Homology groups of the Mapping Torus 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. 
 A: 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.    
