Note: I have actually done the thing I am trying to do. My question is whether there is an easier way to do it.

Question: Let $f:X\to X$ be a map of topological spaces, and define the mapping torus of $f$ to be the space $$ T_f = X\times [0,1] /\sim $$ where $\sim$ is the equivalence relation $(x,0)\sim (f(x),1)$ for all $x\in X$. Using the Mayer-Vietoris Theorem, show that there is a long exact sequence $$ \ldots \to H_n(X) \overset{1 - f_*}{\to} H_n(X) \to H_n(T_f) \to H_{n-1}(X) \to \ldots. $$

My attempt: Let $A$ and $B$ be the subspaces of $T_f$ defined by $$ A = X \times [1/5, 4/5] / \sim, \quad B = (X\times [0, 2/5] \sqcup X\times [3/5,1])/\sim, $$ as illustrated in the following picture:

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Then the usual Mayer-Vietoris sequence gives us a long exact sequence $$ \ldots \to H_n(X)\oplus H_n(X) \to H_n(X)\oplus H_n(X) \to H_n(T_f) \to H_{n-1}(X) \oplus H_{n-1}(X) \to\ldots, $$ where the first map has matrix $$ \begin{pmatrix} 1 & 1 \\ -1 & -f_* \end{pmatrix}. $$ I haven't really justified this, but it's not hard to do, and I'm happy it's true. Now from here, we can kind of check by hand that restricting to the second component of each $H_n(X)\oplus H_n(X)$ gives the desired long exact sequence, but this is a bit fiddly and involves some case-work. Is there an easier way to get from the Mayer-Vietoris Theorem to the desired result?


1 Answer 1


I will try to answer the question in a way which you probably did not expect: Rather than giving an alternative quicker solutition I am trying to convince you that your solution is rather clear.

First of all, I would define $A=X\times ([0,1/3)\cup (1/3,1])$ and $A=X\times ([0,2/3)\cup (2/3,1])$ and then choose the orientation of the homology to be induced by the inclusion of $X\times \{1/2\}$. For me, this makes it even more obvious why the map $H_*(X)\oplus H_*(X)$ should be given by $F=\begin{pmatrix} 1&1\\1 & f_*\end{pmatrix}$. (Note: The sign in the last column differs from your solution, but this is up to personal preference of the orientation of the homology groups.)

By changing the basis and writing $(x_1,x_2)\in H_*(X)\oplus H_*(X)$ as $(x_1+x_2,x_2)$ and $(y_1,y_2)\in H_*(X)\oplus H_*(X)$ as $(y_1,y_1-y_2)$ one then gets $F$ written with respect to the new basis as $F'=\begin{pmatrix} 1&0\\0&1-f_*\end{pmatrix}$. Because the sequence is exact, we then can omit the first copy in each of the two terms of the l.e.s. and obtain the desired result.

In other words, I believe your solution is already the correct one. However, sometimes it is even harder to understand why a solution is quite simple than it is to write it down in the first place.

  • $\begingroup$ That's very elegant - thank you! $\endgroup$ Commented Feb 16, 2021 at 13:47

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