclusion and $\pi$ is the projection.

Now, by means of the Kunneth formula we can compute the reduced homology of the smash $M(\mathbb{Z}_p,m)\wedge M(\mathbb{Z}_q,n)$. We find that it disappears, since the tensor product $\mathbb{Z}_p\otimes\mathbb{Z}_q$ is trivial, as is the torsion product $Tor(\mathbb{Z}_p,\mathbb{Z}_q)$. Thus the inclusion $$k:M(\mathbb{Z}_p,m)\vee M(\mathbb{Z}_q,n)\hookrightarrow M(\mathbb{Z}_p,m)\times M(\mathbb{Z}_q,n)$$ induces an isomorphism on homology groups. Since both domain are codomain are simply connected, this map is a weak equivalence by the homological Whitehead Theorem, and thus a homotopy equivalence, since everything is CW. (Of course we have $M(\mathbb{Z}_p,m)\wedge M(\mathbb{Z}_q,n)\simeq\ast$ but we don't use this explicitly).

Note that the composite of $k$ with the inclusion $i$ is exactly the inclusion $j$. Also, the composite of $j$ with the projection $\pi$ is exactly the pinch map $\xi$.

The conclusion is that the cofibration sequence and the fibration sequence above are the same sequence.

Tyrone
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  • Isn't the Moore space characterized by the fact that it's a simply-connected space with the prescribed homology groups ? (I'm not sure, just asking : there might be an obstruction to having a perfect fundamental groups and only one nontrivial homolgoy group that I don't know about) – Maxime Ramzi Mar 07 '20 at 11:48
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    @Max I agree, and I've added the hypothesis to my answer (which I am implictly assuming, anyway). Like you suggest, wedging with an acyclic space (say, the $2$-skeleton of the Poincaré homology sphere) will retain the connectedness and give you the same homology, but yield a different space in general. – Tyrone Mar 07 '20 at 13:14
  • Ok good, I was afraid I was missing something obvious. Your answer was really nice anyways, I was trying to fiddle around with Moore spaces too but didn't get there before seeing your answer – Maxime Ramzi Mar 07 '20 at 14:18
  • what do you mean by $* = n$? – Intuition Apr 09 '20 at 15:36
  • @Secretly I am using $\ast$ as the placeholder for an integer and $n$ is one such. – Tyrone Apr 09 '20 at 15:40
  • could you include the details of using Kunneth in calculating the reduced homology of the smash ? I think there are many Kunneth formulas … am I correct? – Intuition Apr 09 '20 at 15:44
  • and why the calculation is for the smash and not the wedge that we already have? – Intuition Apr 09 '20 at 15:45
  • @Secretly the calculation for the wedge follows from the axioms of homology (i.e. additivity) and the definition of the Moore space. What is really needed is the calculation for the product, and one way to do this is to use the exact sequence induced by the cofibration and then show that the smash is contractible, which is the line of reasoning I adopted. – Tyrone Apr 09 '20 at 16:03