Five lemma in HoTop* and arbitrary pointed model categories

Question

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.

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.

Answer 1

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 $F$.

Answer 2

Here are some comments on Tyler's nice example, which answers the question of whether a map from a CW complex $Z$ to itself is a homotopy equivalence if it restricts to a homotopy equivalence from a subcomplex $A$ to itself and it induces a homotopy equivalence $Z/A\to Z/A$. The example shows the answer is no in general, but it is yes if $Z$ is simply-connected since the hypotheses imply that the map induces an isomorphism on the homology of $Z$ so Whitehead's theorem applies. More generally the answer is yes if the fundamental group of $Z$ is abelian and acts trivially on all the higher homotopy groups of $Z$ since Whitehead's theorem also holds in this generality. (For a textbook proof see Proposition 4.74 of my algebraic topology book.)

Tyler's example generalizes in the following way. Let $Z=S^1 \vee S^n$ for $n>1$, so $\pi_nZ=\{\mathbb Z}[t,t^{-1}]$, the Laurent polynomials, with $\pi_1Z$ acting by multiplication by powers of $t$. (Look in the universal cover of $Z$ to see this.) Let $g:Z\to Z$ be the identity on $S^1$ and on $S^n$ let it represent a Laurent polynomial $p(t)$ such that $p(1)=\pm 1$, for example $p(t)=2t-1$. Then $g$ is a homotopy equivalence on $A=S^1$ and on the quotient $Z/A=S^n$, but $g$ need not induce an isomorphism on $\pi_nZ$ so $g$ need not be a homotopy equivalence, for example when $p(t)=2t-1$ since in this case the cokernel of $g$ on $\pi_n$ is the quotient module ${\mathbb Z}[t,t^{-1}]/(2t-1)$, which is just the dyadic rationals.