Show that either $X$ or $Z$ is homotopy equivalent to a point. 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.
 A: 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.
