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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.

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  • $\begingroup$ 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? $\endgroup$
    – William
    Mar 6, 2020 at 15:55
  • $\begingroup$ it is not an excercise from a particular book but we are working from "AT". And "Modern Classical Homotopy Theory" by Jeffery Strom. @William $\endgroup$
    – Emptymind
    Mar 6, 2020 at 16:50
  • $\begingroup$ Which sections of these books have you covered so far? It would be helpful to know what tools you have available. $\endgroup$
    – William
    Mar 6, 2020 at 17:10
  • $\begingroup$ @William the exercise may also be on the second cube theorem .... I do not know. $\endgroup$
    – Emptymind
    Mar 7, 2020 at 0:47
  • $\begingroup$ 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 $\endgroup$
    – Emptymind
    Mar 7, 2020 at 1:07

1 Answer 1

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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.

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  • $\begingroup$ 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) $\endgroup$ Mar 7, 2020 at 11:48
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    $\begingroup$ @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. $\endgroup$
    – Tyrone
    Mar 7, 2020 at 13:14
  • $\begingroup$ 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 $\endgroup$ Mar 7, 2020 at 14:18
  • $\begingroup$ what do you mean by $* = n$? $\endgroup$
    – Intuition
    Apr 9, 2020 at 15:36
  • $\begingroup$ @Secretly I am using $\ast$ as the placeholder for an integer and $n$ is one such. $\endgroup$
    – Tyrone
    Apr 9, 2020 at 15:40

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