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Question 2.2.30 of Hatcher: For the mapping torus $T_f$ of a map $f: X \to X$, we constructed in Example 2.48 a long exact sequence $\cdots \rightarrow H_n(X) \xrightarrow{ 1 - f_{\ast} } H_n(X) \longrightarrow H_n(T_f) \longrightarrow H_{n-1}(X) \longrightarrow \cdots.$ Use this to compute the homology of the mapping tori of the following maps:

(a) A reflection $S^2 \to S^2$.

So obviously $$H_n(S^2) = \begin{cases} \mathbb{Z}, & n=0,2 \\ 0, & \text{else}. \end{cases}$$

Moreover, we have that in this case, $$T_f = \frac{S^2 \times I}{(x,0) \sim (-x,1)}.$$

I'm unsure of how to proceed, particular due to the fact that I'm unclear as to what the map $1-f_{ast} : H_n(X) \to H_n(X)$ is defined to be.

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    You don't know what $f_*: S^2 \to S^2$ is, where $f$ is a reflection? –  Oct 11 '16 at 05:29
  • Elliot, I posted this answer last evening, and I would like some feedback: I added all the sources one could dream of, even if the problem was not that hard. If there's something you don't understand, just ask me. – Luca Oct 28 '16 at 17:48

1 Answers1

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I think the answer is:

$$ H_3 (T_f) = 0, \qquad H_2 (T_f) = \mathbb Z_2, \qquad H_1 (T_f) = \mathbb Z, \qquad H_0 (T_f) = \mathbb Z. $$ The only thing we have to care about is finding the effect of $\ f_*\ $ on the homology groups $\quad H_2 (S^2) = \mathbb Z \quad $ and $\quad H_0 (S^2) = \mathbb Z. \quad $

Now the $0$-th induced morphism is just the identity: of course $f$ maps the sphere to itself, and we know that the $0$-th homology group is just the free $\mathbb Z$-module generated by the connected components of our space (cfr. E.H. Spanier "$Algebraic \ Topology$" Chapter 4, pag 155).

So, in our exact sequence $$ H_1 (S^2)\longrightarrow H_1 (T_f) \longrightarrow H_0 (S^2)\xrightarrow{ 1 - f_* } H_0 (S^2) \longrightarrow H_0 (T_f) \longrightarrow 0 $$ we can substitute $$ H_1 (S^2) = 0, \qquad H_0 (S^2) = \mathbb Z, \qquad 1-f_* = 0, \qquad $$

to obtain

$$ 0 \longrightarrow H_1 (T_f) \longrightarrow \mathbb Z \xrightarrow{\ \ 0 \ \ } \mathbb Z \longrightarrow H_0 (T_f) \longrightarrow 0. $$ Hence, $$ H_1 (T_f) = \mathbb Z, \qquad H_0 (T_f) = \mathbb Z. $$ It remains to see $$ ... \longrightarrow H_3 (S^2)\longrightarrow H_3 (T_f) \longrightarrow H_2 (S^2)\xrightarrow{ 1 - f_* } H_2 (S^2) \longrightarrow H_2 (T_f) \longrightarrow H_1 (S^2) \longrightarrow ... $$

Of course $$ H_3 (S^2) = H_1 (S^2) = 0. $$

We need to know what is $$f_*: H_2 (S^2)\rightarrow H_2 (S^2). $$

Now $\quad 1 \in H_2 (S^2) \quad $ can be tought of as the orientation class of $S^2$ (see, ad example https://en.wikipedia.org/wiki/Orientability#Homology_and_the_orientability_of_general_manifolds). For $n$ even the antipodal map is orientation-reversing (M. P. Do Carmo $Riemannian \ Manifolds$, Chpt 0, page 20), and, since it is a cover of order one, it has to induce the map $-1$ in homology.

This is also directly stated in E.H. Spanier "$Algebraic \ Topology$" (Chapter 4, pag 196, section 7, points 9-10), which should be a credible enough reference.

So, since $\ 1 - (-1) = 2\ $ (just kidding), and $\ H_2 (S^2) = \mathbb Z $ $$ 0 \longrightarrow H_3 (T_f) \longrightarrow \mathbb Z\xrightarrow{\quad 2 \times \quad } \mathbb Z \longrightarrow H_2 (T_f) \longrightarrow 0 $$

we have $H_3 (T_f) = \ker (2 \times ) = 0 \ $ and $\ H_2 (T_f) = \mathbb Z / 2 \mathbb Z = \mathbb Z_2.$

A couple comments. The result is quite clear: $f$ is orientation-reversing, so one should expect $T_f$ to be a non-orientable manifold, i. e. $H_3 (T_f) = 0. \ $ $H_0$ and $H_1$ are obvious, since the mapping torus is a $S^2$-fibration over $S^1$. As for $H_2$, you see that two copies of $S^2$ will cancel each other in $T_f$: just think one as the inverted copy of the other.

Second thing: if you want to learn algebraic topology, you really have to study on the textbook by E. H. Spanier. It's old but gold.

Luca
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  • "Now the 0-th induced morphism is just the identity: of course f f maps the sphere to itself, and we know that the 0 0-th homology group is just the free ℤ-module generated by the connected components of our space " - How does this allow you to conclude that $f$ is simply the identity map? –  Oct 29 '16 at 01:41
  • Also, with regard to the second $f_{\ast}$ map, where you deduce that $1- f_{\ast}$ is multiplication by 2, can you do this with degrees? I'm aware that $\deg f = -1$ if $f$ is a reflection. See Hatcher page 134. –  Oct 29 '16 at 01:51
  • Because the $0$-th simplices are points (dimension zero) and the sphere is path connected, so: $f (P) = - P$, but $-P$ and $P$ can be connected by a path, hence they are bordant, so belong to the same equivalence class in $H_0$.
    • – Luca Oct 29 '16 at 20:46
  • If you know that deg $f = - 1$, then you know that $f_* $ is multiplication by $-1$ in $H_n$. Page 196 Spanier. I can also give a direct proof, if you just know that $1 \in H_n $ is a triangulation of the manifold. You could take a tetrahedral one. Ref: Sato "A. T. An intuitive approach." Excuse me, I don't have Hatcher, but, imho, it was not a good book: makes things too long and doesn't say things you need
    • – Luca Oct 29 '16 at 20:53
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    Really, study on Spanier – Luca Oct 29 '16 at 20:55
  • Hi from the future! I wonder what you meant by "$H_1$ is obvious"... I get the argument, but did you mean that it's obvious just because the fibre $\mathbb{S}^2$ is simply connected? – ah-- Sep 07 '21 at 01:51
  • What is $ \pi_1(T_f) $? Is it virtually cyclic? – Ian Gershon Teixeira Nov 30 '21 at 01:16