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 $[\Sigma\alpha]^\*: [\Sigma X^{\prime},-]\to [\Sigma X,-]$. (**edit: this is wrong -- see below**) 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](https://mathoverflow.net/questions/11977/equivariant-map-preserves-stabilizer).


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 $[Y,W]$ if and only if they lie in the same $[\Sigma X,W]$-orbit. 


From (2) and the commutativity of the middle 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 middle 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?


**FINAL EDIT**


I made a mistake in proving that for any morphism of triangles $(\alpha,\beta,\gamma)$ the morphism $\gamma^\*$ is equivariant with respect to $(\Sigma\alpha)^\*$. This is wrong.


So *what remains* is the question on how to fix the proof of theorem 6.5.3 in Hovey's book. Any suggestions?


Thank you.



Five lemma in HoTop* and arbitrary pointed model categories
This is false for spaces.


Let $X = S^0, Y = S^1$, and $f:X \to Y$ be the trivial map. Then $Z = Cf$ is $S^1 \vee S^1$. Then $[X,Y]$ is trivial, so then the the truth of this statement would imply: If you have a map $g: Z \to Z$ which is the identity on the first circle, and such that the induced map $S^1 \to S^1$ after collapsing the first circle is homotopic to the identity, then $g$ is a homotopy equivalence.


The map $g$ is a based map from a wedge of two circles to itself, which has fundamental group $F$, the free group on two generators, with generators $x, y$ corresponding to the two circle factors. A self-map is determined up to homotopy by a pair of elements in $x', y' \in F$. The condition that the first circle is mapped by the identity says $x = x'$, and the condition that the induced map on quotients is homotopic to the identity says that the image of $y'$ in $F/\langle x \rangle$ is the same as the image of $y$.


So: Suppose you have a pair of elements $x$ and $y'$ in $F = \langle x,y \rangle$ such that $y \equiv y'$ in $F/\langle x \rangle$. Then do $x$ and $y'$ freely generate $F$?


And the answer is no. For example, if $y' = (y x)^3 y^{-2}$, then $y' \equiv y$ after taking the quotient, but the pair $x, (y x)^3 y^{-2}$ don't generate the free group. This is believeable, but rather than being vague let's give a proof for completeness.


There is a group homomorphism $F \to S\_3$ sending $x$ to the 2-cycle $(1 2)$ and $y$ to $(1 3)$. This homomorphism is surjective because these generate the group, but $y'$ maps to the trivial element because the image of $yx$ has order 3 and the image of $y$ has order 2. Therefore, $x$ and $y'$ don't generate $S\_3$, and so they can't generate $F$.


