Homology of mapping torus of degree 2 map

Question

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.

Answer 1

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).

Answer 2

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.