Let $\textbf{HoTop}^*$ be the homotopy category of pointed topological spaces. In the following, the word "isomorphism" shall always mean isomorphism in $\textbf{HoTop}^*$, i.e. pointed homotopy equivalence. All constructions like cone or suspensions are pointed/reduced.
A triangle $X\to Y\to Z\to \Sigma X$ is called distinguished if it is isomorphic in $\textbf{HoTop}^*$ to a triangle of the form $X\stackrel{f}{\to} Y\hookrightarrow\text{C}f\to\Sigma X$, where $\text{C}f\to\Sigma X$ is the map collapsing $Y$ to a point.
Problem:
Let $\ \ \matrix{X & \to & Y & \to & Z & \to & \Sigma X\cr\downarrow\alpha &&\downarrow\beta&&\downarrow\gamma &&\downarrow&\Sigma\alpha\cr X^{\prime} & \to & Y^{\prime} & \to & Z^{\prime} & \to & \Sigma X^{\prime}}\ \ $ be a morphism of distinguished triangles such that $\alpha$ and $\beta$ are isomorphisms. Is it true that $\gamma$ is an isomorphism, too?
Suggestions:
For a morphism of triangles as above (where $\alpha$ and $\beta$ are not necessarily isomorphisms), the morphism $\gamma^*: [Z^{\prime},-]\to [Z,-]$ is equivariant with respect to $[\Sigma\alpha]^*: [\Sigma X^{\prime},-]\to [\Sigma X,-]$. Therefore, I thought one could apply theorem 6.5.3 in Hoveys book on Model Categories. Unfortunately, there seems to be a gap at the end of the proof, as already pointed out here.
Therefore, I have the following
Questions:
(1) Am I misunderstanding something in Hovey's proof of 6.5.3(b), or is there really a gap in it? If it is a gap: Do you have any suggestions on how to fix the proof?
(2) If the proof can't be fixed in this generality: Do you have suggestions on how to prove the statement above only for $\textbf{HoTop}^*$?
Edit:
(1) The usual proof of this fact for triangulated categories does not work here, because there one uses the fact that $[X,-]$ is abelian-group valued for any $X$ and uses the classical five lemma together with Yoneda to conclude that $\gamma$ is an isomorphism. This doesn't seem to work here.
(2) Since partial morphisms of distinguished triangles in $\textbf{HoTop}^*$ can always be completed to morphisms of triangles, we can reduce to the case where $\alpha$ and $\beta$ both equal the identity. Therefore, we have a commutative diagram (in $\textbf{HoTop}^*$, i.e. a homotopy commutative diagram in $\textbf{Top}^*$)
$\matrix{X & \to & Y & \to & Z & \to & \Sigma X\cr\downarrow & \text{id}_X &\downarrow & \text{id}_Y&\downarrow&\gamma&\downarrow&\text{id}_{\Sigma X}\cr X & \to & Y & \to & Z& \to & \Sigma X}$
and we have to prove that $\gamma$ is a homotopy equivalence.
Hovey's proof
The way Hovey proceeds in his proof is as follows: We know the following things:
(1) $\gamma^*: [Z,-]\to [Z,-]$ is $[\Sigma X,-]$-equivariant
(2) Two maps $c,d\in[Z,W]$ are equal in $[\Sigma X,W]$$[Y,W]$ if and only if they lie in the same $[\Sigma X,W]$-orbit.
From (2) and the commutativity of the right handmiddle square it follows that for any $h\in [Z,W]$ there is some $\rho\in[\Sigma X,W]$ such that $\gamma^*(h)=h.\rho$; in other words $\gamma^*$ doesn't change the $[\Sigma X,-]$-orbit.
Now, suppose there are $g,h\in [Z,W]$ such that $\gamma^*(h)=\gamma^*(g)$. Then, again by the commutativity of the rightmiddle square, there is some $\alpha\in [\Sigma X,W]$ such that $g = h.\alpha$. Thus, by (1), $\gamma^*(g) = \gamma^*(h).\alpha = \gamma^*(g).\alpha$, and so $\alpha\in\text{Stab}(\gamma^*(g))$.
The point is that Hovey now wants to show that $\text{Stab}(\gamma^*(g))=\text{Stab}(g)$; this would imply $\alpha\in\text{Stab}(g)$, and thus $h = g.\alpha^{-1} = g$ as required. The inclusion $\text{Stab}(\gamma^*(g))\supset\text{Stab}(g)$ is obvious. For the other inclusion, I have no idea how to prove it.
Do you see how one can fix the proof?
Remark
You can find some details on triangles in $\textbf{HoTop}^*$ I have used above on my webpage (the above question is part of a big "christmas-exercise" for the topology III lecture that should introduce the students to triangulated categories).
Thank you.
