Homology of quotient space where points are identified along homeomorphism Let $X$ be topological space and $f:X \rightarrow X$ be a homeomorphism. Define $M=X \times [0,1] / \sim$, where $\sim$ is the equivalence relation given by identifying each $(x,1)$ to $(f(x),0)$.
 So for example, if $X=S^1$ and $f$ is the identity map, then $M$ is the torus  and if $f$ is a reflection, then $M$ is the Klein bottle.
 I would like to compute the homology groups of $M$.
 Writing $M$ as the union $M_1 \cup M_2$, where $M_1=X \times [0,\frac{1}{2}]$ and $M_2=X \times [\frac{1}{2},1]$ (so $M_1 \cap M_2$ is two disjoint copies of $X$), applying Mayer-Vietoris, we obtain a long exact sequence
$$\dots \rightarrow H_{n+1}(M) \rightarrow H_n(M_1 \cap M_2) \rightarrow H_n(M_1) \oplus H_n(M_2) \rightarrow H_n(M) \rightarrow H_{n-1}(M_1 \cap M_2) \rightarrow  \dots$$
I know I need to express the homomorphism $H_n(M_1 \cap M_2) \rightarrow H_n(M_1) \oplus H_n(M_2)$ in terms of the induced homomorphism $H_n(f):H_n(X) \rightarrow H_n(X)$, but I am not sure how to proceed next.
 A: The space $M$ is homeomorphic to the pushout of the triad
$$X\xleftarrow{(1,f)}X\sqcup X\xrightarrow{(in_0,in_1)}X\times I$$
where $(in_0,in_1):X\sqcup X\hookrightarrow X\times I$ includes each of the factors into a different end of the cylinder. This map is a cofibration, so the resulting pushout square is excisive and you get a Mayer-Vietoris sequence
$$\dots\rightarrow H_{*+1}(M)\rightarrow H^*(X\sqcup X)\xrightarrow{(1,f)_*+(in_0,in_1)_*}H_*(X)\oplus H_*(X\times I)\rightarrow H_*(M)\rightarrow\dots$$
This is of course exactly the sequence you have already obtained.
Now the two inclusions $X\xrightarrow{i_1}X\sqcup X\xleftarrow{i_2}X$ induce an isomorphism
$$i_{1*}+i_{2*}: H_*(X)\oplus H_*(X)\xrightarrow\cong H^*(X\sqcup X).$$
Similarly the projection $pr_X:X\times I\rightarrow X$ is a homotopy equivalence which induces an isomorphism $H_*(X\times I)\xrightarrow\cong H_*(X)$. Note that $pr_X\circ(in_0,in_1)=(id_X,id_X)$, and that $(in_0,in_1)\circ i_1=in_0$ and $(in_0,in_1)\circ i_2=in_1$.
When we use these isomorphisms to identify the groups in the Mayer-Vietoris sequence above we see that it becomes
$$\dots\rightarrow H_{*+1}(M)\rightarrow H_*(X)\oplus H_*(X)\xrightarrow{\theta}H_*(X)\oplus H_*(X)\rightarrow H_*(M)\rightarrow\dots$$
where $\theta$ is the map
$$\theta(x,y)=(x+y,f_*x+y).$$
The kernel of $\theta$ is isomorphic to the subgroup $Fix(f_*)_*=\{x\in H_*(X)\mid f_*(x)=x\}\subseteq H_*(X)$ which consists of the fixed points of $f_*$. The cokernel of $\theta$ is isomorphic to the quotient of $H_*(X)$ by the subgroup $\{x+f_*x\in H_*(X)\mid x\in H_*(X)\}$.
Thus we obtain a long exact sequence
$$\dots\rightarrow H_{*+1}(M)\rightarrow H_*(X)\xrightarrow{f_*-1_*}H_*(X)\xrightarrow{j} H_*(M)\rightarrow\dots$$
where the map $j$ is induced by the inclusion of $X$ into $M$ at level $0$. The other map is the composite $H_{n+1}(M)\rightarrow H_n(X)\oplus H_n(X)\xrightarrow{pr_1}H_nX$ and is more difficult to describe explicitly.
You may find it more convenient to split this up into short exact sequences
$$0\rightarrow \text{coker}\left(H_n(X)\xrightarrow{(f_*-1_*)}H_n(X)\right)\rightarrow H_n(M)\rightarrow\ker\left(H_{n-1}(X)\xrightarrow{(f_*-1_*)}H_{n-1}(X)\right)\rightarrow 0$$
