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I'm working on problem 2.2.30 in Hatcher's Algebraic Topology, which asks us to find the homology of the mapping torus $T_f$ of a degree-$2$ map $f : S^2 \to S^2$. That is, the induced map on homology $f_* : H_2(S^2) \to H_2(S^2)$ is the doubling map $1 \mapsto 2$. I computed that all homology groups $H_n(T_f) = 0$ for all $n$, and that doesn't seem right. Is there a flaw in my solution?

We get a long exact sequence of (relative) homology groups $$\cdots \to \tilde H_n(S^2) \xrightarrow[]{\mathbb{1}-f_*} \tilde H_n(S^2) \to \tilde H_n(T_f) \to \tilde H_{n-1}(S^2) \to \cdots$$ where $\mathbb1$ is the identity map. Because $\tilde H_n(S^2) = \mathbb Z$ for $n = 2$ and $=0$ otherwise, this gives us $(\mathbb1 - f_*)(1) = -1$. Therefore the nontrivial part of this long exact sequence is $$0 \to \tilde H_3(T_f) \to \mathbb Z \xrightarrow[]{-\mathbb1} \mathbb Z \to \tilde H_2(T_f) \to 0 \to 0 \to \tilde H_1(T_f) \to 0.$$ Because this sequence is exact, we immediately get $\tilde H_1(T_f) = 0$. Also because the map $-\mathbb 1 : \mathbb Z \to \mathbb Z$ is injective, the map $\tilde H_3(T_f) \to \mathbb Z$ is $0$, which further implies by exactness that $\tilde H_3(T_f) = 0$. Finally, exactness tells us the map $\mathbb Z \to \tilde H_2(T_f)$ is surjective, and the kernel of this map is $\mathrm{Im}(-\mathbb1) = \mathbb Z$, which implies $\tilde H_2(T_f) = 0$.

Is this correct? It seems strange that all homology groups are $0$; in particular, a priori, this doesn't seem like it should be a contractible space.

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In general, for any continuous self-map $f : X \to X$ of a path connected space $X$, the mapping torus $$T_f = X \times [0,1] \, / \, (x,1) \sim (f(x),0) $$ always has an infinite cyclic subgroup in its first homology group, a generator of which is represented by any closed path of the form $\gamma_1 * \gamma_2$ where $\gamma_1$ is the image in the quotient space of the path $x \times [0,1]$ (for arbitrary $x \in X$) and where $\gamma_2$ is the image of a path in $X$ that connects $f(x)$ to $x$.

Your error is that you are using reduced homology. If you use unreduced homology, which is what one uses for this long exact sequence, you'll see that the map $$\mathbb{1}-f_* : \mathbb{Z} = H_0(X) \to H_0(X) \to \mathbb{Z} $$ is always the zero map, which produces an infinite cyclic subgroup in $H_1(T_f)$ (one can also show, by diagram chasing, that there is such a subgroup with a generator as described earlier).

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    $\begingroup$ Yes, I see my error now. One can then use the fact that $\mathbb 1 - f_* = 0$ to show that parts of the sequence of unreduced homology splits, I suppose. Thank you! (edit) As an aside, is there a good rule of thumb to know when you are and are not allowed to use reduced homology? It makes some things much easier but evidently trivializes some arguments that should not be trivial. $\endgroup$
    – D Ford
    Jul 30 '17 at 0:33
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    $\begingroup$ @DomVeconi Let's look at the proof of the LES in Hatcher. One key step is to observe that $H_n(X \times \partial I)$ is isomorphic to $H_n(X) \oplus H_n(X)$, and the kernel of the map $H_n(X \times \partial I)\overset{i_\star}{ \to} H_n(X \times I)$ consists of elements of the form $(\alpha, - \alpha) \in H_n(X) \oplus H_n(X)$. If we use reduced homology, this step clearly fails in degree $n = 0$. $\endgroup$
    – Kenny Wong
    Jul 30 '17 at 9:24
  • $\begingroup$ Oh, got it. Thank you! $\endgroup$
    – D Ford
    Jul 30 '17 at 21:04
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I do not believe this is correct. Your ideas and sequences seem fine, it's just the computation and understanding the maps.

We know that $H_1(S^2)=H_3(S^2)=0$ and $H_0(S^2)=H_2(S^2)=\mathbb{Z}$. We also know that $1-f_*=0$ on this homology degree so that we have

$$0 \longrightarrow H_1(T_f) \longrightarrow \mathbb Z \xrightarrow{ \ \ 0 \ \ } \mathbb Z \longrightarrow H_0(T_f) \longrightarrow 0$$

Which gives that $H_0(T_f)=H_1(T_f)=\mathbb{Z}$. Note that we expect $H_3(T_f)=0$ since we 'imagine' that $T_f$ will be non-orientable. To compute the other homology groups, we look at the rest of our sequence $$\cdots \longrightarrow H_3(S^2)\longrightarrow H_3(T_f) \longrightarrow H_2(S^2)\xrightarrow{1-f_*} H_2(S^2) \longrightarrow H_2(T_f) \longrightarrow H_1(S^2) \longrightarrow \cdots$$ Using what we know above, this gives us $$ 0 \longrightarrow H_3(T_f) \longrightarrow \mathbb{Z} \xrightarrow{ \ \ 2 \ \ } \mathbb{Z} \longrightarrow H_2(T_f) \longrightarrow 0 $$ The middle map is multiplication by $2$ as we can interpret the element $1 \in H_2(S^2)$ to be the orientation class of $S^2$ and since we are looking at the $2$-sphere so the antipodal map is orientation reversing we must have $f_*$ induce the negation map in homology. Then $1-f_*$ gives the stated multiplication. Then the calculation using the chain complex gives $H_3(T_f)=0$ and $H_2(T_f)= \mathbb{Z}/2\mathbb{Z}$.

EDIT: As stated by Kenny Wong, I have looked at (a), which is the antipodal map, but it is routine to make the appropriate edits for the case of a degree $2$ map.

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  • $\begingroup$ Hi! I like your clear explanation! (+1) But I believe the $f$ in the original question is the degree-two map from $S^2 $ to $S^2$, not the antipodal map. So the middle map is an isomorphism. Still, the logic is very similar in the two cases. $\endgroup$
    – Kenny Wong
    Jul 29 '17 at 18:11
  • $\begingroup$ @KennyWong Thank you and thank you for pointing it out. I saw the Hatcher reference and pulled out my copy to the problem but ended up looking at (a) rather than (b), which is the stated problem! $\endgroup$ Jul 29 '17 at 23:10

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